Error 3734 failed to add the foreign key constraint

Hi there sorry if this is a stupid question but I’m new to sql, getting the error 3734

CREATE TABLE orderDetails (
orderNumber INT (10) NOT NULL,
quantityOrdered int(10) NOT NULL,
priceEach decimal (10,2) NOT NULL,
orderLineNumber INT (10) NOT NULL,
productCode VARCHAR(100) NOT NULL,
PRIMARY KEY (orderNumber),
KEY productCode (productCode),
CONSTRAINT orderdetails_asdf_1 FOREIGN KEY (productCode) 
REFERENCES product (productCode)
ON UPDATE CASCADE
);

CREATE TABLE product (
productCode VARCHAR(100) NOT NULL,
orderNumber INT NOT NULL,
orderDate DATE NOT NULL,
requiredDate DATE NOT NULL,
shippedDate DATE NOT NULL,
orderStatus varchar(10) NOT NULL,
Comments VARCHAR(100) NOT NULL,
customerNumber INT NOT NULL,
PRIMARY KEY (orderNumber),
KEY productCode (productCode),
CONSTRAINT orderdetails_asdf_1 FOREIGN KEY (productCode) 
REFERENCES orderDetails (productCode)
ON UPDATE CASCADE
) ;

Error code 3734, Failed to add the foreign key constraint. Missing column productCode for constraint orderDetails_asdf_1 in the referenced table ‘product’

Я пытаюсь передовой инженер свой ERD, который я сделал, но процесс форвард-инженера выдает мне ошибку. ERROR: Error 3734: Failed to add the foreign key constraint. Missing column ‘straatid’ for constraint ‘verzinzelf3’ in the referenced table ‘locatie’.

Но это очень странно, потому что имя столбца straatid на самом деле находится в таблице Locatie. Связь между таблицей Team и Locatie точно такая же, как и связь между Locatie и Plant, но кажется, что я получаю только ошибка между отношениями таблицы Locatie и Plant. Я не могу понять, как решить эту ошибку. Может кто-нибудь помочь мне с этой проблемой ?.

Это обзор SQL скрипт, который должен быть выполнен:

-- MySQL Workbench Forward Engineering

CREATE SCHEMA IF NOT EXISTS `mydb` DEFAULT CHARACTER SET utf8 ;

USE `mydb` ;

-- -----------------------------------------------------
-- Table `mydb`.`table1`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`table1` (
  `StraatID` VARCHAR(45) NOT NULL,
  `Straat` VARCHAR(45) NULL,
  `Latitude` VARCHAR(45) NULL,
  `Longitude` VARCHAR(45) NULL,
  PRIMARY KEY (`StraatID`))
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `mydb`.`Locatie`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Locatie` (
  `Huisnr` VARCHAR(45) NOT NULL,
  `StraatID` VARCHAR(45) NOT NULL,
  `Toevoeging` VARCHAR(45) NULL,
  PRIMARY KEY (`Huisnr`, `StraatID`),
  UNIQUE INDEX `StraatID_UNIQUE` (`StraatID` ASC) VISIBLE,
  UNIQUE INDEX `Huisnr_UNIQUE` (`Huisnr` ASC) VISIBLE,
  INDEX `verzinzelf2_idx` (`StraatID` ASC) VISIBLE,
  CONSTRAINT `verzinzelf2`
    FOREIGN KEY (`StraatID`)
    REFERENCES `mydb`.`table1` (`StraatID`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `mydb`.`Team`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Team` (
  `Teamnr` INT NOT NULL,
  `StraatID` VARCHAR(45) NOT NULL,
  `Huisnr` VARCHAR(45) NOT NULL,
  `StraatID1` VARCHAR(45) NOT NULL,
  PRIMARY KEY (`Teamnr`),
  INDEX `verzinzelf5_idx` (`Huisnr` ASC, `StraatID1` ASC) VISIBLE,
  CONSTRAINT `verzinzelf5`
    FOREIGN KEY (`Huisnr` , `StraatID1`)
    REFERENCES `mydb`.`Locatie` (`Huisnr` , `StraatID`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `mydb`.`Eigenaar`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Eigenaar` (
  `Eigenaar` INT NOT NULL,
  PRIMARY KEY (`Eigenaar`))
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `mydb`.`Plant`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Plant` (
  `Plant beschrijving` VARCHAR(45) NOT NULL,
  `Kaartlokatie` VARCHAR(45) NOT NULL,
  `Oppervlakte` VARCHAR(45) NOT NULL,
  `Eigenaar` INT NOT NULL,
  `Huisnr` VARCHAR(45) NOT NULL,
  `StraatID` VARCHAR(45) NOT NULL,
  PRIMARY KEY (`Plant beschrijving`, `Kaartlokatie`, `Oppervlakte`),
  INDEX `verzinzelf4_idx` (`Eigenaar` ASC) VISIBLE,
  INDEX `verzinzelf3_idx` (`Huisnr` ASC, `StraatID` ASC) VISIBLE,
  CONSTRAINT `verzinzelf4`
    FOREIGN KEY (`Eigenaar`)
    REFERENCES `mydb`.`Eigenaar` (`Eigenaar`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `verzinzelf3`
    FOREIGN KEY (`Huisnr` , `StraatID`)
    REFERENCES `mydb`.`Locatie` (`Huisnr` , `StraatID`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `mydb`.`Meting`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`Meting` (
  `Teamnr` INT NOT NULL,
  `Plant` VARCHAR(45) NOT NULL,
  `Meetsoort` VARCHAR(45) NULL,
  PRIMARY KEY (`Teamnr`, `Plant`),
  INDEX `verzinzelf1_idx` (`Plant` ASC) VISIBLE,
  INDEX `verzinzelf_idx` (`Teamnr` ASC) VISIBLE,
  CONSTRAINT `verzinzelf`
    FOREIGN KEY (`Teamnr`)
    REFERENCES `mydb`.`Team` (`Teamnr`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `verzinzelf1`
    FOREIGN KEY (`Plant`)
    REFERENCES `mydb`.`Plant` (`Plant beschrijving`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

И полное сообщение об ошибке:

Executing SQL script in server

ERROR: Error 3734: Failed to add the foreign key constraint. Missing column 'straatid' for constraint 'verzinzelf3' in the referenced table 'locatie'
SQL Code:
        -- -----------------------------------------------------
        -- Table `mydb`.`Plant`
        -- -----------------------------------------------------
        CREATE TABLE IF NOT EXISTS `mydb`.`Plant` (
          `Plant beschrijving` VARCHAR(45) NOT NULL,
          `Kaartlokatie` VARCHAR(45) NOT NULL,
          `Oppervlakte` VARCHAR(45) NOT NULL,
          `Eigenaar` INT NOT NULL,
          `Huisnr` VARCHAR(45) NOT NULL,
          `StraatID` VARCHAR(45) NOT NULL,
          PRIMARY KEY (`Plant beschrijving`, `Kaartlokatie`, `Oppervlakte`),
          INDEX `verzinzelf4_idx` (`Eigenaar` ASC) VISIBLE,
          INDEX `verzinzelf3_idx` (`Huisnr` ASC, `StraatID` ASC) VISIBLE,
          CONSTRAINT `verzinzelf4`
            FOREIGN KEY (`Eigenaar`)
            REFERENCES `mydb`.`Eigenaar` (`Eigenaar`)
            ON DELETE NO ACTION
            ON UPDATE NO ACTION,
          CONSTRAINT `verzinzelf3`
            FOREIGN KEY (`Huisnr` , `StraatID`)
            REFERENCES `mydb`.`Locatie` (`Huisnr` , `StraatID`)
            ON DELETE NO ACTION
            ON UPDATE NO ACTION)
        ENGINE = InnoDB

SQL script execution finished: statements: 9 succeeded, 1 failed

Fetching back view definitions in final form.
Nothing to fetch

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.