General error 1215 cannot add foreign key constraint sql alter table

I just wanted to add this case as well for VARCHAR foreign key relation. I spent the last week trying to figure this out in MySQL Workbench 8.0 and was finally able to fix the error.

Short Answer:
The character set and collation of the schema, the table, the column, the referencing table, the referencing column and any other tables that reference to the parent table have to match.

Long Answer:
I had an ENUM datatype in my table. I changed this to VARCHAR and I can get the values from a reference table so that I don’t have to alter the parent table to add additional options. This foreign-key relationship seemed straightforward but I got 1215 error. arvind’s answer and the following link suggested the use of

SHOW ENGINE INNODB STATUS;

On using this command I got the following verbose description for the error with no additional helpful information

Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
Note that the internal storage type of ENUM and SET changed in
tables created with >= InnoDB-4.1.12, and such columns in old tables
cannot be referenced by such columns in new tables.
Please refer to http://dev.mysql.com/doc/refman/8.0/en/innodb-foreign-key-constraints.html for correct foreign key definition.

After which I used SET FOREIGN_KEY_CHECKS=0; as suggested by Arvind Bharadwaj and the link here:

This gave the following error message:

Error Code: 1822. Failed to add the foreign key constraint. Missing
index for constraint

At this point, I ‘reverse engineer’-ed the schema and I was able to make the foreign-key relationship in the EER diagram. On ‘forward engineer’-ing, I got the following error:

Error 1452: Cannot add or update a child row: a foreign key constraint
fails

When I ‘forward engineer’-ed the EER diagram to a new schema, the SQL script ran without issues. On comparing the generated SQL from the attempts to forward engineer, I found that the difference was the character set and collation. The parent table, child table and the two columns had utf8mb4 character set and utf8mb4_0900_ai_ci collation, however, another column in the parent table was referenced using CHARACTER SET = utf8 , COLLATE = utf8_bin ; to a different child table.

For the entire schema, I changed the character set and collation for all the tables and all the columns to the following:

CHARACTER SET = utf8mb4 COLLATE = utf8mb4_general_ci;

This finally solved my problem with 1215 error.

Side Note:
The collation utf8mb4_general_ci works in MySQL Workbench 5.0 or later. Collation utf8mb4_0900_ai_ci works just for MySQL Workbench 8.0 or higher. I believe one of the reasons I had issues with character set and collation is due to MySQL Workbench upgrade to 8.0 in between. Here is a link that talks more about this collation.

I am running into the same problem here. I have a migration that adds a foreign key, referencing the id column on the users table and after upgrading to Laravel 5.8 this no longer works. This is the up of my original migration:

public function up()
{
    Schema::create('two_factor_auths', function (Blueprint $table) {
        $table->string('id')->nullable();
        $table->unsignedInteger('user_id');
        $table->foreign('user_id')->references('id')->on('users')->onDelete('cascade');
        $table->timestamps();
    });
}

But then running php artisan migrate -v on a freshly created db (InnoDB) results in:

  IlluminateDatabaseQueryException  : SQLSTATE[HY000]: General error: 1215 Cannot add foreign key constraint (SQL: alter table `two_factor_auths` add constraint `two_factor_auths_user_id_foreign` foreign key (`user_id`) references `users` (`id`) on delete cascade)

  at /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Connection.php:664
    660|         // If an exception occurs when attempting to run a query, we'll format the error
    661|         // message to include the bindings with SQL, which will make this exception a
    662|         // lot more helpful to the developer instead of just the database's errors.
    663|         catch (Exception $e) {
  > 664|             throw new QueryException(
    665|                 $query, $this->prepareBindings($bindings), $e
    666|             );
    667|         }
    668| 

  Exception trace:

  1   PDOException::("SQLSTATE[HY000]: General error: 1215 Cannot add foreign key constraint")
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Connection.php:458

  2   PDOStatement::execute()
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Connection.php:458

  3   IlluminateDatabaseConnection::IlluminateDatabase{closure}("alter table `two_factor_auths` add constraint `two_factor_auths_user_id_foreign` foreign key (`user_id`) references `users` (`id`) on delete cascade", [])
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Connection.php:657

  4   IlluminateDatabaseConnection::runQueryCallback("alter table `two_factor_auths` add constraint `two_factor_auths_user_id_foreign` foreign key (`user_id`) references `users` (`id`) on delete cascade", [], Object(Closure))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Connection.php:624

  5   IlluminateDatabaseConnection::run("alter table `two_factor_auths` add constraint `two_factor_auths_user_id_foreign` foreign key (`user_id`) references `users` (`id`) on delete cascade", [], Object(Closure))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Connection.php:459

  6   IlluminateDatabaseConnection::statement("alter table `two_factor_auths` add constraint `two_factor_auths_user_id_foreign` foreign key (`user_id`) references `users` (`id`) on delete cascade")
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Schema/Blueprint.php:97

  7   IlluminateDatabaseSchemaBlueprint::build(Object(IlluminateDatabaseMySqlConnection), Object(IlluminateDatabaseSchemaGrammarsMySqlGrammar))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Schema/Builder.php:264

  8   IlluminateDatabaseSchemaBuilder::build(Object(IlluminateDatabaseSchemaBlueprint))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Schema/Builder.php:165

  9   IlluminateDatabaseSchemaBuilder::create("two_factor_auths", Object(Closure))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Support/Facades/Facade.php:237

  10  IlluminateSupportFacadesFacade::__callStatic("create")
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/michaeldzjap/twofactor-auth/src/database/migrations/2017_05_26_102832_create_two_factor_auths_table.php:21

  11  CreateTwoFactorAuthsTable::up()
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Migrations/Migrator.php:360

  12  IlluminateDatabaseMigrationsMigrator::IlluminateDatabaseMigrations{closure}()
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Migrations/Migrator.php:367

  13  IlluminateDatabaseMigrationsMigrator::runMigration(Object(CreateTwoFactorAuthsTable), "up")
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Migrations/Migrator.php:178

  14  IlluminateDatabaseMigrationsMigrator::runUp("/home/vagrant/code/laravel-two-factor-authentication-example/vendor/michaeldzjap/twofactor-auth/src/database/migrations/2017_05_26_102832_create_two_factor_auths_table.php")
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Migrations/Migrator.php:147

  15  IlluminateDatabaseMigrationsMigrator::runPending([])
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Migrations/Migrator.php:96

  16  IlluminateDatabaseMigrationsMigrator::run([])
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Database/Console/Migrations/MigrateCommand.php:71

  17  IlluminateDatabaseConsoleMigrationsMigrateCommand::handle()
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Container/BoundMethod.php:32

  18  call_user_func_array([])
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Container/BoundMethod.php:32

  19  IlluminateContainerBoundMethod::IlluminateContainer{closure}()
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Container/BoundMethod.php:90

  20  IlluminateContainerBoundMethod::callBoundMethod(Object(IlluminateFoundationApplication), Object(Closure))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Container/BoundMethod.php:34

  21  IlluminateContainerBoundMethod::call(Object(IlluminateFoundationApplication), [])
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Container/Container.php:580

  22  IlluminateContainerContainer::call()
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Console/Command.php:183

  23  IlluminateConsoleCommand::execute(Object(SymfonyComponentConsoleInputArgvInput), Object(IlluminateConsoleOutputStyle))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/symfony/console/Command/Command.php:255

  24  SymfonyComponentConsoleCommandCommand::run(Object(SymfonyComponentConsoleInputArgvInput), Object(IlluminateConsoleOutputStyle))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Console/Command.php:170

  25  IlluminateConsoleCommand::run(Object(SymfonyComponentConsoleInputArgvInput), Object(SymfonyComponentConsoleOutputConsoleOutput))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/symfony/console/Application.php:901

  26  SymfonyComponentConsoleApplication::doRunCommand(Object(IlluminateDatabaseConsoleMigrationsMigrateCommand), Object(SymfonyComponentConsoleInputArgvInput), Object(SymfonyComponentConsoleOutputConsoleOutput))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/symfony/console/Application.php:262

  27  SymfonyComponentConsoleApplication::doRun(Object(SymfonyComponentConsoleInputArgvInput), Object(SymfonyComponentConsoleOutputConsoleOutput))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/symfony/console/Application.php:145

  28  SymfonyComponentConsoleApplication::run(Object(SymfonyComponentConsoleInputArgvInput), Object(SymfonyComponentConsoleOutputConsoleOutput))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Console/Application.php:90

  29  IlluminateConsoleApplication::run(Object(SymfonyComponentConsoleInputArgvInput), Object(SymfonyComponentConsoleOutputConsoleOutput))
      /home/vagrant/code/laravel-two-factor-authentication-example/vendor/laravel/framework/src/Illuminate/Foundation/Console/Kernel.php:122

  30  IlluminateFoundationConsoleKernel::handle(Object(SymfonyComponentConsoleInputArgvInput), Object(SymfonyComponentConsoleOutputConsoleOutput))
      /home/vagrant/code/laravel-two-factor-authentication-example/artisan:37

Changing from $table->unsignedInteger('user_id'); to $table->bigIncrements('user_id'); in my migration makes it work. I can then successfully run the migration. Maybe this will help.

How to fix MySQL error 1215 Cannot add foreign key constraint

Posted on Nov 13, 2021

Learn how to fix MySQL error 1215 Cannot add foreign key constraint with this tutorial

When you try to create a foreign key constraint between two tables, you may encounter the MySQL error 1215 that says Cannot add foreign key constraint .

For example, suppose we have a table called cities with the following data:

Then, suppose we want to create a table named users with a foreign key constraint, referencing the id column from the cities table.

Here’s how we might do it:

The response from MySQL may look like this:

Unfortunately, there are many issues that could cause this error.

This tutorial will list the most common cause for ERROR 1215 and give you suggestions on how to fix them.

Make sure that you are using the correct syntax

The first thing to do is to make sure that you are using the correct syntax for creating the FOREIGN KEY constraint.

The syntax to add a foreign key on CREATE TABLE statement must follow this pattern:

You must replace [target_column_name] next to the FOREIGN KEY syntax with the column name in the current table, while [origin_table_name] and [origin_column_name] must refer to the table and column name of an existing table.

Once you have the correct syntax, make sure that there’s no typo in [target_column_name] , [origin_table_name] , and [origin_column_name] or you may trigger the same error.

Once you are sure you have the correct syntax, let’s check the engine used by your tables next.

Make sure your tables are using InnoDB engine

You need to check whether the existing table and the table you want to create are using InnoDB engine.

This is because the MyISAM engine doesn’t support adding foreign key constraints, so when you try to add a foreign key constraint to the table, it will trigger the ERROR 1215 .

To check the engine of your existing table, you need to run the SHOW TABLE STATUS statement like this:

If you’re using the mysql command line client, then add a G next to the table name to organize the output as lists instead of a table.

Here’s an example output from the command line client:

As you can see from the highlighted line, the cities table is using the MyISAM engine.

You can change the engine of your MySQL table by using the ALTER TABLE statement as follows:

Once you altered the table engine, you can try to add the foreign key constraint to the new table again.

The default engine used for CREATE TABLE statement should be InnoDB , but you can add the engine explicitly as shown below:

If the error still happens, then it’s time to check the data type of the two columns.

Make sure the two columns are using the same data type

When adding foreign key constraints, the referenced column and the referencing column must both have the same data type.

An important tip here is to look at the full specification of your column using the DESCRIBE statement.

As you can see from the output above, the field id has the data type of int unsigned , but the referencing column city_id on the CREATE TABLE statement has the int type:

Keep in mind that the two columns type for the foreign key constraint must exactly match ( int signed with int signed , or int unsigned with int unsigned ).

You need to fix this issue by either altering the referenced column or the referencing column until they have the same type

Now that you have the same type for the two columns, you can try adding the foreign key again.

Adding ON DELETE / UPDATE SET NULL clause on a NOT NULL column

One more thing that could cause this error is when you add the ON DELETE SET NULL clause to the FOREIGN KEY constraint while the actual column is set to NOT NULL

Take a look at the highlighted lines on the example below:

While the city_id column is specified as NOT NULL , the ON DELETE SET NULL clause on the FOREIGN KEY constraint will cause the same error.

You need to either set the column as DEFAULT NULL :

Or you need to remove the ON DELETE SET NULL clause.

The same thing also happens when you add the ON UPDATE SET NULL clause to the FOREIGN KEY constraint.

For VARCHAR columns, make sure you have the same collation for both tables

When you’re adding a foreign key constraint with columns of VARCHAR types, you need to make sure that both tables are using the same collation.

Just like the engine type, you can check the table collation using the SHOW TABLE STATUS statement.

Here’s an example output from my database:

Then, you can check the Collation and Charset you need to use in your CREATE TABLE statement by running the SHOW COLLATION statement as follows:

The result for collation utf8mb4_0900_ai_ci is as follows:

In your CREATE TABLE statement, add the COLLATE and CHARSET options as shown below:

That should allow you to add foreign key constraints with columns of VARCHAR type.

Conclusion

Through this tutorial, you’ve learned five things that you can check to resolve the MySQL error 1215 Cannot add foreign key constraint .

This error message is not helpful when trying to find the cause, and in recent MySQL versions, the error has been replaced with more descriptive ones.

For example, when you type the wrong table name, you’ll have ERROR 1824 saying Failed to open the referenced table as shown below:

The error message above directly points you to the problem with the syntax.

In another example, different column data types will make MySQL throw ERROR 3780 saying the columns are incompatible:

Unfortunately, I wasn’t able to pinpoint the exact MySQL version that updates the error messages.

I have updated MySQL to the latest version 8.0.27 , so if you have some free time, you might want to upgrade your MySQL version to at least version 8 so that it gives more helpful error messages.

Good luck in resolving the error! 👍

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General error 1215 cannot add foreign key constraint sql alter table

General error: 1215 Cannot add foreign key constraint n

I made a search in google and i found out that the problem is because I’m trying to reference a foreign key from a table that not exist. I try to change the execution sort of each table so i can execute first the payment system table before i create the payment_currencies, by change the time of the file, and put it earlier than the payment_currencies, but it didn’t work. can anyone suggest a better solution to change the order of migration without the need of creating new migration files? n

The issue was that either mysql didn’t want foreign keys during table creation, or laravel was issuing them in the wrong order. n

In short, this didn’t work: n

Schema::create(‘cart_items’, function(Blueprint $table)n <n $table->engine = ‘InnoDB’;n $table->increments(‘id’);n $table->string(‘code’, 16)->index();nn $table->integer(‘user_id’)->unsigned();n $table->foreign(‘user_id’)->references(‘id’)->on(‘users’)->onDelete(‘cascade’);n $table->integer(‘product_id’)->unsigned();n $table->foreign(‘product_id’)->references(‘id’)->on(‘products’)->onDelete(‘cascade’);nn $table->integer(‘quantity’)->unsigned();n $table->timestamps();n >);n n

Schema::create(‘cart_items’, function(Blueprint $table)n <n $table->engine = ‘InnoDB’;n $table->increments(‘id’);n $table->string(‘code’, 16)->index();n $table->integer(‘user_id’)->unsigned();n $table->integer(‘product_id’)->unsigned();n $table->integer(‘quantity’)->unsigned();n $table->timestamps();n >);n Schema::table(‘cart_items’, function(Blueprint $table)n <n $table->foreign(‘user_id’)->references(‘id’)->on(‘users’)->onDelete(‘cascade’);n $table->foreign(‘product_id’)->references(‘id’)->on(‘products’)->onDelete(‘cascade’);n >);n n»,»bodyInMarkdown»:»@giwrgos I had the same issue when running L5.0 application on homestead (whether other laravel versions and environments are affected is unknown).rnrnThe issue was that either mysql didn’t want foreign keys during table creation, or laravel was issuing them in the wrong order.rnrnIn short, this didn’t work:rn«`phprnttSchema::create(‘cart_items’, function(Blueprint $table)rntt<rnttt$table->engine = ‘InnoDB’;rnttt$table->increments(‘id’);rnttt$table->string(‘code’, 16)->index();rnrnttt$table->integer(‘user_id’)->unsigned();rnttt$table->foreign(‘user_id’)->references(‘id’)->on(‘users’)->onDelete(‘cascade’);rnttt$table->integer(‘product_id’)->unsigned();rnttt$table->foreign(‘product_id’)->references(‘id’)->on(‘products’)->onDelete(‘cascade’);rnrnttt$table->integer(‘quantity’)->unsigned();rnttt$table->timestamps();rntt>);rn«`rnrnThis worked:rn«`phprnttSchema::create(‘cart_items’, function(Blueprint $table)rntt<rnttt$table->engine = ‘InnoDB’;rnttt$table->increments(‘id’);rnttt$table->string(‘code’, 16)->index();rnttt$table->integer(‘user_id’)->unsigned();rnttt$table->integer(‘product_id’)->unsigned();rnttt$table->integer(‘quantity’)->unsigned();rnttt$table->timestamps();rntt>);rnttSchema::table(‘cart_items’, function(Blueprint $table)rntt<rnttt$table->foreign(‘user_id’)->references(‘id’)->on(‘users’)->onDelete(‘cascade’);rnttt$table->foreign(‘product_id’)->references(‘id’)->on(‘products’)->onDelete(‘cascade’);rntt>);rn«`»,»replies»:[],»user»:<«id»:17861,»username»:»constb»,»avatar»:»//unavatar.io/github/constb»,»experience»:<«award_count»:»19″,»level»:6,»points»:»26,005″,»pointsUntilNextLevel»:»3,995″>,»achievements»:[],»reported»:null,»staff»:false,»subscribed»:false,»profile»:<«full_name»:null,»bio»:null,»twitter»:»constb»,»github»:»constb»,»website»:»»,»employment»:»http://designstudio.ag/»,»job_title»:»»,»location»:»Tomsk»,»flag»:»ru»,»available_for_hire»:0>,»dateSegments»:<«created_diff»:»7 years ago»>,»settings»:<«show_profile»:true>,»links»:<«profile»:»https://laracasts.com/@constb»>>,»likes»:[«sambellerose»,»vadykoo»],»created_at»:»2015-06-30T05:16:18.000000Z»,»links»:<«delete»:»/discuss/replies/76274″,»like»:»/discuss/replies/76274/likes»,»best_answer»:»/discuss/conversations/14696/best»>,»best_answer»:false,»dateSegments»:<«createdDiff»:»7 years ago»>>,<«id»:76275,»conversation_id»:14696,»body»:»

Here are my files n

filename: 2015_06_25_153736_payment_systems.php n

public function up()n<n Schema::create(‘payment_systems’, function(Blueprint $table)n <n $table->bigIncrements(‘id_payment_system’);n $table->string(«title»,100)->collate(‘utf8_general_ci’);n $table->string(«code»,10)->collate(‘utf8_general_ci’);n $table->string(«type»,4)->collate(‘utf8_general_ci’);n $table->boolean(«active»)->default(true);n $table->timestamps();n $table->softDeletes();n $table->unique(‘title’);n $table->unique(‘code’);n >);n>n n

filename: 2015_06_25_153743_payment_currencies.phpncontains: n

public function up()n<n Schema::create(‘payment_currencies’, function(Blueprint $table)n <n $table->bigIncrements(‘id_payment_currencies’);n $table->integer(‘payment_system_id_fn’)->unsigned();n $table->integer(‘opening_balance_id_fn’)->unsigned();n $table->boolean(«active»)->default(true);n $table->timestamps();n $table->softDeletes();n //$table->foreign(‘payment_system_id_fn’)->references(‘id_payment_system’)->on(‘payment_systems’);n //$table->foreign(‘opening_balance_id_fn’)->references(‘id_opening_balance’)->on(‘opening_balance’);nn >);nn Schema::table(‘payment_currencies’, function(Blueprint $table)n <n $table->foreign(‘payment_system_id_fn’)->references(‘id_payment_system’)->on(‘payment_systems’);n >);n>n n»,»bodyInMarkdown»:»@constb I tried it but is not working,rnrnHere are my files rnrnfilename: 2015_06_25_153736_payment_systems.phprnrnContains: rnrn public function up()rn <rn Schema::create(‘payment_systems’, function(Blueprint $table)rn <rn $table->bigIncrements(‘id_payment_system’);rn $table->string(»title»,100)->collate(‘utf8_general_ci’);rn $table->string(»code»,10)->collate(‘utf8_general_ci’);rn $table->string(»type»,4)->collate(‘utf8_general_ci’);rn $table->boolean(»active»)->default(true);rn $table->timestamps();rn $table->softDeletes();rn $table->unique(‘title’);rn $table->unique(‘code’);rn >);rn >rnrnrnrnfilename: 2015_06_25_153743_payment_currencies.phprncontains:rnrnt public function up()rn <rn Schema::create(‘payment_currencies’, function(Blueprint $table)rn <rn $table->bigIncrements(‘id_payment_currencies’);rn $table->integer(‘payment_system_id_fn’)->unsigned();rn $table->integer(‘opening_balance_id_fn’)->unsigned();rn $table->boolean(»active»)->default(true);rn $table->timestamps();rn $table->softDeletes();rn //$table->foreign(‘payment_system_id_fn’)->references(‘id_payment_system’)->on(‘payment_systems’);rn //$table->foreign(‘opening_balance_id_fn’)->references(‘id_opening_balance’)->on(‘opening_balance’);rnrn >);rnrn Schema::table(‘payment_currencies’, function(Blueprint $table)rn <rn $table->foreign(‘payment_system_id_fn’)->references(‘id_payment_system’)->on(‘payment_systems’);rn >);rn >»,»replies»:[],»user»:<«id»:15653,»username»:»giwrgos»,»avatar»:»//unavatar.io/github/giwrgos88″,»experience»:<«award_count»:»0″,»level»:2,»points»:»8,070″,»pointsUntilNextLevel»:»1,930″>,»achievements»:[],»reported»:null,»staff»:false,»subscribed»:false,»profile»:<«full_name»:null,»bio»:null,»twitter»:»giwrgos_panayi»,»github»:»giwrgos88″,»website»:»gpanayi.com»,»employment»:»»,»job_title»:»Senior Web Developer»,»location»:»Nicosia»,»flag»:»cy»,»available_for_hire»:1>,»dateSegments»:<«created_diff»:»7 years ago»>,»settings»:<«show_profile»:true>,»links»:<«profile»:»https://laracasts.com/@giwrgos»>>,»likes»:[],»created_at»:»2015-06-30T05:20:19.000000Z»,»links»:<«delete»:»/discuss/replies/76275″,»like»:»/discuss/replies/76275/likes»,»best_answer»:»/discuss/conversations/14696/best»>,»best_answer»:false,»dateSegments»:<«createdDiff»:»7 years ago»>>,<«id»:76293,»conversation_id»:14696,»body»:»

I have changed it but now I’m getting a different error message n

[Illuminate\Database\QueryException]
nSQLSTATE[42000]: Syntax error or access violation: 1075 Incorrect table def
ninition; there can be only one auto column and it must be defined as a key
n(SQL: create table platforms ( id_platform_details int unsigned not null
nauto_increment primary key, title varchar(100) not null, host varchar( n

n
not null, port int not null auto_increment primary key, active tin
nyint(1) not null default ‘0’, created_at timestamp default 0 not null, u pdated_at timestamp default 0 not null, deleted_at timestamp null) defau
nlt character set utf8 collate utf8_unicode_ci engine = InnoDB) n n

where my function is this one n

public function up()n<nSchema::create(‘platforms’, function(Blueprint $table)n<n$table->engine = ‘InnoDB’;n$table->increments(‘id_platform_details’);n$table->string(«title»,100)->collate(‘utf8_general_ci’);n$table->string(«host»,100)->collate(‘utf8_general_ci’);n$table->integer(«port»,5);n$table->boolean(«active»)->default(false);n$table->timestamps();n$table->softDeletes();n>);n> n

does the migration keeps any cache files? n»,»bodyInMarkdown»:»I have changed it but now I’m getting a different error messagernrn [Illuminate\Database\QueryException] rn SQLSTATE[42000]: Syntax error or access violation: 1075 Incorrect table def rn inition; there can be only one auto column and it must be defined as a key rn (SQL: create table `platforms` (`id_platform_details` int unsigned not null rn auto_increment primary key, `title` varchar(100) not null, `host` varchar( rn 100) not null, `port` int not null auto_increment primary key, `active` tin rn yint(1) not null default ‘0’, `created_at` timestamp default 0 not null, `u rn pdated_at` timestamp default 0 not null, `deleted_at` timestamp null) defau rn lt character set utf8 collate utf8_unicode_ci engine = InnoDB) rnrnwhere my function is this onern rnpublic function up()rn <rn Schema::create(‘platforms’, function(Blueprint $table)rn <rn $table->engine = ‘InnoDB’;rn $table->increments(‘id_platform_details’);rn $table->string(»title»,100)->collate(‘utf8_general_ci’);rn $table->string(»host»,100)->collate(‘utf8_general_ci’);rn $table->integer(»port»,5);rn $table->boolean(»active»)->default(false);rn $table->timestamps();rn $table->softDeletes();rn >);rn >rnrndoes the migration keeps any cache files?»,»replies»:[],»user»:<«id»:15653,»username»:»giwrgos»,»avatar»:»//unavatar.io/github/giwrgos88″,»experience»:<«award_count»:»0″,»level»:2,»points»:»8,070″,»pointsUntilNextLevel»:»1,930″>,»achievements»:[],»reported»:null,»staff»:false,»subscribed»:false,»profile»:<«full_name»:null,»bio»:null,»twitter»:»giwrgos_panayi»,»github»:»giwrgos88″,»website»:»gpanayi.com»,»employment»:»»,»job_title»:»Senior Web Developer»,»location»:»Nicosia»,»flag»:»cy»,»available_for_hire»:1>,»dateSegments»:<«created_diff»:»7 years ago»>,»settings»:<«show_profile»:true>,»links»:<«profile»:»https://laracasts.com/@giwrgos»>>,»likes»:[],»created_at»:»2015-06-30T06:18:40.000000Z»,»links»:<«delete»:»/discuss/replies/76297″,»like»:»/discuss/replies/76297/likes»,»best_answer»:»/discuss/conversations/14696/best»>,»best_answer»:false,»dateSegments»:<«createdDiff»:»7 years ago»>>,<«id»:76304,»conversation_id»:14696,»body»:»

there can be only one auto column and it must be defined as a key n n

id_platform_details int unsigned not null auto_increment primary key n n

port int not null auto_increment primary key n n

I understand what the error message says but as you can see the port is only integern$table->integer(«port»,5); n

i had this problem to. just solved it.nadd ->unsigned(); n

Well there are 3 things I did when I encountered this issue:nSay I have two tables users and clients n

n
1. I changed the timestamp of my users table to be earlier than clients table n
2. Just to be sure everything is okay, I ran composer dump-autoload (since I use composer) n
3. changed the type of the foreign key in my clients table to integer since the users id is also integer. n n

Finding out why Foreign key creation fail

When MySQL is unable to create a Foreign Key, it throws out this generic error message:

ERROR 1215 (HY000): Cannot add foreign key constraint

– The most useful error message ever.

Fortunately, MySQL has this useful command that can give the actual reason about why it could not create the Foreign Key.

mysql> SHOW ENGINE INNODB STATUS;

That will print out lots of output but the part we are interested in is under the heading ‘LATEST FOREIGN KEY ERROR’:

------------------------
LATEST FOREIGN KEY ERROR
------------------------
2020-08-29 13:40:56 0x7f3cb452e700 Error in foreign key constraint of table test_database/my_table:
there is no index in referenced table which would contain
the columns as the first columns, or the data types in the
referenced table do not match the ones in table. Constraint:
,
CONSTRAINT idx_name FOREIGN KEY (employee_id) REFERENCES employees (id)
The index in the foreign key in table is idx_name
Please refer to http://dev.mysql.com/doc/refman/5.7/en/innodb-foreign-key-constraints.html for correct foreign key definition.

This output could give you some clue about the actual reason why MySQL could not create your Foreign Key

Reason #1 – Missing unique index on the referenced table

This is probably the most common reason why MySQL won’t create your Foreign Key constraint. Let’s look at an example with a new database and new tables:

In the all below examples, we’ll use a simple ‘Employee to Department” relationship:

mysql> CREATE DATABASE foreign_key_1;
Query OK, 1 row affected (0.00 sec)

mysql> USE foreign_key_1;
Database changed

mysql> CREATE TABLE employees(
    ->     id int,
    ->     name varchar(20),
    ->     department_id int
    -> );
Query OK, 0 rows affected (0.08 sec)

mysql> CREATE TABLE departments(
    ->     id int,
    ->     name varchar(20)
    -> );
Query OK, 0 rows affected (0.07 sec)

As you may have noticed, we have not created the table with PRIMARY KEY or unique indexes. Now let’s try to create Foreign Key constraint between employees.department_id column and departments.id column:

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
ERROR 1215 (HY000): Cannot add foreign key constraint

Let’s look at the detailed error:

mysql> SHOW ENGINE INNODB STATUS;

------------------------
LATEST FOREIGN KEY ERROR
------------------------
2020-08-31 09:25:13 0x7fddc805f700 Error in foreign key constraint of table foreign_key_1/#sql-5ed_49b:
FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id):
Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
Note that the internal storage type of ENUM and SET changed in
tables created with >= InnoDB-4.1.12, and such columns in old tables
cannot be referenced by such columns in new tables.
Please refer to http://dev.mysql.com/doc/refman/5.7/en/innodb-foreign-key-constraints.html for correct foreign key definition.

This is because we don’t have any unique index on the referenced table i.e. departments. We have two ways of fixing this:

Option 1: Primary Keys

Let’s fix this by adding a primary key departments.id

mysql> ALTER TABLE departments ADD PRIMARY KEY (id);
Query OK, 0 rows affected (0.20 sec)
Records: 0  Duplicates: 0  Warnings: 0

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
Query OK, 0 rows affected (0.19 sec)
Records: 0  Duplicates: 0  Warnings: 0

Option 2: Unique Index

mysql> CREATE UNIQUE INDEX idx_department_id ON departments(id);
Query OK, 0 rows affected (0.13 sec)
Records: 0  Duplicates: 0  Warnings: 0

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
Query OK, 0 rows affected (0.21 sec)
Records: 0  Duplicates: 0  Warnings: 0

Reason #2 – Different data types on the columns

MySQL requires the columns involved in the foreign key to be of the same data types.

mysql> CREATE DATABASE foreign_key_1;
Query OK, 1 row affected (0.00 sec)

mysql> USE foreign_key_1;
Database changed

mysql> CREATE TABLE employees(
    ->     id int,
    ->     name varchar(20),
    ->     department_id int,
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.06 sec)

mysql> CREATE TABLE departments(
    ->     id char(20),
    ->     name varchar(20),
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.07 sec)

You may have noticed that employees.department_id is int while departments.id is char(20). Let’s try to create a foreign key now:

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
ERROR 1215 (HY000): Cannot add foreign key constraint

Let’s fix the type of departments.id and try to create the foreign key again:

mysql> ALTER TABLE departments MODIFY id INT;
Query OK, 0 rows affected (0.18 sec)
Records: 0  Duplicates: 0  Warnings: 0

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
Query OK, 0 rows affected (0.26 sec)
Records: 0  Duplicates: 0  Warnings: 0

It works now!

Reason #3 – Different collation/charset type on the table

This is a surprising reason and hard to find out. Let’s create two tables with different collation (or also called charset):

Let’s start from scratch to explain this scenario:

mysql> CREATE DATABASE foreign_key_1;                                                                                        Query OK, 1 row affected (0.00 sec)

mysql> USE foreign_key_1;                                                                                                    Database changed

mysql> CREATE TABLE employees(
    ->     id int,
    ->     name varchar(20),
    ->     department_id int,
    ->     PRIMARY KEY (id)
    -> ) ENGINE=InnoDB CHARACTER SET=utf8;
Query OK, 0 rows affected (0.06 sec)

mysql> CREATE TABLE departments(
    ->     id int,
    ->     name varchar(20),
    ->     PRIMARY KEY (id)
    -> ) ENGINE=InnoDB CHARACTER SET=latin1;
Query OK, 0 rows affected (0.08 sec)

You may notice that we are using a different character set (utf8 and latin1` for both these tables. Let’s try to create the foreign key:

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
ERROR 1215 (HY000): Cannot add foreign key constraint

It failed because of different character sets. Let’s fix that.

mysql> SET foreign_key_checks = 0; ALTER TABLE departments CONVERT TO CHARACTER SET utf8 COLLATE utf8_general_ci; SET foreign_key_checks = 1;
Query OK, 0 rows affected (0.00 sec)

Query OK, 0 rows affected (0.18 sec)
Records: 0  Duplicates: 0  Warnings: 0

Query OK, 0 rows affected (0.00 sec)

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
Query OK, 0 rows affected (0.20 sec)
Records: 0  Duplicates: 0  Warnings: 0

If you have many tables with a different collation/character set, use this script to generate a list of commands to fix all tables at once:

mysql --database=your_database -B -N -e "SHOW TABLES" | awk '{print "SET foreign_key_checks = 0; ALTER TABLE", $1, "CONVERT TO CHARACTER SET utf8 COLLATE utf8_general_ci; SET foreign_key_checks = 1; "}'

Reason #4 – Different collation types on the columns

This is a rare reason, similar to reason #3 above but at a column level.

Let’s try to reproduce this from scratch:

mysql> CREATE DATABASE foreign_key_1;                                                                                        Query OK, 1 row affected (0.00 sec)

mysql> USE foreign_key_1;                                                                                                    Database changed

mysql> CREATE TABLE employees(
    ->     id int,
    ->     name varchar(20),
    ->     department_id char(26) CHARACTER SET utf8,
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.07 sec)

mysql> CREATE TABLE departments(
    ->     id char(26) CHARACTER SET latin1,
    ->     name varchar(20),
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.08 sec)

We are using a different character set for employees.department_id and departments.id (utf8 and latin1). Let’s check if the Foreign Key can be created:

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
ERROR 1215 (HY000): Cannot add foreign key constraint

Nope, as expected. Let’s fix that by changing the character set of departments.id to match with employees.department_id:

mysql> ALTER TABLE departments MODIFY id CHAR(26) CHARACTER SET utf8;
Query OK, 0 rows affected (0.20 sec)
Records: 0  Duplicates: 0  Warnings: 0

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
Query OK, 0 rows affected (0.20 sec)
Records: 0  Duplicates: 0  Warnings: 0

It works now!

Reason #5 -Inconsistent data

This would be the most obvious reason. A foreign key is to ensure that your data remains consistent between the parent and the child table. So when you are creating the foreign key, the existing data is expected to be already consistent.

Let’s setup some inconsistent data to reproduce this problem:

mysql> CREATE DATABASE foreign_key_1;                                                                                        Query OK, 1 row affected (0.00 sec)

mysql> USE foreign_key_1;                                                                                                    Database changed

mysql> CREATE TABLE employees(
    ->     id int,
    ->     name varchar(20),
    ->     department_id int,
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.06 sec)

mysql> CREATE TABLE departments(
    ->     id int,
    ->     name varchar(20),
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.08 sec)

Let’s insert a department_id in employees table that will not exist in departments.id:

mysql> INSERT INTO employees VALUES (1, 'Amber', 145);
Query OK, 1 row affected (0.01 sec)

Let’s create a foreign key now and see if it works:

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);

ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint fails (`foreign_key_1`.`#sql-5ed_49b`, CONSTRAINT `fk_department_id` FOREIGN KEY (`department_id`) REFERENCES `departments` (`id`))

This error message is atleast more useful. We can fix this in two ways. Either by adding the missing department in departments table or by deleting all the employees with the missing department. We’ll do the first option now:

mysql> INSERT INTO departments VALUES (145, 'HR');
Query OK, 1 row affected (0.00 sec)

Let’s try to create the Foreign Key again:

mysql> ALTER TABLE employees ADD CONSTRAINT fk_department_id FOREIGN KEY idx_employees_department_id (department_id) REFERENCES departments(id);
Query OK, 1 row affected (0.24 sec)
Records: 1  Duplicates: 0  Warnings: 0

It worked this time.

So we have seen 5 different ways a Foreign Key creation can fail and possible solutions of how we can fix them. If you have encountered a reason not listed above, add them in the comments.

If you are using MySQL 8.x, the error message will be a little different:

SQLSTATE[HY000]: General error: 3780 Referencing column 'column' and referenced column 'id' in foreign key constraint 'idx_column_id' are incompatible. 

In this blog, we’ll look at how to resolve MySQL error code 1215: “Cannot add foreign key constraint”.

Our Support customers often come to us with things like “My database deployment fails with error 1215”, “Am trying to create a foreign key and can’t get it working,” or “Why am I unable to create a constraint?” To be honest, the error message doesn’t help much. You just get the following line:

ERROR 1215 (HY000): Cannot add foreign key constraint

But MySQL never tells you exactly WHY it failed. There’s actually a multitude of reasons this can happen. This blog post is a compendium of the most common reasons why you can get ERROR 1215, how to diagnose your case to find which one is affecting you and potential solutions for adding the foreign key.

(Note: be careful when applying the proposed solutions, as many involve ALTERing the parent table and that can take a long time blocking the table, depending on your table size, MySQL version and the specific ALTER operation being applied; in many cases, using pt-online-schema-change will be likely a good idea).

So, onto the list:

1) The Table or Index the Constraint Refers to Does Not Exist yet (Usual When Loading Dumps)

How to diagnose: Run SHOW TABLES or SHOW CREATE TABLE for each of the parent tables. If you get error 1146 for any of them, it means tables are being created in wrong order.

How to fix: Run the missing CREATE TABLE and try again, or temporarily disable foreign-key-checks. This is especially needed during backup restores where circular references might exist. Simply run:

SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS;
SET FOREIGN_KEY_CHECKS=0;
SOURCE /backups/mydump.sql; -- restore your backup within THIS session
SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS;

Example:

mysql> CREATE TABLE child (
    ->   id INT(10) NOT NULL PRIMARY KEY,
    ->   parent_id INT(10),
    ->   FOREIGN KEY (parent_id) REFERENCES `parent`(`id`)
    -> ) ENGINE INNODB;
ERROR 1215 (HY000): Cannot add foreign key constraint
# We check for the parent table and is not there.
mysql> SHOW TABLES LIKE 'par%';
Empty set (0.00 sec)
# We go ahead and create the parent table (we’ll use the same parent table structure for all other example in this blogpost):
mysql> CREATE TABLE parent (
    ->   id INT(10) NOT NULL PRIMARY KEY,
    ->   column_1 INT(10) NOT NULL,
    ->   column_2 INT(10) NOT NULL,
    ->   column_3 INT(10) NOT NULL,
    ->   column_4 CHAR(10) CHARACTER SET utf8 COLLATE utf8_bin,
    ->   KEY column_2_column_3_idx (column_2, column_3),
    ->   KEY column_4_idx (column_4)
    -> ) ENGINE INNODB;
Query OK, 0 rows affected (0.00 sec)
# And now we re-attempt to create the child table
mysql> CREATE TABLE child (
    ->   id INT(10) NOT NULL PRIMARY KEY,drop table child;
    ->   parent_id INT(10),
    ->   FOREIGN KEY (parent_id) REFERENCES `parent`(`id`)
    -> ) ENGINE INNODB;
Query OK, 0 rows affected (0.01 sec)

2) The Table or Index in the Constraint References Misuses Quotes

How to diagnose: Inspect each FOREIGN KEY declaration and make sure you either have no quotes around object qualifiers, or that you have quotes around the table and a SEPARATE pair of quotes around the column name.

How to fix: Either don’t quote anything, or quote the table and the column separately.

Example:

# wrong; single pair of backticks wraps both table and column
ALTER TABLE child  ADD FOREIGN KEY (parent_id) REFERENCES `parent(id)`;
# correct; one pair for each part
ALTER TABLE child ADD FOREIGN KEY (parent_id) REFERENCES `parent`(`id`);
# also correct; no backticks anywhere
ALTER TABLE child ADD FOREIGN KEY (parent_id) REFERENCES parent(id);
# also correct; backticks on either object (in case it’s a keyword)
ALTER TABLE child ADD FOREIGN KEY (parent_id) REFERENCES parent(`id`);

3) The Local Key, Foreign Table or Column in the Constraint References Have a Typo

How to diagnose: Run SHOW TABLES and SHOW COLUMNS and compare strings with those in your REFERENCES declaration.

How to fix: Fix the typo once you find it.

Example:

# wrong; Parent table name is ‘parent’
ALTER TABLE child ADD FOREIGN KEY (parent_id) REFERENCES pariente(id);
# correct
ALTER TABLE child ADD FOREIGN KEY (parent_id) REFERENCES parent(id);

4) The Column the Constraint Refers to Is Not of the Same Type or Width as the Foreign Column

How to diagnose: Use SHOW CREATE TABLE parent to check that the local column and the referenced column both have same data type and width.

How to fix: Edit your DDL statement such that the column definition in the child table matches that of the parent table.

Example:

# wrong; id column in parent is INT(10)
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_id BIGINT(10) NOT NULL,
  FOREIGN KEY (parent_id) REFERENCES `parent`(`id`)
) ENGINE INNODB;
# correct; id column matches definition of parent table
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_id INT(10) NOT NULL,
  FOREIGN KEY (parent_id) REFERENCES `parent`(`id`)
) ENGINE INNODB;

5) The Foreign Object Is Not a KEY of Any Kind

How to diagnose: Use SHOW CREATE TABLE parent to check that if the REFERENCES part points to a column, it is not indexed in any way.

How to fix: Make the column a KEY, UNIQUE KEY or PRIMARY KEY on the parent.

Example:

# wrong; column_1 is not indexed in our example table
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_column_1 INT(10),
  FOREIGN KEY (parent_column_1) REFERENCES `parent`(`column_1`)
) ENGINE INNODB;
# correct; we first add an index and then re-attempt creation of child table
ALTER TABLE parent ADD INDEX column_1_idx(column_1);
# and then re-attempt creation of child table
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_column_1 INT(10),
  FOREIGN KEY (parent_column_1) REFERENCES `parent`(`column_1`)
) ENGINE INNODB;

6) The Foreign Key Is a Multi-Column PK or UK, Where the Referenced Column Is Not the Leftmost One

How to diagnose: Do a SHOW CREATE TABLE parent to check if the REFERENCES part points to a column that is present in some multi-column index(es), but is not the leftmost one in its definition.

How to fix: Add an index on the parent table where the referenced column is the leftmost (or only) column.

Example:

# wrong; column_3 only appears as the second part of an index on parent table
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_column_3 INT(10),
  FOREIGN KEY (parent_column_3) REFERENCES `parent`(`column_3`)
) ENGINE INNODB;
# correct; create a new index for the referenced column
ALTER TABLE parent ADD INDEX column_3_idx (column_3);
# then re-attempt creation of child
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_column_3 INT(10),
  FOREIGN KEY (parent_column_3) REFERENCES `parent`(`column_3`)
) ENGINE INNODB;

7) Different Charsets/Collations Among the Two Table/Columns

How to diagnose: Run SHOW CREATE TABLE parent and compare that the child column (and table) CHARACTER SET and COLLATE parts match those of the parent table.

How to fix: Modify the child table DDL so that it matches the character set and collation of the parent table/column (or ALTER the parent table to match the child’s wanted definition.

Example:

# wrong; the parent table uses utf8/utf8_bin for charset/collation
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_column_4 CHAR(10) CHARACTER SET utf8 COLLATE utf8_unicode_ci,
  FOREIGN KEY (parent_column_4) REFERENCES `parent`(`column_4`)
) ENGINE INNODB;
# correct; edited DDL so COLLATE matches parent definition
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_column_4 CHAR(10) CHARACTER SET utf8 COLLATE utf8_bin,
  FOREIGN KEY (parent_column_4) REFERENCES `parent`(`column_4`)
) ENGINE INNODB;

The Parent Table Is Not Using InnoDB

How to diagnose: Run SHOW CREATE TABLE parent and verify if ENGINE=INNODB or not.

How to fix: ALTER the parent table to change the engine to InnoDB.

Example:

# wrong; the parent table in this example is MyISAM:
CREATE TABLE parent (
  id INT(10) NOT NULL PRIMARY KEY
) ENGINE MyISAM;
# correct: we modify the parent’s engine
ALTER TABLE parent ENGINE=INNODB;

9) Using Syntax Shorthands to Reference the Foreign Key

How to diagnose: Check if the REFERENCES part only mentions the table name. As explained by ex-colleague Bill Karwin in http://stackoverflow.com/questions/41045234/mysql-error-1215-cannot-add-foreign-key-constraint, MySQL doesn’t support this shortcut (even though this is valid SQL).

How to fix: Edit the child table DDL so that it specifies both the table and the column.

Example:

# wrong; only parent table name is specified in REFERENCES
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  column_2 INT(10) NOT NULL,
  FOREIGN KEY (column_2) REFERENCES parent
) ENGINE INNODB;
# correct; both the table and column are in the REFERENCES definition
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  column_2 INT(10) NOT NULL,
  FOREIGN KEY (column_2) REFERENCES parent(column_2)
) ENGINE INNODB;

10) The Parent Table Is Partitioned

How to diagnose: Run SHOW CREATE TABLE parent and find out if it’s partitioned or not.
How to fix: Removing the partitioning (i.e., merging all partitions back into a single table) is the only way to get it working.

Example:

# wrong: the parent table we see below is using PARTITIONs
CREATE TABLE parent (
  id INT(10) NOT NULL PRIMARY KEY
) ENGINE INNODB
PARTITION BY HASH(id)
PARTITIONS 6;
#correct: ALTER parent table to remove partitioning
ALTER TABLE parent REMOVE PARTITIONING;

11) Referenced Column Is a Generated Virtual Column (This Is Only Possible With 5.7 and Newer)

How to diagnose: Run SHOW CREATE TABLE parent and verify that the referenced column is not a virtual column.

How to fix: CREATE or ALTER the parent table so that the column will be stored and not generated.

Example:

# wrong; this parent table has a generated virtual column
CREATE TABLE parent (
  id INT(10) NOT NULL PRIMARY KEY,
  column_1 INT(10) NOT NULL,
  column_2 INT(10) NOT NULL,
  column_virt INT(10) AS (column_1 + column_2) NOT NULL,
  KEY column_virt_idx (column_virt)
) ENGINE INNODB;
# correct: make the column STORED so it can be used as a foreign key
ALTER TABLE parent DROP COLUMN column_virt, ADD COLUMN column_virt INT(10) AS (column_1 + column_2) STORED NOT NULL;
# And now the child table can be created pointing to column_virt
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_virt INT(10) NOT NULL,
  FOREIGN KEY (parent_virt) REFERENCES parent(column_virt)
) ENGINE INNODB;

12) Using SET DEFAULT for a Constraint Action

How to diagnose: Check your child table DDL and see if any of your constraint actions (ON DELETE, ON UPDATE) try to use SET DEFAULT

How to fix: Remove or modify actions that use SET DEFAULT from the child table CREATE or ALTER statement.

Example:

# wrong; this parent table has a generated virtual column
CREATE TABLE parent (
  id INT(10) NOT NULL PRIMARY KEY,
  column_1 INT(10) NOT NULL,
  column_2 INT(10) NOT NULL,
  column_virt INT(10) AS (column_1 + column_2) NOT NULL,
  KEY column_virt_idx (column_virt)
) ENGINE INNODB;
# correct: make the column STORED so it can be used as a foreign key
ALTER TABLE parent DROP COLUMN column_virt, ADD COLUMN column_virt INT(10) AS (column_1 + column_2) STORED NOT NULL;
# And now the child table can be created pointing to column_virt
CREATE TABLE child (
  id INT(10) NOT NULL PRIMARY KEY,
  parent_virt INT(10) NOT NULL,
  FOREIGN KEY (parent_virt) REFERENCES parent(column_virt)
) ENGINE INNODB;

I realize many of the solutions are not what you might desire, but these are limitations in MySQL that must be overcome on the application side for the time being. I do hope the list above gets shorter by the time 8.0 is released!

If you know other ways MySQL can fail with ERROR 1215, let us know in the comments!

More information regarding Foreign Key restrictions can be found here: https://dev.mysql.com/doc/refman/5.7/en/innodb-foreign-key-constraints.html.

Дата: 2.12.2016

Автор: Василий Лукьянчиков , vl (at) sqlinfo (dot) ru

Функционирование внешних ключей в MySQL имеет много нюансов и ограничений из-за чего существует немало возможностей получить ошибку при работе с ними. Одна из проблем состоит в том, что сообщения об ошибках содержат мало полезной информации и не указывают причину возникновения ошибки. В данной статье дается объяснение как получить дополнительную информацию об ошибке и приведен полный список причин возникновения ошибок внешних ключей. Каждая причина снабжена уникальным буквенно-цифровым кодом (А4, Б1, ..), использующимся в сводной таблице в конце статьи, которая поможет вам быстро диагностировать проблему.

Внешний ключ — это поле (или набор полей) в таблице, называемой дочерней, которое ссылается на поле (или набор полей) в таблице, называемой родительской. Дочерняя и родительская таблицы могут совпадать, т.е. таблица будет ссылаться на саму себя. Внешние ключи позволяют связать записи в двух таблицах по определенным полям так, что при обновлении поля в родительской автоматически происходит изменение записи в дочерней таблице.

В MySQL внешние ключи не реализованы на уровне сервера, их поддержка зависит от используемого хранилища данных. Содержание статьи справедливо для InnoDB (в том числе и для XtraDB).

Как получить больше данных об ошибке

После получения ошибки выполните SHOW ENGINE INNODB STATUS и смотрите содержимое секции LATEST FOREIGN KEY ERROR. Этот способ имеет следующие недостатки:

• требует привилегии SUPER
• содержит информацию о последней ошибке, связанной с внешними ключами, из-за чего нужно выполнять SHOW ENGINE INNODB STATUS сразу после возникновения ошибки, что не всегда удобно/возможно
• используются внутренние имена таблиц (например, ‘test.#sql-d88_b’), что затрудняет диагностику
• порой содержит мало полезной информации или таковая вообще отсутствует.

Альтернатива: использовать MariaDB версий больше 5.5.45 и 10.0.21, в которых сообщения об ошибках значительно улучшены и указывают причину возникновения ошибки.

Errno 150

Если в сообщении об ошибке содержится errno 150 (или errno 121), значит парсер MySQL не смог распознать ошибку и передал команду (create/alter) на выполнение в InnoDB. В этом разделе перечислены ситуации, приводящие к ошибкам, содержащим errno 150.

А1. Нет индекса в родительской таблице. Набор полей, на которые ссылается дочерняя таблица, должен быть проиндексирован (или являться левой частью другого индекса). Порядок полей в индексе должен быть таким же как в определении внешнего ключа. Сюда же относится случай отсутствия нужной колонки в родительской таблице (нет колонки, нет и индекса).

Неочевидный момент: на колонке родительской таблицы есть индекс — полнотекстовый (fulltext). Но внешний ключ всё равно не создается и сервер ругается на отсутствие индекса. Это происходит потому, что индекс должен быть обычным (btree).

Другой неочевидный момент: на колонке родительской таблицы есть индекс — префиксный. Но внешний ключ всё равно не создается и сервер ругается на отсутствие индекса. Это происходит потому, что индекс должен быть определен на всей длине колонки.

Строго говоря, поля в дочерней таблице тоже должны быть проиндексированы, но если нет подходящего индекса, MySQL автоматически его создаст при добавлении внешнего ключа (в совсем уж древних версиях требовалось предварительное создание индекса).

А2. Родительская таблица не найдена в словаре данных InnoDB. Это означает, что родительская таблица должна существовать и быть постоянной InnoDB таблицей. Не временной InnoDB таблицей, так как информация о временных таблицах не сохраняется в словаре данных InnoDB. И уж тем более не представлением.

А3. Синтаксическая ошибка. Внешние ключи реализованы на уровне хранилища, и в старых версиях парсер сервера MySQL не распознавал синтаксические ошибки внешних ключей, из-за чего их было трудно идентифицировать.

А4. Несовпадение типов данных. Столбцы дочерней таблицы, входящие в определение внешнего ключа, должны иметь такие же типы данных, что и столбцы родительской таблицы, на которые они ссылаются, вплоть до атрибутов: знак и кодировка/сопоставление.

А5. Некорректно задано действие внешнего ключа. Если в определении внешнего ключа указано ON UPDATE SET NULL и/или ON DELETE SET NULL, то соответствующие столбцы дочерней таблицы не должны быть определены как NOT NULL.

А6. Дочерняя таблица является временной InnoDB таблицей. Внешние ключи можно создавать только в постоянной, несекционированной InnoDB таблице.

А7. Родительская таблица является секционированной таблицей. На данный момент (MySQL 5.7 и MariaDB 10.1) внешние ключи не поддерживаются для секционированных таблиц (partitioned tables). Иными словами, ни родительская, ни дочерняя таблица не должны иметь секции. В случае, когда внешний ключ ссылается на секционированную таблицу диагностика ошибки затруднена ошибкой вывода show engine innodb status:

Errno 121

Такой результат возникает только в одном случае.

Б1. Неуникальное имя ограничения. Обратите внимание: речь не о имени внешнего ключа. Если при создании внешнего ключа вы указываете не обязательное ключевое слово CONSTRAINT, то идущий после него идентификатор должен быть уникальным в пределах базы данных.

Нет ошибок

Внешний ключ не создается, и нет никаких ошибок. Это может происходить по следующим причинам:

В1. Дочерняя таблица не является InnoDB таблицей. В этом случае для совместимости с другими субд парсер MySQL просто проигнорирует конструкцию внешнего ключа.

В2. Не соответствует синтаксису MySQL. Стандарт SQL разрешает указывать внешний ключ сразу при объявлении колонки с помощью конструкции REFERENCES (например, … a int references t1(a), …), однако MySQL игнорирует такую форму записи. Единственный способ создать в нем внешний ключ — это использовать отдельный блок FOREIGN KEY:

Несоответствие данных

В этой части собраны ошибки, которые возникают из-за нарушения ссылочной целостности, т.е. наличие в дочерней таблице записей, которым нет соответствия в родительской таблице.

Г1. Удаление родительской таблицы. Нельзя удалить родительскую таблицу при наличии внешнего ключа.

Удаление следует понимать в расширенном варианте как удаление из множества InnoDB таблиц. Например, если мы сменим (alter table) движок родительской таблицы на MyISAM, то с точки зрения ограничения внешнего ключа родительская таблица перестанет существовать (т.к. она должна быть постоянной innodb таблицей):

alter table t1 engine=myisam;
ERROR 1217 (23000): Cannot delete or update a parent row: a foreign key constraint fails

Сначала нужно удалить внешний ключ (или всю дочернюю таблицу, что удалит в том числе и внешний ключ). Если вы не знаете какие таблицы являются дочерними для заданной таблицы, то это можно определить через запрос к information_schema:

select table_name from information_schema.key_column_usage
where table_schema = «test» and references_table_name = «t1»;

Г2. Изменение данных в родительской таблице. Если в определении внешнего ключа не задано действие при update/delete, то такие операции над родительской таблицей могут привести к несогласованности данных, т.е. появлению в дочерней таблице записей не имеющих соответствия в родительской таблице.

Г3. Изменение данных в дочерней таблице. Если insert/update записи в дочерней таблицы приводит к несогласованности данных, то

Г4. Добавление внешнего ключа на не пустую таблицу. При попытке добавить внешний ключ на таблицу, в которой есть записи, не удовлетворяющие условию внешнего ключа (т.е. не имеющие соответствия в родительской таблице), будет ошибка:

Г5. Не уникальный ключ в родительской таблице. По стандарту SQL набор полей, на которые ссылается внешний ключ, должен быть уникальным. Однако, реализация внешних ключей в InnoDB позволяет иметь несколько «родителей». Из-за этого возникает трудно диагностируемая ошибка:

Сводная таблица

По вертикали расположены коды ошибок MySQL, которые возникают при работе с внешними ключами («нет ошибок» соответствует ситуации, когда сервер не генерирует ошибку, но и не создает внешний ключ). По горизонтали — идентификаторы причин, которые могут привести к ошибке. Плюсы на пересечении указывают какие причины приводят к той или иной ошибке.

P.S. Если ваш случай не рассмотрен в статье, то задавайте вопрос на форуме SQLinfo. Вам ответят, а статья будет расширена.

Дата публикации: 2.12.2016

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