Ошибка 1452 mysql workbench

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?

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

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.

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;

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;

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;

Ответы:

SET FOREIGN_KEY_CHECKS=0;

SET GLOBAL FOREIGN_KEY_CHECKS=0;

SELECT child_table.* 
FROM child_table 
LEFT JOIN parent_table 
  ON parent_table.referenced_column = child_table.referencing_column 
WHERE parent_table.referenced_column IS NULL

У меня немного странная проблема. Я пытаюсь добавить внешний ключ к одной таблице, которая ссылается на другую, но по какой-то причине она не работает. Имея мои ограниченные знания MySQL, единственное, что может быть подозрительным, это наличие внешнего ключа в другой таблице, ссылающейся на ту, которую я пытаюсь ссылаться.

Вот изображение моих связей в таблице, сгенерированное с помощью phpMyAdmin:
Отношения

Я выполнил запрос SHOW CREATE TABLE для обеих таблиц, sourcecodes_tags — это таблица с внешним ключом, sourcecodes — ссылочная таблица.

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

Было бы здорово, если бы кто-нибудь мог рассказать мне, что здесь происходит, у меня не было формального обучения или чего-то еще с MySQL:)

Спасибо.

Изменить: Это код, который генерирует ошибку:

ALTER TABLE sourcecodes_tags ADD FOREIGN KEY (sourcecode_id) REFERENCES sourcecodes (id) ON DELETE CASCADE ON UPDATE CASCADE