Ошибка 1452 mysql как исправить

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, CONSTRAINT Ordrelinje_fk FOREIGN KEY (Ordre) REFERENCES Ordre (OrdreID))

I’ve seen the other posts on this topic, but no luck.
Am I overseeing something or any idea what to do?

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.

PREVENT YOUR SERVER FROM CRASHING!

Never again lose customers to poor server speed! Let us help you.

Our server experts will monitor & maintain your server 24/7 so that it remains lightning fast and secure.

GET STARTED

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_CHECKS in 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! 👍

I am trying to insert values into my comments table and I am getting a error. Its saying that I can not add or update child row and I have no idea what that means.

my schema looks something like this

-- ----------------------------
-- Table structure for `comments`
-- ----------------------------
DROP TABLE IF EXISTS `comments`;
CREATE TABLE `comments` (
  `id` varchar(36) NOT NULL,
  `project_id` varchar(36) NOT NULL,
  `user_id` varchar(36) NOT NULL,
  `task_id` varchar(36) NOT NULL,
  `data_type_id` varchar(36) NOT NULL,
  `data_path` varchar(255) DEFAULT NULL,
  `message` longtext,
  `created` datetime DEFAULT NULL,
  `modified` datetime DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `fk_comments_users` (`user_id`),
  KEY `fk_comments_projects1` (`project_id`),
  KEY `fk_comments_data_types1` (`data_type_id`),
  CONSTRAINT `fk_comments_data_types1` FOREIGN KEY (`data_type_id`) REFERENCES `data_types` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `fk_comments_projects1` FOREIGN KEY (`project_id`) REFERENCES `projects` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `fk_comments_users` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=utf32;

-- ----------------------------
-- Records of comments
-- ----------------------------

-- ----------------------------
-- Table structure for `projects`
-- ----------------------------
DROP TABLE IF EXISTS `projects`;
CREATE TABLE `projects` (
  `id` varchar(36) NOT NULL,
  `user_id` varchar(36) NOT NULL,
  `title` varchar(45) DEFAULT NULL,
  `description` longtext,
  `created` datetime DEFAULT NULL,
  `modified` datetime DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `fk_projects_users1` (`user_id`),
  CONSTRAINT `fk_projects_users1` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=utf32;

-- ----------------------------
-- Records of projects
-- ----------------------------
INSERT INTO `projects` VALUES ('50dcbc72-3410-4596-8b71-0e80ae7aaee3', '50dcbc5c-d684-40bf-9715-0becae7aaee3', 'Brand New Project', 'This is a brand new project', '2012-12-27 15:24:02', '2012-12-27 15:24:02');

and the mysql statement I am trying to do looks something like this

INSERT INTO `anthonyl_fbpj`.`comments` (`project_id`, `user_id`, `task_id`, `data_type_id`, `message`, `modified`, `created`, `id`) 
VALUES ('50dc845a-83e4-4db3-8705-5432ae7aaee3', '50dcbc5c-d684-40bf-9715-0becae7aaee3', '1', '50d32e5c-abdc-491a-a0ef-25d84e9f49a8', 'this is a test', '2012-12-27 19:20:46', '2012-12-27 19:20:46', '50dcf3ee-8bf4-4685-aa45-4eb4ae7aaee3')

the error I get looks like this

SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot add or
update a child row: a foreign key constraint fails
(anthonyl_fbpj.comments, CONSTRAINT fk_comments_projects1
FOREIGN KEY (project_id) REFERENCES projects (id) ON DELETE NO
ACTION ON UPDATE NO ACTION)

MySQL is a potent tool for designing and building a website, but you often make severe errors when you approach this tool. Among them, “MySQL ERROR 1452 a foreign key constraint fails” is one of the most common errors. Check out this article to find out how to deal with it.

Why does the error “MySQL ERROR 1452 a foreign key constraint fails” occur?

In MySQL, there is a property called Foreign Key. The Foreign Key is used to increase referentiality in the MySQL database. A Foreign Key means that a value in one table must appear in another. The reference table is called the parent table, and the table containing the foreign key is called the child table.

For example, I have a class table containing the name and id of those classes. Then I create a table students regarding that class table with a foreign key of class_id, and the student table will look like this:

CREATE TABLE students (
firstName varchar(255) NOT NULL,
class_id int unsigned NOT NULL,
PRIMARY KEY (firstName),
CONSTRAINT students _ibfk_1
FOREIGN KEY (class_id) REFERENCES class (id)
)

Continue, I try to insert a value that is not in the id column into the class_id column, and it throws an error like this:

ERROR 1452 (23000): Cannot add or update a child row:
a foreign key constraint fails
(test_db.students, CONSTRAINT students _ibfk_1
FOREIGN KEY (class_id) REFERENCES class (id))

The cause of the error “MySQL ERROR 1452 a foreign key constraint fails” is when you put a value in the table, but the value is not available in the reference table. In other words, you enter an invalid value when Foreign Key is constrained.

Option 1: Add value to the referenced table

The easiest way to fix this error is to add a value to your referencing table. In the above example, in the class table, add the id value that you need to use in the students table with the following syntax:

INSERT INTO table_name VALUES (value1,value2,value3);

Parameter:

  • value1, value2, and value3: are the values ​​of column 1, column 2, and column 3, respectively.

Or, more explicitly:

INSERT INTO table_name (column1,column2,column3)
VALUES (value1,value2,value3);

Parameter:

  •  column1, column2, and column3: are the names of column 1, column 2, and column 3, respectively.
  •  value1, value2, and value3: are the values ​​of column 1, column 2, and column 3, respectively.

Note: For this syntax, you must ensure the number of columns equals the number of values. Otherwise, when running the command will get an error.

Option 2: Disable the FOREIGN_KEY_CHECKS variable in the MySQL server

MySQL supports checking foreign key variables and allows us to turn them on or off.

Check as follows:

SHOW GLOBAL VARIABLES LIKE 'FOREIGN_KEY_CHECKS'

Output:

— +——————–+——-+
— | Variable_name | Value |
— +——————–+——-+
— | foreign_key_checks | ON |
— +——————–+——-+

Next, turn off the operation of the foreign key and proceed to insert whatever data you want without causing an error:

SET GLOBAL FOREIGN_KEY_CHECKS=0;

Finally, return the foreign key’s state to its original state:

SET GLOBAL FOREIGN_KEY_CHECKS=1;

So you can insert more data into the reference table without causing the error “MySQL ERROR 1452 a foreign key constraint fails”.

Summary

MySQL is a pretty intuitive tool to use, and while it’s not too difficult, it’s not easy to master either. The error “MySQL ERROR 1452 a foreign key constraint fails” is an example above. Let’s start learning from the basics and fixing common mistakes to master them.

Maybe you are interested:

  • MySQL Command Not Found
  • How To MySQL Split String By Delimiter

Tom

My name is Tom Joseph, and I work as a software engineer. I enjoy programming and passing on my experience. C, C++, JAVA, and Python are my strong programming languages that I can share with everyone. In addition, I have also developed projects using Javascript, html, css.

Job: Developer
Name of the university: UTC
Programming Languages: C, C++, Javascript, JAVA, python, html, css

Добавить комментарий