I have created tables in MySQL Workbench as shown below :
ORDRE table:
CREATE TABLE Ordre (
OrdreID INT NOT NULL,
OrdreDato DATE DEFAULT NULL,
KundeID INT DEFAULT NULL,
CONSTRAINT Ordre_pk PRIMARY KEY (OrdreID),
CONSTRAINT Ordre_fk FOREIGN KEY (KundeID) REFERENCES Kunde (KundeID)
)
ENGINE = InnoDB;
PRODUKT table:
CREATE TABLE Produkt (
ProduktID INT NOT NULL,
ProduktBeskrivelse VARCHAR(100) DEFAULT NULL,
ProduktFarge VARCHAR(20) DEFAULT NULL,
Enhetpris INT DEFAULT NULL,
CONSTRAINT Produkt_pk PRIMARY KEY (ProduktID)
)
ENGINE = InnoDB;
and ORDRELINJE table:
CREATE TABLE Ordrelinje (
Ordre INT NOT NULL,
Produkt INT NOT NULL,
AntallBestilt INT DEFAULT NULL,
CONSTRAINT Ordrelinje_pk PRIMARY KEY (Ordre, Produkt),
CONSTRAINT Ordrelinje_fk FOREIGN KEY (Ordre) REFERENCES Ordre (OrdreID),
CONSTRAINT Ordrelinje_fk1 FOREIGN KEY (Produkt) REFERENCES Produkt (ProduktID)
)
ENGINE = InnoDB;
so when I try to insert values into ORDRELINJE table i get:
Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails (
srdjank.Ordrelinje, CONSTRAINTOrdrelinje_fkFOREIGN KEY (Ordre) REFERENCESOrdre(OrdreID))
I’ve seen the other posts on this topic, but no luck.
Am I overseeing something or any idea what to do?
I’m still fairly new to SQL. I’m updating a DB and I came across this message. The problem is, I’ve already executed this insert before but had to delete it due to me entering the same address 3 times instead of once.
Can anybody help me, I don’t understand what is wrong:
> insert into ort
(plz, name) values
('4900', 'Langenthal')
;
>insert into adresse
(strasse, strassennr, ortID) values
('Eisenbahnstrasse', '7', (select oid from ort where name = 'Langenthal' and plz='4900'))
;
>
insert into liegenschaft
(liegenschafttypid, adressid) values
((select ltypid from liegenschaft_typ where name = 'Wohnhaus / Firma'), (select oid from ort where name = 'Langenthal' and plz = '4900'))
;
I keep on getting this message:
> 0 16 14:09:25 insert into liegenschaft (liegenschafttypid, adressid) values
((select ltypid from liegenschaft_typ where name = 'Wohnhaus / Firma'), (select oid from ort where name = 'Langenthal' and plz = '4900')) Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails (`parking`.`liegenschaft`, CONSTRAINT `FK_adresse` FOREIGN KEY (`adressID`) REFERENCES `adresse` (`AID`)) 0.015 sec
The MySQL ERROR 1452 happens when you try to execute a data manipulation query into a table that has one or more failing foreign key constraints.
The cause of this error is the values you’re trying to put into the table are not available in the referencing (parent) table.
Let’s see an example of this error with two MySQL tables.
Suppose you have a Cities table that contains the following data:
+----+------------+
| id | city_name |
+----+------------+
| 1 | York |
| 2 | Manchester |
| 3 | London |
| 4 | Edinburgh |
+----+------------+
Then, you create a Friends table to keep a record of people you know who lives in different cities.
You reference the id column of the Cities table as the FOREIGN KEY of the city_id column in the Friends table as follows:
CREATE TABLE `Friends` (
`firstName` varchar(255) NOT NULL,
`city_id` int unsigned NOT NULL,
PRIMARY KEY (`firstName`),
CONSTRAINT `friends_ibfk_1`
FOREIGN KEY (`city_id`) REFERENCES `Cities` (`id`)
)
In the code above, a CONSTRAINT named friends_ibfk_1 is created for the city_id column, referencing the id column in the Cities table.
This CONSTRAINT means that only values recoded in the id column can be inserted into the city_id column.
(To avoid confusion, I have omitted the id column from the Friends table. In real life, You may have an id column in both tables, but a FOREIGN KEY constraint will always refer to a different table.)
When I try to insert 5 as the value of the city_id column, I will trigger the error as shown below:
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('John', 5);
The response from MySQL:
ERROR 1452 (23000): Cannot add or update a child row:
a foreign key constraint fails
(`test_db`.`friends`, CONSTRAINT `friends_ibfk_1`
FOREIGN KEY (`city_id`) REFERENCES `cities` (`id`))
As you can see, the error above even describes which constraint you are failing from the table.
Based on the Cities table data above, I can only insert numbers between 1 to 4 for the city_id column to make a valid INSERT statement.
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('John', 1);
-- Query OK, 1 row affected (0.00 sec)
The same error will happen when I try to update the Friends row with a city_id value that’s not available.
Take a look at the following example:
UPDATE `Friends` SET city_id = 5 WHERE `firstName` = 'John';
-- ERROR 1452 (23000): Cannot add or update a child row
There are two ways you can fix the ERROR 1452 in your MySQL database server:
- You add the value into the referenced table
- You disable the
FOREIGN_KEY_CHECKSin your server
The first option is to add the value you need to the referenced table.
In the example above, I need to add the id value of 5 to the Cities table:
INSERT INTO `Cities` VALUES (5, 'Liverpool');
-- Cities table:
+----+------------+
| id | city_name |
+----+------------+
| 1 | York |
| 2 | Manchester |
| 3 | London |
| 4 | Edinburgh |
| 5 | Liverpool |
+----+------------+
Now I can insert a new row in the Friends table with the city_id value of 5:
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('Susan', 5);
-- Query OK, 1 row affected (0.00 sec)
Disabling the foreign key check
The second way you can fix the ERROR 1452 issue is to disable the FOREIGN_KEY_CHECKS variable in your MySQL server.
You can check whether the variable is active or not by running the following query:
SHOW GLOBAL VARIABLES LIKE 'FOREIGN_KEY_CHECKS';
-- +--------------------+-------+
-- | Variable_name | Value |
-- +--------------------+-------+
-- | foreign_key_checks | ON |
-- +--------------------+-------+
This variable causes MySQL to check any foreign key constraint added to your table(s) before inserting or updating.
You can disable the variable for the current session only or globally:
-- set for the current session:
SET FOREIGN_KEY_CHECKS=0;
-- set globally:
SET GLOBAL FOREIGN_KEY_CHECKS=0;
Now you can INSERT or UPDATE rows in your table without triggering a foreign key constraint fails:
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('Natalia', 8);
-- Query OK, 1 row affected (0.01 sec)
UPDATE `Friends` SET city_id = 17 WHERE `firstName` = 'John';
-- Query OK, 1 row affected (0.00 sec)
-- Rows matched: 1 Changed: 1 Warnings: 0
After you’re done with the manipulation query, you can set the FOREIGN_KEY_CHECKS active again by setting its value to 1:
-- set for the current session:
SET FOREIGN_KEY_CHECKS=1;
-- set globally:
SET GLOBAL FOREIGN_KEY_CHECKS=1;
But please be warned that turning off your FOREIGN_KEY_CHECKS variable will cause the city_id column to reference a NULL column in the cities table.
It may cause problems when you need to perform a JOIN query later.
Now you’ve learned the cause of ERROR 1452 and how to resolve this issue in your MySQL database server. Great work! 👍
Wondering how to resolve MySQL Error 1452? We can help you.
At Bobcares, we offer solutions for every query, big and small, as a part of our Microsoft SQL Server Support Services.
Let’s take a look at how our Support Team is ready to help customers with MySQL Error 1452.
How to resolve MySQL Error 1452?
Usually, this error occurs when we try to execute a data manipulation query into a table that has one or more failing foreign key constraints.
What causes MySQL Error 1452?
The cause of this error is the values we are trying to put into the table are not available in the referencing (parent) table.
When a column of a table is referenced from another table, it is called Foreign Key.
For example, consider a table City that contains the name of a city and its ID.
Also, there is another table Buddies to keep a record of people that we know who lives in different cities.
We have to reference the id column of the City table as the FOREIGN KEY of the city_id column in the friends table as follows:
CREATE TABLE friends (
firstName varchar(255) NOT NULL,
city_id int unsigned NOT NULL,
PRIMARY KEY (firstName),
CONSTRAINT friends_ibfk_1
FOREIGN KEY (city_id) REFERENCES City (id)
)In the code above, a CONSTRAINT named buddies_ibfk_1 is created for the city_id column, referencing the id column in the City table.
This CONSTRAINT means that only values in the id column can be inserted into the city_id column.
If we try to insert a value that is not present in id column into the city_id column, it will trigger the error as shown below:
ERROR 1452 (23000): Cannot add or update a child row:
a foreign key constraint fails
(test_db.friends, CONSTRAINT friends_ibfk_1
FOREIGN KEY (city_id) REFERENCES city (id))How to resolve it?
Today, let us see the steps followed by our Support Techs to resolve it:
There are two ways to fix the ERROR 1452 in MySQL database server:
1. Firstly, add the value into the referenced table
2. Then, disable the FOREIGN_KEY_CHECKS in the server
1. Add the value into the referenced table
The first option is to add the value we need to the referenced table.
In the example above, add the required id value to the City table.
Now we can insert a new row in the Buddies table with the city_id value that we inserted.
Disabling the foreign key check
2. Disable the FOREIGN_KEY_CHECKS variable in MySQL server.
We can check whether the variable is active or not by running the following query:
SHOW GLOBAL VARIABLES LIKE ‘FOREIGN_KEY_CHECKS’;— +——————–+——-+ — | Variable_name | Value | — +——————–+——-+ — | foreign_key_checks | ON | — +——————–+——-+
This variable causes MySQL to check any foreign key constraint added to our table(s) before inserting or updating.
We can disable the variable for the current session only or globally:
— set for the current session:
SET FOREIGN_KEY_CHECKS=0;— set globally:
SET GLOBAL FOREIGN_KEY_CHECKS=0;Now we can INSERT or UPDATE rows in our table without triggering a foreign key constraint fails.
After we are done with the manipulation query, we can set the FOREIGN_KEY_CHECKS active again by setting its value to 1:
— set for the current session:
SET FOREIGN_KEY_CHECKS=1;— set globally:
SET GLOBAL FOREIGN_KEY_CHECKS=1;Turning off FOREIGN_KEY_CHECKS variable will cause the city_id column to reference a NULL column in the City table.
It may cause problems when we need to perform a JOIN query later.
[Looking for a solution to another query? We are just a click away.]
Conclusion
To sum up, our skilled Support Engineers at Bobcares demonstrated how to resolve MySQL Error 1452.
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I’m having a bit of a strange problem. I’m trying to add a foreign key to one table that references another, but it is failing for some reason. With my limited knowledge of MySQL, the only thing that could possibly be suspect is that there is a foreign key on a different table referencing the one I am trying to reference.
I’ve done a SHOW CREATE TABLE query on both tables, sourcecodes_tags is the table with the foreign key, sourcecodes is the referenced table.
CREATE TABLE `sourcecodes` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) unsigned NOT NULL,
`language_id` int(11) unsigned NOT NULL,
`category_id` int(11) unsigned NOT NULL,
`title` varchar(40) CHARACTER SET utf8 NOT NULL,
`description` text CHARACTER SET utf8 NOT NULL,
`views` int(11) unsigned NOT NULL,
`downloads` int(11) unsigned NOT NULL,
`time_posted` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`),
KEY `language_id` (`language_id`),
KEY `category_id` (`category_id`),
CONSTRAINT `sourcecodes_ibfk_3` FOREIGN KEY (`language_id`) REFERENCES `languages` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_1` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_2` FOREIGN KEY (`category_id`) REFERENCES `categories` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1
CREATE TABLE `sourcecodes_tags` (
`sourcecode_id` int(11) unsigned NOT NULL,
`tag_id` int(11) unsigned NOT NULL,
KEY `sourcecode_id` (`sourcecode_id`),
KEY `tag_id` (`tag_id`),
CONSTRAINT `sourcecodes_tags_ibfk_1` FOREIGN KEY (`tag_id`) REFERENCES `tags` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1
This is the code that generates the error:
ALTER TABLE sourcecodes_tags ADD FOREIGN KEY (sourcecode_id) REFERENCES sourcecodes (id) ON DELETE CASCADE ON UPDATE CASCADE
I’m having a bit of a strange problem. I’m trying to add a foreign key to one table that references another, but it is failing for some reason. With my limited knowledge of MySQL, the only thing that could possibly be suspect is that there is a foreign key on a different table referencing the one I am trying to reference.
I’ve done a SHOW CREATE TABLE query on both tables, sourcecodes_tags is the table with the foreign key, sourcecodes is the referenced table.
CREATE TABLE `sourcecodes` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) unsigned NOT NULL,
`language_id` int(11) unsigned NOT NULL,
`category_id` int(11) unsigned NOT NULL,
`title` varchar(40) CHARACTER SET utf8 NOT NULL,
`description` text CHARACTER SET utf8 NOT NULL,
`views` int(11) unsigned NOT NULL,
`downloads` int(11) unsigned NOT NULL,
`time_posted` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`),
KEY `language_id` (`language_id`),
KEY `category_id` (`category_id`),
CONSTRAINT `sourcecodes_ibfk_3` FOREIGN KEY (`language_id`) REFERENCES `languages` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_1` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_2` FOREIGN KEY (`category_id`) REFERENCES `categories` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1
CREATE TABLE `sourcecodes_tags` (
`sourcecode_id` int(11) unsigned NOT NULL,
`tag_id` int(11) unsigned NOT NULL,
KEY `sourcecode_id` (`sourcecode_id`),
KEY `tag_id` (`tag_id`),
CONSTRAINT `sourcecodes_tags_ibfk_1` FOREIGN KEY (`tag_id`) REFERENCES `tags` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1
This is the code that generates the error:
ALTER TABLE sourcecodes_tags ADD FOREIGN KEY (sourcecode_id) REFERENCES sourcecodes (id) ON DELETE CASCADE ON UPDATE CASCADE
The MySQL ERROR 1452 happens when you try to execute a data manipulation query into a table that has one or more failing foreign key constraints.
The cause of this error is the values you’re trying to put into the table are not available in the referencing (parent) table.
Let’s see an example of this error with two MySQL tables.
Suppose you have a Cities table that contains the following data:
+----+------------+
| id | city_name |
+----+------------+
| 1 | York |
| 2 | Manchester |
| 3 | London |
| 4 | Edinburgh |
+----+------------+
Then, you create a Friends table to keep a record of people you know who lives in different cities.
You reference the id column of the Cities table as the FOREIGN KEY of the city_id column in the Friends table as follows:
CREATE TABLE `Friends` (
`firstName` varchar(255) NOT NULL,
`city_id` int unsigned NOT NULL,
PRIMARY KEY (`firstName`),
CONSTRAINT `friends_ibfk_1`
FOREIGN KEY (`city_id`) REFERENCES `Cities` (`id`)
)
In the code above, a CONSTRAINT named friends_ibfk_1 is created for the city_id column, referencing the id column in the Cities table.
This CONSTRAINT means that only values recoded in the id column can be inserted into the city_id column.
(To avoid confusion, I have omitted the id column from the Friends table. In real life, You may have an id column in both tables, but a FOREIGN KEY constraint will always refer to a different table.)
When I try to insert 5 as the value of the city_id column, I will trigger the error as shown below:
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('John', 5);
The response from MySQL:
ERROR 1452 (23000): Cannot add or update a child row:
a foreign key constraint fails
(`test_db`.`friends`, CONSTRAINT `friends_ibfk_1`
FOREIGN KEY (`city_id`) REFERENCES `cities` (`id`))
As you can see, the error above even describes which constraint you are failing from the table.
Based on the Cities table data above, I can only insert numbers between 1 to 4 for the city_id column to make a valid INSERT statement.
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('John', 1);
-- Query OK, 1 row affected (0.00 sec)
The same error will happen when I try to update the Friends row with a city_id value that’s not available.
Take a look at the following example:
UPDATE `Friends` SET city_id = 5 WHERE `firstName` = 'John';
-- ERROR 1452 (23000): Cannot add or update a child row
There are two ways you can fix the ERROR 1452 in your MySQL database server:
- You add the value into the referenced table
- You disable the
FOREIGN_KEY_CHECKSin your server
The first option is to add the value you need to the referenced table.
In the example above, I need to add the id value of 5 to the Cities table:
INSERT INTO `Cities` VALUES (5, 'Liverpool');
-- Cities table:
+----+------------+
| id | city_name |
+----+------------+
| 1 | York |
| 2 | Manchester |
| 3 | London |
| 4 | Edinburgh |
| 5 | Liverpool |
+----+------------+
Now I can insert a new row in the Friends table with the city_id value of 5:
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('Susan', 5);
-- Query OK, 1 row affected (0.00 sec)
Disabling the foreign key check
The second way you can fix the ERROR 1452 issue is to disable the FOREIGN_KEY_CHECKS variable in your MySQL server.
You can check whether the variable is active or not by running the following query:
SHOW GLOBAL VARIABLES LIKE 'FOREIGN_KEY_CHECKS';
-- +--------------------+-------+
-- | Variable_name | Value |
-- +--------------------+-------+
-- | foreign_key_checks | ON |
-- +--------------------+-------+
This variable causes MySQL to check any foreign key constraint added to your table(s) before inserting or updating.
You can disable the variable for the current session only or globally:
-- set for the current session:
SET FOREIGN_KEY_CHECKS=0;
-- set globally:
SET GLOBAL FOREIGN_KEY_CHECKS=0;
Now you can INSERT or UPDATE rows in your table without triggering a foreign key constraint fails:
INSERT INTO `Friends` (`firstName`, `city_id`) VALUES ('Natalia', 8);
-- Query OK, 1 row affected (0.01 sec)
UPDATE `Friends` SET city_id = 17 WHERE `firstName` = 'John';
-- Query OK, 1 row affected (0.00 sec)
-- Rows matched: 1 Changed: 1 Warnings: 0
After you’re done with the manipulation query, you can set the FOREIGN_KEY_CHECKS active again by setting its value to 1:
-- set for the current session:
SET FOREIGN_KEY_CHECKS=1;
-- set globally:
SET GLOBAL FOREIGN_KEY_CHECKS=1;
But please be warned that turning off your FOREIGN_KEY_CHECKS variable will cause the city_id column to reference a NULL column in the cities table.
It may cause problems when you need to perform a JOIN query later.
Now you’ve learned the cause of ERROR 1452 and how to resolve this issue in your MySQL database server. Great work! 👍
Я создал таблицы в MySQL Workbench, как показано ниже:
Таблица ORDRE:
Таблица PRODUKT:
и таблица ORDRELINJE:
поэтому, когда я пытаюсь вставить значения в ORDRELINJE таблицу, я получаю:
Код ошибки: 1452. Не удается добавить или обновить дочернюю строку: ограничение внешнего ключа не выполняется ( srdjank . Ordrelinje , ОГРАНИЧЕНИЕ Ordrelinje_fk FOREIGN KEY ( Ordre ) REFERENCES Ordre ( OrdreID ))
Я видел другие сообщения по этой теме, но безуспешно. Я что-то контролирую или есть идеи, что делать?
Взаимосвязи внешнего ключа включают родительскую таблицу, которая содержит центральные значения данных, и дочернюю таблицу с идентичными значениями, указывающими на ее родительскую. Предложение FOREIGN KEY указано в дочерней таблице.
Он отклонит любую операцию INSERT или UPDATE, которая пытается создать значение внешнего ключа в дочерней таблице, если в родительской таблице нет подходящего значения ключа кандидата.
Таким образом, ваша ошибка по Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails существу означает, что вы пытаетесь добавить строку в свою Ordrelinje таблицу, для которой в таблице нет соответствующей строки (OrderID) Ordre .
Вы должны сначала вставить строку в свою Ordre таблицу.
Вы получаете эту проверку ограничения, потому что Ordre таблица не имеет ссылки OrdreID в команде вставки.
Чтобы вставить значение в Ordrelinje , вы сначала должны ввести значение в Ordre таблицу и использовать его OrdreID в Orderlinje таблице.
Или вы можете удалить ненулевое ограничение и вставить в него значение NULL.
Вы должны удалить данные в дочерней таблице, которая не имеет соответствующего значения внешнего ключа для первичного ключа родительской таблицы. Или удалите все данные из дочерней таблицы, а затем вставьте новые данные, имеющие то же значение внешнего ключа, что и первичный ключ в родительской таблице. , Это должно сработать. Здесь также видео на YouTube
Проблема связана с ограничением FOREIGN KEY. По умолчанию (SET FOREIGN_KEY_CHECKS = 1). Параметр FOREIGN_KEY_CHECKS указывает, следует ли проверять ограничения внешнего ключа для таблиц InnoDB. MySQL — УСТАНОВИТЬ FOREIGN_KEY_CHECKS
Мы можем отключить проверку внешнего ключа перед запуском запроса. Отключить внешний ключ .
Выполните одну из этих строк перед запуском вашего запроса, тогда вы сможете успешно выполнить свой запрос. 🙂
1) Для сеанса (рекомендуется)
Эта ошибка обычно возникает из-за того, что у нас есть некоторые значения в поле ссылки дочерней таблицы, которые не существуют в поле ссылки / кандидата родительской таблицы.
Иногда мы можем получить эту ошибку, когда применяем ограничения внешнего ключа к существующим таблицам (таблицам), в которых уже есть данные . В некоторых других ответах предлагается полностью удалить данные из дочерней таблицы, а затем применить ограничение. Однако это не вариант, когда у нас уже есть рабочие / производственные данные в дочерней таблице. В большинстве сценариев нам потребуется обновить данные в дочерней таблице (вместо их удаления).
Теперь мы можем использовать Left Join для поиска всех этих строк в дочерней таблице, которая не имеет совпадающих значений в родительской таблице. Следующий запрос был бы полезен для получения этих несовпадающих строк:
Теперь вы обычно можете выполнить один (или несколько) из следующих шагов, чтобы исправить данные.
- Основываясь на вашей «бизнес-логике», вам нужно будет обновить / сопоставить эти несоответствующие значения с существующими значениями в родительской таблице. Иногда вам также может потребоваться их установить null .
- Удалите эти строки с несоответствующими значениями.
- Добавьте новые строки в родительскую таблицу, соответствующие несоответствующим значениям в дочерней таблице.
Как только данные зафиксированы, мы можем применить ограничение внешнего ключа, используя ALTER TABLE синтаксис.
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