CODEWITHSUNDEEP
mysqlsql
steven
steven

Reputation: 13537

Cannot delete or update a parent row: a foreign key constraint fails

When doing:

DELETE FROM `jobs` WHERE `job_id` =1 LIMIT 1

It errors:

#1451 - Cannot delete or update a parent row: a foreign key constraint fails 
(paymesomething.advertisers, CONSTRAINT advertisers_ibfk_1 FOREIGN KEY 
(advertiser_id) REFERENCES jobs (advertiser_id))

Here are my tables:

CREATE TABLE IF NOT EXISTS `advertisers` (
  `advertiser_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(255) NOT NULL,
  `password` char(32) NOT NULL,
  `email` varchar(128) NOT NULL,
  `address` varchar(255) NOT NULL,
  `phone` varchar(255) NOT NULL,
  `fax` varchar(255) NOT NULL,
  `session_token` char(30) NOT NULL,
  PRIMARY KEY (`advertiser_id`),
  UNIQUE KEY `email` (`email`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;


INSERT INTO `advertisers` (`advertiser_id`, `name`, `password`, `email`, `address`, `phone`, `fax`, `session_token`) VALUES
(1, 'TEST COMPANY', '', '', '', '', '', '');

CREATE TABLE IF NOT EXISTS `jobs` (
  `job_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `advertiser_id` int(11) unsigned NOT NULL,
  `name` varchar(255) NOT NULL,
  `shortdesc` varchar(255) NOT NULL,
  `longdesc` text NOT NULL,
  `address` varchar(255) NOT NULL,
  `time_added` int(11) NOT NULL,
  `active` tinyint(1) NOT NULL,
  `moderated` tinyint(1) NOT NULL,
  PRIMARY KEY (`job_id`),
  KEY `advertiser_id` (`advertiser_id`,`active`,`moderated`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;


INSERT INTO `jobs` (`job_id`, `advertiser_id`, `name`, `shortdesc`, `longdesc`, `address`, `active`, `moderated`) VALUES
(1, 1, 'TEST', 'TESTTEST', 'TESTTESTES', '', 0, 0);

ALTER TABLE `advertisers`
  ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) REFERENCES `jobs` (`advertiser_id`);

Upvotes: 273

Views: 966552

Answers (19)

Kipruto
Kipruto

Reputation: 857

Go to phpmyadmin copy your SQL query and paste into the insert query box.Uncheck the enable foreign key check before pressing go. When dropping you can Uncheck the box and proceed to drop the table. Should workenter image description here

Upvotes: 1

GreenCat
GreenCat

Reputation: 79

This error can still when working in Symfony with Doctrine Query Language, i added onDelete in Entity file

/**
 * @ORM\ManyToOne(targetEntity=Pricelist::class)
 * @ORM\JoinColumn(name="pricelist_id", referencedColumnName="id", onDelete="SET NULL")
 */

Upvotes: 0

Tariq Kamal
Tariq Kamal

Reputation: 496

This happened to me as well and due to a dependency and reference from other tables, I could not remove the entry. What I did is, I added a delete column (of type boolean) to the table. The value in that field showed whether the item is marked for deletion or not. If marked for deletion, then it would not be fetched or used, otherwise, it would be used.

Upvotes: -2

Alan
Alan

Reputation: 10125

Disable the foreign key check and make the changes then re-enable foreign key check.

SET FOREIGN_KEY_CHECKS=0; -- to disable them
DELETE FROM `jobs` WHERE `job_id` = 1 LIMIT 1 
SET FOREIGN_KEY_CHECKS=1; -- to re-enable them

Upvotes: 41

Moh .S
Moh .S

Reputation: 2090

I tried the solution mentioned by @Alino Manzi but it didn't work for me on the WordPress related tables using wpdb.

then I modified the code as below and it worked

SET FOREIGN_KEY_CHECKS=OFF; //disabling foreign key

//run the queries which are giving foreign key errors

SET FOREIGN_KEY_CHECKS=ON; // enabling foreign key

Upvotes: 24

Rashmi Pandit
Rashmi Pandit

Reputation: 23788

If there are more than one job having the same advertiser_id, then your foreign key should be:

ALTER TABLE `jobs`
ADD CONSTRAINT `advertisers_ibfk_1` 
FOREIGN KEY (`advertiser_id`) 
REFERENCES `advertisers` (`advertiser_id`);

Otherwise (if its the other way round in your case), if you want the rows in advertiser to be automatically deleted if the row in job is deleted add the 'ON DELETE CASCADE' option to the end of your foreign key:

ALTER TABLE `advertisers`
ADD CONSTRAINT `advertisers_ibfk_1` 
FOREIGN KEY (`advertiser_id`) 
REFERENCES `jobs` (`advertiser_id`)
ON DELETE CASCADE;

Check out Foreign Key constraints

Upvotes: 6

Ng Sek Long
Ng Sek Long

Reputation: 4786

The main problem with this erorr Error Code: 1451. Cannot delete or update a parent row: a foreign key constraint fails is that it doesn't let you know which table contains the FK failure, so it is difficult to solve the conflict.

If you use MySQL or similar, I found out that you can create an ER diagram for your database, then you can review and safely remove any conflicts triggering the error.

  1. Use MySQL workbench
  2. Click on Database -> Reverse Engineering
  3. Select a correct connection
  4. Next till the end, remember to select database & tables that need examine
  5. Now you have the ER diagram, you can see which table have FK conflict

Upvotes: 2

Patch92
Patch92

Reputation: 1094

You could create a trigger to delete the referenced rows in before deleting the job.

    DELIMITER $$
    CREATE TRIGGER before_jobs_delete 
        BEFORE DELETE ON jobs
        FOR EACH ROW 
    BEGIN
        delete from advertisers where advertiser_id=OLD.advertiser_id;
    END$$
    DELIMITER ;

Upvotes: 1

Amin
Amin

Reputation: 433

I had this problem in laravel migration too
the order of drop tables in down() method does matter

Schema::dropIfExists('groups');
Schema::dropIfExists('contact');

may not work, but if you change the order, it works.

Schema::dropIfExists('contact');
Schema::dropIfExists('groups');

Upvotes: 2

OMG Ponies
OMG Ponies

Reputation: 332501

As is, you must delete the row out of the advertisers table before you can delete the row in the jobs table that it references. This:

ALTER TABLE `advertisers`
  ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) 
      REFERENCES `jobs` (`advertiser_id`);

...is actually the opposite to what it should be. As it is, it means that you'd have to have a record in the jobs table before the advertisers. So you need to use:

ALTER TABLE `jobs`
  ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) 
      REFERENCES `advertisers` (`advertiser_id`);

Once you correct the foreign key relationship, your delete statement will work.

Upvotes: 160

Marius Cucuruz
Marius Cucuruz

Reputation: 165

How about this alternative I've been using: allow the foreign key to be NULL and then choose ON DELETE SET NULL.

Personally I prefer using both "ON UPDATE CASCADE" as well as "ON DELETE SET NULL" to avoid unnecessary complications, but on your set up you may want a different approach. Also, NULL'ing foreign key values may latter lead complications as you won't know what exactly happened there. So this change should be in close relation to how your application code works.

Hope this helps.

Upvotes: 4

Abin John
Abin John

Reputation: 381

If you want to drop a table you should execute the following query in a single step

SET FOREIGN_KEY_CHECKS=0; DROP TABLE table_name;

Upvotes: 19

Oleksii Kyslytsyn
Oleksii Kyslytsyn

Reputation: 2426

if you need to support client as soon as possible, and do not have access to 

FOREIGN_KEY_CHECKS

so that data integrity can be disabled:

1) delete foreign key

ALTER TABLE `advertisers` 
DROP FOREIGN KEY `advertisers_ibfk_1`;

2) activate your deleting operation thruogh sql or api

3) add the foreign key back to schema

ALTER TABLE `advertisers`
  ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) REFERENCES `jobs` (`advertiser_id`);

however, it is a hot-fix, so it is on your own risk, because the main flaw of such approach is that it is needed afterwards to keep the data integrity manually.

Upvotes: 1

Asaph
Asaph

Reputation: 162761

Under your current (possibly flawed) design, you must delete the row out of the advertisers table before you can delete the row in the jobs table that it references.

Alternatively, you could set up your foreign key such that a delete in the parent table causes rows in child tables to be deleted automatically. This is called a cascading delete. It looks something like this:

ALTER TABLE `advertisers`
ADD CONSTRAINT `advertisers_ibfk_1`
FOREIGN KEY (`advertiser_id`) REFERENCES `jobs` (`advertiser_id`)
ON DELETE CASCADE;

Having said that, as others have already pointed out, your foreign key feels like it should go the other way around since the advertisers table really contains the primary key and the jobs table contains the foreign key. I would rewrite it like this:

ALTER TABLE `jobs`
ADD FOREIGN KEY (`advertiser_id`) REFERENCES `advertisers` (`advertiser_id`);

And the cascading delete won't be necessary.

Upvotes: 56

Ran Adler
Ran Adler

Reputation: 3711

You need to delete it by order There are dependency in the tables

Upvotes: 4

Alino Manzi
Alino Manzi

Reputation: 4419

The simple way would be to disable the foreign key check; make the changes then re-enable foreign key check.

SET FOREIGN_KEY_CHECKS=0; -- to disable them
SET FOREIGN_KEY_CHECKS=1; -- to re-enable them

Upvotes: 393

Quy Le
Quy Le

Reputation: 2379

When you create database or create tables

You should add that line at top script create database or table

SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=1;

Now you want to delete records from table? then you write as

SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=1;
DELETE FROM `jobs` WHERE `job_id` =1 LIMIT 1

Good luck!

Upvotes: 2

UFOman
UFOman

Reputation:

Maybe you should try ON DELETE CASCADE

Upvotes: -1

Tom H
Tom H

Reputation: 47444

I think that your foreign key is backwards. Try:

ALTER TABLE 'jobs'
ADD CONSTRAINT `advertisers_ibfk_1` FOREIGN KEY (`advertiser_id`) REFERENCES `advertisers` (`advertiser_id`)

Upvotes: 7

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