Foreign key relationships employ a parent table that contains the central data values, and a child table with identical values pointing back to its parent. The FOREIGN KEY clause is fixed in the child table.
It will refuse any INSERT or UPDATE operation that tries to make a foreign key value in a child table in case there is no a matching candidate key value in the parent table.
Hence your error
Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails originally refers that, you are attempting to include a row to your
Ordrelinje table for which no matching row (OrderID) is present in
You should first insert the row to your
You are getting this constraint check since
Ordre table does not have reference
OrdreID provided in insert command.
To insert value in
Ordrelinje, you first should enter value in
Ordre table and employ same
Or you can remove not null constraint and insert a NULL value.
You should delete data in the child table which does not have any corresponding foreign key value to the parent table primary key .Or delete all data from the child table then insert fresh data having the similar foreign key value as the primary key in the parent table .
The issue is with FOREIGN KEY Constraint. By Default . FOREIGN_KEY_CHECKS option secludes whether or not to check foreign key constraints for InnoDB tables. MySQL - SET FOREIGN_KEY_CHECKS
We can fixed foreign key check as disable before running Query. Disable Foreign key.
Perform one of these lines before running your query, then you can run your query successfully.
1) For Session (recommended)
SET GLOBAL FOREIGN_KEY_CHECKS=0;
This error originally happens since we have some values in the referencing field of the child table, which do not subsist in the referenced/candidate field of the parent table.
Occasionnally, we may get this error at the time we are employing Foreign Key constraints to existing table(s), having data in them meanwhile. Few of the other answers are exposing to delete the data fully from child table, and then employ the constraint. But, this is not an option at the time we meanwhile have working/production data in the child table. In most continuity, we will require to update the data in the child table (instead of deleting them).
Currently, we can utilize
Left link to trace all those rows in the child table, which does not have matching values in the parent table. Following query , would be useful to fetch those non-matching rows:
LEFT JOIN parent_table
ON parent_table.referenced_column = child_table.referencing_column
WHERE parent_table.referenced_column IS NULL
Presently, you can originally do one (or more) of the following steps to fix the data.
- Based on your "business logic", you will require to update/match these unmatching value(s), with the existing values in the parent table. You may occasionally requires to set them
null as well.
- Delete these rows having unmatching values.
- Include new rows in your parent table, corresponding to the unmatching values in the child table.
Once the data is set, we can employ the Foreign key constraint using
ALTER TABLE syntax.
sourcecodes_tags table contains
sourcecode_id values that no longer subsists in your
sourcecodes table. You have to exclude those first.
Here's a query that can trace those IDs:
SELECT DISTINCT sourcecode_id FROM
sourcecodes_tags tags LEFT JOIN sourcecodes sc ON tags.sourcecode_id=sc.id
WHERE sc.id IS NULL;
I had the similar problem with my MySQL database however finally, I got a solution which performed for me.
Because in my table everything was fine from the mysql point of view(both tables should use InnoDB engine and the datatype of each column must be of the same type which takes part in foreign key constraint).
The single thing that I did was to disable the foreign key check and later on enabled it after performing the foreign key operation.
Steps that I took:
SET foreign_key_checks = 0;
alter table tblUsedDestination add constraint f_operatorId foreign key(iOperatorId) references tblOperators (iOperatorId); Query
OK, 8 rows affected (0.23 sec) Records: 8 Duplicates: 0 Warnings: 0
SET foreign_key_checks = 1;