Дата: 25.11.2013
Автор: Даниил Каменский , dkamenskiy (at) yandex (dot) ru
При использовании ряда CMS (например, DLE, vBulletin и др.) временами возникает ошибка mysql с номером 1054.
Текст ошибки Unknown column ‘ИМЯ_СТОЛБЦА’ in ‘field list’ в переводе означает «Неизвестный столбец ‘ИМЯ_СТОЛБЦА’ в списке полей.«. Такая ошибка возникает в том случае, если попытаться выбрать (запрос вида select) или изменить (запрос вида update) данные из столбца, которого не существует. Ошибка чаще всего возникает из-за стoронних модулей. Перечислим несколько возможных причин:
- установлен модуль, расчитанный на более новую версию CMS, чем используемая;
- при установке модуля не выполнились операции изменения структуры таблиц;
- после установки сторонних модулей выполнено обновление системы, которое привело к изменению структуры таблиц; при этом модуль не был обновлен на совместимый;
- Из резервной копии восстановлена более старая база данных, а файлы сайта остались в новой версии.
Пример №1:
Имеется таблица сотрудников подразделения.
Поля: id, фамилия, имя, отчество, год рождения, наличие высшего образования.
create table if not exists employee
(
`id` int(11) NOT NULL auto_increment primary key,
`surname` varchar(255) not null,
`name` varchar(255) not null,
`patronymic` varchar(255) not null,
`year_of_birth` int unsigned default 0,
`higher_education` tinyint unsigned default 0
) ENGINE=MyISAM;
Если обратиться к этой таблице с запросом на выборку несуществующего поля, например пола сотрудника по фамилии Власенко, то результатом будет вышеуказанная ошибка:
mysql> select sex from employee where surname=’Власенко’;
ERROR 1054 (42S22): Unknown column ‘sex’ in ‘field list’
Пример №2:
Воспользуемся той же таблицей из примера 1. Если попытаться указать мужской пол у сотрудника по имени Власенко (выяснилось его имя и стало ясно, что это мужчина), то результатом будет та же ошибка:
mysql> update employee set sex=1 where surname=’Власенко’;
ERROR 1054 (42S22): Unknown column ‘sex’ in ‘field list’
Способы борьбы
Самый корректный способ борьбы в устранении причины ошибки. Например, все обновления сайта рекомендуем выполнять сначала на копии сайта и если ошибок нет, то повторять на рабочем сайте. Если при обновлении возникла ошибка, следует найти способ сделать обновление корректно с учетом версий сторонних модулей.
Если по каким-то причинам корректно избежать ошибки не получилось, можно прибегнуть к симптоматическому лечению, которое состоит в простом добавлении недостающих полей в таблицу.
Запрос на добавление:
ALTER TABLE employee ADD COLUMN sex ENUM(‘male’, ‘female’) DEFAULT ‘female’
Что в переводе означает «Изменить таблицу employee, добавив столбец `пол`, назначив ему тип перечисление(мужской/женский) по умолчанию мужской».
При таком добавлении столбца необходимо учитывать, что у всех записей в таблице в столбце sex появится значение по умолчанию. Если добавлять такой столбец как пол (который не может быть равен null и обязательно присутствует у каждого человека), то просто необходимо сразу же
после этого прописать нужное значение во все записи в таблице. В данном случае с добавлением столбца «пол» нужно будет поменять значение на male у всех сотрудников мужского пола.
Трудности могут возникнуть из-за того, что часто нужно самостоятельно определять тип добавляемого столбца.
Примеры:
a) Запрос:
SELECT faqname, faqparent, displayorder, volatile FROM faq where product
IN (», ‘vbulletin’, ‘watermark’, ‘cyb_sfa’, ‘access_post_and_days’);
Ответ сервера:
Invalid SQL: SELECT faqname, faqparent, displayorder, volatile FROM faq where
product IN (», ‘vbulletin’, ‘watermark’, ‘cyb_sfa’, ‘access_post_and_days’);
MySQL Error: Unknown column ‘faqname’ in ‘field list’
Error Number: 1054
Отсутствует столбец faqname, добавим его. Логика подсказывает, что если имя — то это скорее всего символы, а не целое число или тип datetime. Количество символов заранее, конечно, неизвестно, но редко имя бывает больше чем 255 символов. Поэтому добавим столбец faqname с указанием типа varchar(255):
ALTER TABLE faq ADD faqname varchar(255)
б) Запроc:
UPDATE dle_usergroups set group_name=‘Журналисты’, allow_html=‘0’ WHERE id=‘3’;
Ответ сервера:
Invalid SQL: UPDATE dle_usergroups set group_name=’Журналисты’, allow_html=’0′ WHERE id=’3′;
MySQL Error: Unknown column ‘allow_html’ in ‘field list’
Error Number: 1054
Отсутствует столбец allow_html, добавим его. Смотрим на то значение, которое туда пытается вставить запрос, видим 0. Скорее всего этот столбец может принимать два значения — разрешить/не разрешить (1 или 0), то есть однобайтное целое число вполне подойдёт. Поэтому добавим столбец allow_html с указанием типа tinyint:
ALTER TABLE faq ADD allow_html tinyint
Таким образом можно составить шаблон для «лечения» таких проблем: ALTER TABLE [a] ADD [b] [c];, где
a — имя таблицы, откуда выбираются (или где обновляются) данные;
b — имя столбца, который нужно добавить;
c — тип данных.
Примеры (во всех примерах идёт работа с таблицей dle_usergroups):
1) Запрос:
UPDATE dle_usergroups set group_name=‘Журналисты’, allow_html=‘0’ WHERE id=‘3’;
Ответ сервера:
Invalid SQL: UPDATE dle_usergroups set group_name=’Журналисты’, allow_html=’0′ WHERE id=’3′;
MySQL Error: Unknown column ‘allow_html’ in ‘field list’
Error Number: 1054
Решение:
a=dle_usergroups, b=allow_html, c=tinyint, то есть
ALTER TABLE dle_usergroups ADD allow_html tinyint
Для того, чтобы выполнить исправляющий ошибку запрос, необходимо воспользоваться каким-либо mysql-клиентом. В стандартной поставке mysql всегда идёт консольный клиент с названием mysql (в windows mysql.exe). Для того, чтобы подключиться к mysql выполните команду
mysql -hНАЗВАНИЕ_ХОСТА -uИМЯ_ПОЛЬЗОВАТЕЛЯ -pПАРОЛЬ ИМЯ_БАЗЫ_ДАННЫХ,
после чего введите необходимый запрос и точку с запятой после него в появившейся командной строке.
В том случае, если работа происходит на чужом сервере (например, арендуется хостинг) и нет возможности воспользоваться mysql-клиентом из командной строки (не всегда хостеры представляют такую возможность), можно воспользоваться тем инструментом, который предоставляет хостер — например, phpMyAdmin, и в нём ввести нужный sql-запрос.
В то же время наиболее подходящий инструмент для работы с mysql — это MySQL Workbench — разработка создателей mysql с достаточно удобным пользовательским интерфейсом.
Если же нет возможности подключиться к mysql напрямую (например из-за ограничений файрвола), то в ряде случаев возможно удалённо подключиться к MySQL-серверу через SSH-туннель.
2) Запрос:
UPDATE dle_usergroups set group_name=‘Журналисты’, allow_subscribe=‘0’ WHERE id=‘3’;
Ответ сервера:
Invalid SQL: UPDATE dle_usergroups set group_name=’Журналисты’, allow_subscribe=’0′ WHERE id=’3′;
MySQL Error: Unknown column ‘allow_subscribe’ in ‘field list’
Error Number: 1054
Решение:
a=dle_usergroups, b=allow_subscribe, c=tinyint, то есть
ALTER TABLE dle_usergroups ADD allow_subscribe tinyint
3) Запрос:
SELECT faqname, faqparent, displayorder, volatile FROM faq where product IN (», ‘vbulletin’, ‘watermark’, ‘cyb_sfa’, ‘access_post_and_days’);
Oтвет сервера:
InvalidSQL: SELECT faqname, faqparent, displayorder, volatile FROM faq where product IN (», ‘vbulletin’, ‘watermark’, ‘cyb_sfa’, ‘access_post_and_days’);
MySQL Error: Unknown column ‘faqname’ in ‘field list’
Error Number: 1054
Решение:
a= faq, b=faqname, c=varchar(255), то есть
ALTER TABLE faq ADD faqname varchar(255)
Результат
В результате добавления необходимого поля ошибка должна исчезнуть. Однако, существует вероятность того, что в структуре таблиц не хватало несколько столбцов: в этом случае ошибка повторится с указанием другого имени столбца, для которого потребуется повторить процедуру. Помните, что добавление незаполненных столбцов угаданного типа не всегда будет соответствовать задуманной логике приложения и может нарушить часть функциональности.
Источник: webew.ru
Дата публикации: 25.11.2013
© Все права на данную статью принадлежат порталу SQLInfo.ru. Перепечатка в интернет-изданиях разрешается только с указанием автора и прямой ссылки на оригинальную статью. Перепечатка в бумажных изданиях допускается только с разрешения редакции.
Whenever I try to input data into my tblorder I get the error message #1054 — Unknown column ‘FK_Customer_ID’ in ‘field list’.
I have tried breaking my code down and in doing this I found that the error is repeated for FK_Customer_ID and OrderQuantity whereas FK_DVD_ID it will take single data entries. I have tried dropping the table and recreating it, I have dropped the database and recreated it but nothing works. As far as I can tell my code is correct along with my spelling so I’m really stuck.
My tblorder is-
CREATE TABLE tblorder
(
Order_ID INT AUTO_INCREMENT NOT NULL,
FK_Customer_ID INT NOT NULL,
FK_DVD_ID INT NOT NULL,
OrderDate DATETIME NOT NULL DEFAULT NOW(),
OrderQantity INT NOT NULL,
PRIMARY KEY (Order_ID),
FOREIGN KEY (FK_Customer_ID) REFERENCES tblcustomer (Customer_ID),
FOREIGN KEY (FK_DVD_ID) REFERENCES tbldvd (PK_ID)
);
The data I am trying to put in is-
INSERT INTO tblorder
(FK_Customer_ID, FK_DVD_ID, OrderQuantity)
VALUES
(1, 3, 2),
(1, 5, 1),
(1, 10, 4),
(1, 15, 3),
(2, 5, 4),
(2, 17, 3),
(3, 15, 1),
(3, 16, 1),
(3, 17, 1);
FK_Customer_ID is addressing —
CREATE TABLE tblcustomer
(
Customer_ID INT AUTO_INCREMENT NOT NULL,
FirstName VARCHAR(50) NOT NULL,
LastName VARCHAR(50) NOT NULL,
Age INT NOT NULL,
PRIMARY KEY (Customer_ID)
);
FK_DVD_ID is addressing —
CREATE TABLE tblDVD
(
PK_ID INT AUTO_INCREMENT NOT NULL,
Title VARCHAR(100) NOT NULL,
DIrector VARCHAR(100) NOT NULL,
Genre VARCHAR(40) NOT NULL,
dvd_Year YEAR NOT NULL,
Price FLOAT(2) NOT NULL,
Quantity INT NOT NULL,
PRIMARY KEY (PK_ID)
);
Any help in fixing the will be greatly appreciated as it will help me with my A2 computing lesson!
When you execute a MySQL statement, you may sometimes encounter ERROR 1054 as shown below:
mysql> SELECT user_name FROM users;
ERROR 1054 (42S22): Unknown column 'user_name' in 'field list'
The ERROR 1054 in MySQL occurs because MySQL can’t find the column or field you specified in your statement.
This error can happen when you execute any valid MySQL statements like a SELECT
, INSERT
, UPDATE
, or ALTER TABLE
statement.
This tutorial will help you fix the error by adjusting your SQL statements.
Let’s start with the SELECT
statement.
Fix ERROR 1054 on a SELECT statement
To fix the error in your SELECT
statement, you need to make sure that the column(s) you specified in your SQL statement actually exists in your database table.
Because the error above says that user_name
column is unknown, let’s check the users
table and see if the column exists or not.
To help you check the table in question, you can use the DESCRIBE
or EXPLAIN
statement to show your table information.
The example below shows the output of EXPLAIN
statement for the users
table:
mysql> EXPLAIN users;
+--------------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+--------------+-------------+------+-----+---------+-------+
| username | varchar(25) | NO | | | |
| display_name | varchar(50) | NO | | | |
| age | int | YES | | NULL | |
| comments | text | YES | | NULL | |
+--------------+-------------+------+-----+---------+-------+
From the result above, you can see that the users
table has no user_name
field (column)
Instead, it has the username
column without the underscore.
Knowing this, I can adjust my previous SQL query to fix the error:
SELECT username FROM users;
That should fix the error and your SQL query should show the result set.
Fix ERROR 1054 on an INSERT statement
When you specify column names in an INSERT
statement, then the error can be triggered on an INSERT
statement because of a wrong column name, just like in the SELECT
statement.
First, you need to check that you have the right column names in your statement.
Once you are sure, the next step is to look at the VALUES()
you specified in the statement.
For example, when I ran the following statement, I triggered the 1054 error:
mysql> INSERT INTO users(username, display_name)
-> VALUES ("jackolantern", Jack);
ERROR 1054 (42S22): Unknown column 'Jack' in 'field list'
The column names above are correct, and the error itself comes from the last entry in the VALUES()
function.
The display_name
column is of VARCHAR
type, so MySQL expects you to insert a VARCHAR
value into the column.
But Jack
is not a VARCHAR
value because it’s not enclosed in a quotation mark. MySQL considers the value to be a column name.
To fix the error above, simply add a quotation mark around the value. You can use both single quotes or double quotes as shown below:
INSERT INTO users(username, display_name)
VALUES ("jackolantern", 'Jack');
Now the INSERT
statement should run without any error.
Fix ERROR 1054 on an UPDATE statement
To fix the 1054 error caused by an UPDATE
statement, you need to look into the SET
and WHERE
clauses of your statement and make sure that the column names are all correct.
You can look at the error message that MySQL gave you to identify where the error is happening.
For example, the following SQL statement:
UPDATE users
SET username = "jackfrost", display_name = "Jack Frost"
WHERE user_name = "jackolantern";
Produces the following error:
ERROR 1054 (42S22): Unknown column 'user_name' in 'where clause'
The error clearly points toward the user_name
column in the WHERE
clause, so you only need to change that.
If the error points toward the field_list
as shown below:
ERROR 1054 (42S22): Unknown column 'displayname' in 'field list'
Then you need to check on the SET
statement and make sure that:
- You have the right column names
- Any
string
type values are enclosed in a quotation mark
You can also check on the table name that you specified in the UPDATE
statement and make sure that you’re operating on the right table.
Next, let’s look at how to fix the error on an ALTER TABLE
statement
Fix ERROR 1054 on an ALTER TABLE statement
The error 1054 can also happen on an ALTER TABLE
statement.
For example, the following statement tries to rename the displayname
column to realname
:
ALTER TABLE users
RENAME COLUMN displayname TO realname;
Because there’s no displayname
column name in the table, MySQL will respond with the ERROR 1054 message.
Conclusion
In short, ERROR 1054 means that MySQL can’t find the column name that you specified in your SQL statements.
It doesn’t matter if you’re writing an INSERT
, SELECT
, or UPDATE
statement.
There are only two things you need to check to fix the error:
- Make sure you’ve specified the right column name in your statement
- Make sure that any value of
string
type in your statement is surrounded by a quotation mark
You can check on your table structure using the DESCRIBE
or EXPLAIN
statement to help you match the column name and type with your statement.
And that’s how you fix the MySQL ERROR 1054 caused by your SQL statements.
I hope this tutorial has been useful for you 🙏
If you’re getting an error that reads something like “ERROR 1054 (42S22): Unknown column ‘tab.ColName’ in ‘on clause”” in MariaDB, here are three likely causes:
- The column doesn’t exist.
- You’re trying to reference an aliased column by its column name.
- Or it could be the other way around. You could be referencing the column with an alias that was never declared.
If a column has an alias, then you must use that alias when referencing it in any ON
clause when doing a join against two or more tables. Conversely, if you reference a column by an alias, then you need to ensure that the alias is actually declared in the first place.
Example 1
Here’s an example of code that produces the error:
SELECT
c.CatId,
c.CatName
FROM Cats c
INNER JOIN Dogs d
ON c.DogName = d.DogName;
Result:
ERROR 1054 (42S22): Unknown column 'c.DogName' in 'on clause'
Here I accidentally used c.DogName
in the ON
clause when I meant to use c.CatName
.
In this case, the fix is simple. Use the correct column name:
SELECT
c.CatId,
c.CatName
FROM Cats c
INNER JOIN Dogs d
ON c.CatName = d.DogName;
Example 2
Here’s another example of code that produces the error:
SELECT
CatId,
CatName
FROM Cats
INNER JOIN Dogs d
ON c.CatName = d.DogName;
Result:
ERROR 1054 (42S22): Unknown column 'c.CatName' in 'on clause'
Here I referenced a non-existent alias in the ON
clause. I used c.CatName
to refer to the CatName
column in the Cats
table. The only problem is that the Cats
table doesn’t have an alias.
To fix this issue, all we have to do is provide an alias for the Cats
table:
SELECT
CatId,
CatName
FROM Cats c
INNER JOIN Dogs d
ON c.CatName = d.DogName;
Alternatively, we could remove all references to the alias, and just use the full table name:
SELECT
CatId,
CatName
FROM Cats
INNER JOIN Dogs
ON Cats.CatName = Dogs.DogName;
One thing I should point out is that, in this example we didn’t prefix the column names in the SELECT
list with the alias. If we had done that, we would have seen the same error, but with a slightly different message:
SELECT
c.CatId,
c.CatName
FROM Cats
INNER JOIN Dogs d
ON c.CatName = d.DogName;
Result:
ERROR 1054 (42S22): Unknown column 'c.CatId' in 'field list'
In this case, it detected the unknown columns in the field list before it found the one in the ON
clause. Either way, the solution is the same.
Example 3
Here’s another example of code that produces the error:
SELECT
c.CatId,
c.CatName
FROM Cats c
INNER JOIN Dogs d
ON Cats.CatName = d.DogName;
Result:
ERROR 1054 (42S22): Unknown column 'Cats.CatName' in 'on clause'
In this case, an alias was declared for the Cats
table, but I didn’t use that alias in the ON
clause.
The solution here, is to use the alias instead of the table name:
SELECT
c.CatId,
c.CatName
FROM Cats c
INNER JOIN Dogs d
ON c.CatName = d.DogName;
Error: Code 1054. Unknown column 'U2.id_naslov' in 'field list'
gets thrown on
this simple query in MySQL Workbench:
UPDATE krneki_1 AS U1, krneki_2 AS U2
SET U1.id_naslov = U2.id_naslov
WHERE (U2.id_zaposlen = U1.id_naslovi_zaposleni)
I have searched and read other posts on the net but nothing helps…
I guess it’s a trivial solution but I just can’t see it.
This kind of error never came up on TSQL (sql server).
Table krneki_2 was created by Mysql workbench via data import (create new table) later on when this error occurred I also altered the number fields to smallint just to see if it helps… but… nothing.
Result of SHOW CREATE TABLE krneki_2
:
Table: krneki_2
Create Table: CREATE TABLE `krneki_2`
( `id` smallint(6) NOT NULL AUTO_INCREMENT,
`id_naslov` smallint(6) NOT NULL,
`id_zaposlen` smallint(6) NOT NULL,
PRIMARY KEY (id)
) ENGINE=InnoDB AUTO_INCREMENT=204 DEFAULT CHARSET=utf8
1 row in set (0.00 sec)
Result of SHOW CREATE TABLE krneki_1
:
Table: krneki_1
Create Table: CREATE TABLE `krneki_1` (
`id_naslovi_zaposleni` smallint(6) NOT NULL AUTO_INCREMENT,
`id_naslov` smallint(6) DEFAULT NULL,
`id_zaposleni` smallint(6) DEFAULT NULL,
`id_aktiven` tinyint(4) DEFAULT '0',
`cas_vnosa` datetime DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`id_naslovi_zaposleni`)
) ENGINE=InnoDB AUTO_INCREMENT=256 DEFAULT CHARSET=utf8
1 row in set (0.00 sec)
Results from information_schema
, specifically from this query suggested in comments:
select
table_catalog, table_schema, table_name, column_name, ordinal_position
from information_schema.columns
where table_name like '%krneki_1%' and column_name like '%naslov%' ;
Results for krneki_1
and naslov
:
+---------------+--------------+-------------+----------------------+------------------+
| table_catalog | table_schema | table_name | column_name | ordinal_position |
+---------------+--------------+-------------+----------------------+------------------+
| def | hq_db | krneki_1 | id_naslovi_zaposleni | 1 |
| def | hq_db | krneki_1 | id_naslov | 2 |
+---------------+--------------+-------------+----------------------+------------------+
2 rows in set (0.00 sec)
Results for krneki_2
and naslov
:
+---------------+--------------+-------------+--------------+------------------+
| table_catalog | table_schema | table_name | column_name | ordinal_position |
+---------------+--------------+-------------+--------------+------------------+
| def | hq_db | krneki_2 | id_naslov | 2 |
+---------------+--------------+-------------+--------------+------------------+
1 row in set (0.00 sec)
Results for krneki_2
and zaposlen
:
+---------------+--------------+-------------+--------------+------------------+
| table_catalog | table_schema | table_name | column_name | ordinal_position |
+---------------+--------------+-------------+--------------+------------------+
| def | hq_db | krneki_2 | id_zaposlen | 3 |
+---------------+--------------+-------------+--------------+------------------+
1 row in set (0.00 sec)
Further digging, as suggested:
select
table_catalog, table_schema, table_name, column_name, ordinal_position,
char_length(column_name) as cl, length(column_name) as l
from information_schema.columns
where table_name = 'krneki_2' ;
Results for krneki_2
:
+-------------+------------+----------+-----------+----------------+---+---+-------------+
|table_catalog|table_schema|table_name|column_name|ordinal_position| cl| l | column_type |
+-------------+------------+----------+-----------+----------------+---+---+-------------+
| def | hq_db | krneki_2 |id | 1 | 2| 2| smallint(6) |
| def | hq_db | krneki_2 |id_naslov | 2 | 10| 12| smallint(6) |
| def | hq_db | krneki_2 |id_zaposlen| 3 | 11| 11| smallint(6) |
+-------------+------------+----------+-----------+----------------+---+---+-------------+
3 rows in set (0.00 sec)
Results for krneki_1
:
+-------------+------------+----------+--------------------+----------------+--+--+-----------+
|table_catalog|table_schema|table_name| column_name |ordinal_position|cl| l|column_type|
+-------------+------------+----------+--------------------+----------------+--+--+-----------+
| def | hq_db | krneki_1 |id_naslovi_zaposleni| 1 |20|20|smallint(6)|
| def | hq_db | krneki_1 |id_naslov | 2 | 9| 9|smallint(6)|
| def | hq_db | krneki_1 |id_zaposleni | 3 |12|12|smallint(6)|
| def | hq_db | krneki_1 |id_aktiven | 4 |10|10|tinyint(4) |
| def | hq_db | krneki_1 |cas_vnosa | 5 | 9| 9|datetime |
+-------------+------------+----------+--------------------+----------------+--+--+-----------+
5 rows in set (0.00 sec)
krneki_2 with HEX:
+-------------+------------+----------+-----------+----------------+--+--+-------------------------+
|table_catalog|table_schema|table_name|column_name|ordinal_position|cl|l | hex |
+-------------+------------+----------+-----------+----------------+--+--+-------------------------+
| def | hq_db | krneki_2 |id | 1 | 2| 2|6964 |
| def | hq_db | krneki_2 |id_naslov | 2 |10|12|EFBBBF69645F6E61736C6F76 |
| def | hq_db | krneki_2 |id_zaposlen| 3 |11|11|69645F7A61706F736C656E |
+-------------+------------+----------+-----------+----------------+--+--+-------------------------+
3 rows in set (0.00 sec)
krneki_1 with HEX:
+-------------+------------+----------+--------------------+----------------+--+--+----------------------------------------+
|table_catalog|table_schema|table_name|column_name |ordinal_position|cl| l|hex |
+-------------+------------+----------+--------------------+----------------+--+--+----------------------------------------+
| def | hq_db | krneki_1 |id_naslovi_zaposleni| 1 |20|20|69645F6E61736C6F76695F7A61706F736C656E69|
| def | hq_db | krneki_1 |id_naslov | 2 | 9| 9|69645F6E61736C6F76 |
| def | hq_db | krneki_1 |id_zaposleni | 3 |12|12|69645F7A61706F736C656E69 |
| def | hq_db | krneki_1 |id_aktiven | 4 |10|10|69645F616B746976656E |
| def | hq_db | krneki_1 |cas_vnosa | 5 | 9| 9|6361735F766E6F7361 |
+-------------+------------+----------+--------------------+----------------+--+--+----------------------------------------+
5 rows in set (0.00 sec)
Joomla is a great way to create websites. It comes as free and requires zero expertise in coding too.
But, often due to bad configuration, Joomla reports error in the website. One such error is Joomla error 1054.
Usually, Joomla shows error 1054 when updating from older versions. And, often it makes Joomla backend unavailable.
That’s why, we often get requests from customers to fix Joomla errors as part of our Technical Support Services.
Today, we’ll see the details of error 1054 and how Bobcares’ Engineers fix it.
When, What and Why of Joomla error 1054 ?
Before proceeding further, we’ll see further details of error.
Usually, the 1054 error happens when upgrading Joomla to latest version from older version.
Recently, one of our customers reported this problem when doing an upgrade from 3.7.5 to 3.8. Unfortunately, the immediate effect is that Joomla back-end becomes inaccessible. And, it reports variation of the error as:
Error 1054 Unknown column 'a.client_id' in 'where clause.'
Surprisingly, the Joomla frontend will work fine. Therefore, it affects customers only while make changes in the Joomla backend. When accessing the Joomla administrator panel, you do not see the usual control panel. Instead, it will show up as completely blank screen.
But, what causes the error?
Here, by this error, Joomla complains about missing columns in Joomla database. Just as site building tools like WordPress, Joomla also stores site configuration in Databases. And, error 1054 means problem with sites Joomla database.
How we fix error 1054 in Joomla?
Ok. The reason for error is screwed up database. Therefore, to make Joomla backend panel working again, we need to fix the database.
Again, fixing database errors need utmost care. We’ll now see how our Dedicated Engineers fix it for the customers.
1. From Joomla Admin panel
When Joomla admin panel is accessible, the error 1054 can be fixed from the panel itself.
For this, we follow the steps as listed below.
We select Extensions → Manage → Database from the drop-down menu of the Joomla! Administrator Panel. It will show errors with the Database. To fix them, our Support Engineers Click on the Fix button.
After a few seconds, it will show the “Database table structure is up to date” message:
That fixes the 1054 Unknown Column ‘client_id’ in ‘field list’ error and admin panel becomes accessible again.
2. From PHPMyAdmin
Unfortunately, in certain cases, Joomla admin panel just shows a blank page. In such cases, PHPMyAdmin helps in fixing database inconsistencies easily. Luckily, running an SQL query in the database can solve the Joomla error.
That’s why, our Dedicated Engineers connect to the websites’ control panel and then to PHPMyAdmin.
We run the following SQL query commands to alter the database table.
ALTER TABLE `xxx_menu_types` ADD COLUMN `client_id` int(11) NOT NULL DEFAULT 0;
UPDATE `xxx_menu` SET `published` = 1 WHERE `menutype` = 'main' OR `menutype` = 'menu';
Here, we substitute ‘xxx’ with the prefix of website’s database tables:
After running this query, the Joomla panel becomes accessible again.
Conclusion
In short, Joomla error 1054 happens due to inconsistencies with website’s database. Today, we saw the details of the error and how our Dedicated Engineers fix it and make Joomla backend working again.
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Let’s see when the #1054 error occurs in MySQL. While inserting a varchar value, if you will forget to add single quotes, then this error will arise.
Following is the error −
mysql> insert into DemoTable798 values(100,Adam); ERROR 1054 (42S22): Unknown column 'Adam' in 'field list'
You need to use single quotes around the string value to fix this error as shown below −
mysql> insert into DemoTable798 values(100,’Adam’);
Let us first create a table −
mysql> create table DemoTable798 ( StudentId int, StudentName varchar(100) ); Query OK, 0 rows affected (0.51 sec)
Insert some records in the table using insert command −
mysql> insert into DemoTable798 values(100,'Adam'); Query OK, 1 row affected (0.16 sec) mysql> insert into DemoTable798 values(101,'Chris'); Query OK, 1 row affected (0.19 sec) mysql> insert into DemoTable798 values(102,'Robert'); Query OK, 1 row affected (0.16 sec) mysql> insert into DemoTable798 values(103,'Carol'); Query OK, 1 row affected (0.16 sec)
Display all records from the table using select statement −
mysql> select *from DemoTable798;
This will produce the following output —
+-----------+-------------+ | StudentId | StudentName | +-----------+-------------+ | 100 | Adam | | 101 | Chris | | 102 | Robert | | 103 | Carol | +-----------+-------------+ 4 rows in set (0.00 sec)
Recently we got request from our customer that something is going wrong with their database and they’re are getting strange errors after each insert or update to specific table. The strangeness caused by «Unknown column ‘column-name’ in ‘field list'» message while this column was existing in this table. Our investigation shown that this was caused by trigger on the table they were trying to do the insert/update. This trigger did the insert to another table where the mentioned column didn’t exist.
Let me show you example:
create test DB and 2 test tables:
create database test; use test; create table t1 (field1 int auto_increment not null, field2 varchar(10), field3 varchar(10), primary key(field1)) engine=innodb; create table t2 (field1 int, field2 varchar(10)) engine=innodb;
Then create a new insert trigger on t1 table:
delimiter ## create trigger insert_on_t1_to_t2 after insert on t1 for each row begin insert into t2 (field1,field2,field3) values (new.field1, new.field2, '1'); end## delimiter ;
Insert to t1 table:
mysql> insert into t1(field2,field3) values("val2","val3"); ERROR 1054 (42S22): Unknown column 'field3' in 'field list'
From the query text and the output above it’s pretty easy to assume that mysql is complaining about `c` column in `t1` table. But in fact it’s complaining about `c` column in `t2` table because this query is running trigger and does insert into `t2` tables and its `(a,b,c)` columns but the latter column doesn’t exist in `t2` table.
One of the simplest way to check if it’s so is show triggers and grep result by table and column name:
mysql> pager grep -E "t1|field3" PAGER set to 'grep -E "t1|field3”' mysql> show triggers from test like 't1'G Trigger: insert_on_t1_to_t2 Table: t1 insert into t2 (field1,field2,field3) values (new.field1, new.field2, '1'); 1 row in set (0.00 sec)
To solve the error I will need either rewrite trigger and remove `field3` column from insert command or alter table `t2` and add column `field3`.
In my test I did rewrite trigger:
drop trigger insert_on_t1_to_t2; delimiter ## create trigger insert_on_t1_to_t2 after insert on t1 for each row begin insert into t2 (field1,field2) values (new.field1, new.field2); end## delimiter ;
Now the insert works:
mysql> select * from t1; +--------+--------+--------+ | field1 | field2 | field3 | +--------+--------+--------+ | 2 | val20 | val30 | | 3 | val21 | val31 | +--------+--------+--------+ 2 rows in set (0.00 sec) mysql> select * from t2; +--------+--------+ | field1 | field2 | +--------+--------+ | 2 | val20 | | 3 | val21 | +--------+--------+ 2 rows in set (0.00 sec)
Summary: if you’re seeing error like this one that some of the column doesn’t exist in the table but in fact it exists then one of the culprit could be a trigger on this table.