CODEWITHSUNDEEP
mysqlforeign-keysmysql-error-1452
Zim
Zim

Reputation: 5523

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

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

Upvotes: 254

Views: 522269

Answers (21)

Shamal Sabah
Shamal Sabah

Reputation: 239

In my case, I created a new table with the same structure, created the relationships with the other tables, then extracted the data in CSV from the old table that has the problem, then imported the CSV to the new table and disabled foreign key checking and disabled import interruption, all my data are inserted to the new table that has no problem successfully, then deleted the old table.

It worked for me.

Upvotes: 0

Milad Rahimi
Milad Rahimi

Reputation: 3834

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.

You can follow these steps:

  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.

Upvotes: 9

nobody
nobody

Reputation: 414

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?

If the answer is no, then follow other instructions.

Upvotes: 2

Michael Helwig
Michael Helwig

Reputation: 225

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.

Upvotes: 14

Mostafa -T
Mostafa -T

Reputation: 740

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.

Upvotes: 5

Prakash
Prakash

Reputation: 1147

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;

Upvotes: 108

zmonteca
zmonteca

Reputation: 2354

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;

Upvotes: 17

Fouad Mekkey
Fouad Mekkey

Reputation: 146

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 enter image description here

hope this soloution fix your problem :)

Upvotes: 1

Kingsley Mitchell
Kingsley Mitchell

Reputation: 2589

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.

Upvotes: 1

Carlo Guevarra
Carlo Guevarra

Reputation: 139

try this

SET foreign_key_checks = 0;

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

SET foreign_key_checks = 1;

Upvotes: 4

akelec
akelec

Reputation: 4003

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!

Upvotes: 0

twigg
twigg

Reputation: 3983

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/

Upvotes: 2

CesareoAguirre
CesareoAguirre

Reputation: 1607

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.

Upvotes: 1

naXa stands with Ukraine
naXa stands with Ukraine

Reputation: 37896

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.

Upvotes: 0

Vijay Rana
Vijay Rana

Reputation: 1079

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

Upvotes: 3

mopsyd
mopsyd

Reputation: 1922

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.

Upvotes: 2

nos
nos

Reputation: 229058

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;

Upvotes: 235

Shankar Narayana Damodaran
Shankar Narayana Damodaran

Reputation: 68446

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.

Upvotes: 23

VHanded
VHanded

Reputation: 2089

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.

Upvotes: 4

fyrye
fyrye

Reputation: 279

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.

Upvotes: 16

RD Ward
RD Ward

Reputation: 6737

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.

Upvotes: 56

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