# Mysql error 1452 – Cannot add or update a child row: a foreign key constraint fails

## The Question :

239 people think this question is useful

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


• could you also post your insert/update command which results in the error?
• try running this query to see if there is any sourcecode_id that is not a real id: SELECT sourcecode_id FROM sourcecodes_tags WHERE sourcecode_id NOT IN (SELECT id FROM sourcecodes AS tmp);
• Thanks Zed, that was the problem one of the tables had data in it. Thinking about it now it makes sense that it was failing because there were things that were referencing non-existing items, but I never would have guessed that. Thanks!
• Why does it fail if table is empty?

227 people think this answer is useful

Quite likely your sourcecodes_tags table contains sourcecode_id values that no longer exists in your sourcecodes table. You have to get rid of those first.

Here’s a query that can find 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;



100 people think this answer is useful

I had the same issue with my MySQL database but finally, I got a solution which worked for me.
Since in my table everything was fine from the mysql point of view(both tables should use InnoDB engine and the datatype of each column should be of the same type which takes part in foreign key constraint).
The only 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;



55 people think this answer is useful

Use NOT IN to find where constraints are constraining:

SELECT column FROM table WHERE column NOT IN
(SELECT intended_foreign_key FROM another_table)



so, more specifically:

SELECT sourcecode_id FROM sourcecodes_tags WHERE sourcecode_id NOT IN
(SELECT id FROM sourcecodes)



EDIT: IN and NOT IN operators are known to be much faster than the JOIN operators, as well as much easier to construct, and repeat.

23 people think this answer is useful

## Truncate the tables and then try adding the FK Constraint.

I know this solution is a bit awkward but it does work 100%. But I agree that this is not an ideal solution to deal with problem, but I hope it helps.

17 people think this answer is useful

For me, this problem was a little different and super easy to check and solve.

You must ensure BOTH of your tables are InnoDB. If one of the tables, namely the reference table is a MyISAM, the constraint will fail.

    SHOW TABLE STATUS WHERE Name =  't1';

ALTER TABLE t1 ENGINE=InnoDB;



15 people think this answer is useful

This also happens when setting a foreign key to parent.id to child.column if the child.column has a value of 0 already and no parent.id value is 0

You would need to ensure that each child.column is NULL or has value that exists in parent.id

And now that I read the statement nos wrote, that’s what he is validating.

14 people think this answer is useful

I had the same problem today. I tested for four things, some of them already mentioned here:

1. Are there any values in your child column that don’t exist in the parent column (besides NULL, if the child column is nullable)

2. Do child and parent columns have the same datatype?

3. Is there an index on the parent column you are referencing? MySQL seems to require this for performance reasons (http://dev.mysql.com/doc/refman/5.5/en/create-table-foreign-keys.html)

4. And this one solved it for me: Do both tables have identical collation?

I had one table in UTF-8 and the other in iso-something. That didn’t work. After changing the iso-table to UTF-8 collation the constraints could be added without problems. In my case, phpMyAdmin didn’t even show the child table in iso-encoding in the dropdown for creating the foreign key constraint.

8 people think this answer is useful

It seems there is some invalid value for the column line 0 that is not a valid foreign key so MySQL cannot set a foreign key constraint for it.

1. Drop the column which you have tried to set FK constraint for.

2. Add it again and set its default value as NULL.

3. Try to set a foreign key constraint for it again.

5 people think this answer is useful

I’d the same problem, I checked rows of my tables and found there was some incompatibility with the value of fields that I wanted to define a foreign key. I corrected those value, tried again and the problem was solved.

4 people think this answer is useful

I end up delete all the data in my table, and run alter again. It works. Not the brilliant one, but it save a lot time, especially your application is still in development stage without any customer data.

4 people think this answer is useful

try this

SET foreign_key_checks = 0;

SET foreign_key_checks = 1;



2 people think this answer is useful

I had this exact same problem about three different times. In each instance it was because one (or more) of my records did not conform to the new foreign key. You may want to update your existing records to follow the syntax constraints of the foreign key before trying to add the key itself. The following example should generally isolate the problem records:

SELECT * FROM (tablename)
WHERE (candidate key) <> (proposed foreign key value)
AND (candidate key) <> (next proposed foreign key value)



repeat AND (candidate key) <> (next proposed foreign key value) within your query for each value in the foreign key.

If you have a ton of records this can be difficult, but if your table is reasonably small it shouldn’t take too long. I’m not super amazing in SQL syntax, but this has always isolated the issue for me.

2 people think this answer is useful

Empty both your tables’ data and run the command. It will work.

2 people think this answer is useful

I was getting this error when using Laravel and eloquent, trying to make a foreign key link would cause a 1452. The problem was lack of data in the linked table.

Please see here for an example: http://mstd.eu/index.php/2016/12/02/laravel-eloquent-integrity-constraint-violation-1452-foreign-key-constraint/

1 people think this answer is useful

I was readying this solutions and this example may help.

My database have two tables (email and credit_card) with primary keys for their IDs. Another table (client) refers to this tables IDs as foreign keys. I have a reason to have the email apart from the client data.

First I insert the row data for the referenced tables (email, credit_card) then you get the ID for each, those IDs are needed in the third table (client).

If you don’t insert first the rows in the referenced tables, MySQL wont be able to make the correspondences when you insert a new row in the third table that reference the foreign keys.

If you first insert the referenced rows for the referenced tables, then the row that refers to foreign keys, no error occurs.

Hope this helps.

1 people think this answer is useful

Make sure the value is in the other table otherwise you will get this error, in the assigned corresponding column.

So if it is assigned column is assigned to a row id of another table , make sure there is a row that is in the table otherwise this error will appear.

1 people think this answer is useful

you can try this exapmple

 START TRANSACTION;
SET foreign_key_checks = 0;
ALTER TABLE job_definers ADD CONSTRAINT job_cities_foreign FOREIGN KEY
(job_cities) REFERENCES drop_down_lists(id) ON DELETE CASCADE ON UPDATE CASCADE;
SET foreign_key_checks = 1;
COMMIT;



Note : if you are using phpmyadmin just uncheck Enable foreign key checks

as example

hope this soloution fix your problem 🙂

1 people think this answer is useful

You just need to answer one question:

Is your table already storing data? (Especially the table included foreign key.)

If the answer is yes, then the only thing you need to do is to delete all the records, then you are free to add any foreign key to your table.

Delete instruction: From child(which include foreign key table) to parent table.

The reason you cannot add in foreign key after data entries is due to the table inconsistency, how are you going to deal with a new foreign key on the former data-filled the table?

0 people think this answer is useful
UPDATE sourcecodes_tags
SET sourcecode_id = NULL
WHERE sourcecode_id NOT IN (
SELECT id FROM sourcecodes);



should help to get rid of those IDs. Or if null is not allowed in sourcecode_id, then remove those rows or add those missing values to the sourcecodes table.

0 people think this answer is useful

I had the same problem and found solution, placing NULL instead of NOT NULL on foreign key column. Here is a query:

ALTER TABLE db.table1
ADD COLUMN col_table2_fk INT UNSIGNED NULL,
ADD INDEX col_table2_fk_idx (col_table2_fk ASC),
ADD CONSTRAINT col_table2_fk1
FOREIGN KEY (col_table2_fk)
REFERENCES db.table2 (table2_id)
ON DELETE NO ACTION
ON UPDATE NO ACTION;



MySQL has executed this query!