Related
Some how, my database has gotten into a bad state. I previously had a table named live_stream. When I tried to drop a foreign key constraint, I got an error that mariadb could not rename #sql-26_e7a to live_stream. Now when I try to run the following statement, I get this error.
Can't create table live_stream (errno: 150 "Foreign key constraint is incorrectly formed")
CREATE TABLE live_stream
(idbigint(20) NOT NULL PRIMARY KEY);
As you can see I don't have any foreign key constraints in the definition. If I try the exact same definition with a different table name, it works. If I try to drop the table, mariadb complains that live_stream doesn't exist. Its like the table or foreign key are stuck in a transaction or something like that.
I am using galara with maria db 10.3.
UPDATE
I believe the problem was introduced when a foreign key and unique index were given the same name. I recreated the scenario, and when I try to drop the index, mariadb prevents it.
* UPDATE 2 *
Here is the output of SHOW ENGINE INNODB STATUS;
* UPDATE3 *
Here are the steps to reproduce.
create table tb1
(
id bigint null,
constraint tb1_pk
primary key (id)
);
create table tb2
(
id bigint null,
tb1_id bigint null,
constraint tb2_pk
primary key (id),
constraint tb2_tb1_id_fk
foreign key (tb1_id) references tb1 (id)
);
ALTER TABLE tb2 ADD CONSTRAINT tb2_tb1_id_fk UNIQUE (tb1_id, tb1_id);
drop index tb2_tb1_id_fk on tb2;
The problem is that the unique constraint has the same name as the foreign key and references the same column twice.
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
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;
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;
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.
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.
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;
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.
I had the same problem today. I tested for four things, some of them already mentioned here:
Are there any values in your child column that don't exist in the parent column (besides NULL, if the child column is nullable)
Do child and parent columns have the same datatype?
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)
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.
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:
Drop the column which you have tried to set FK constraint for.
Add it again and set its default value as NULL.
Try to set a foreign key constraint for it again.
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.
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.
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;
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.
Empty both your tables' data and run the command. It will work.
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/
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.
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.
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.
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 :)
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.
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!
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.
I have a problem during SQL generation of my database diagram from MySQL Workbench (UNIQUE constraints are added in every field from combined primary key).
I have following diagram: https://dl.dropboxusercontent.com/u/3843729/baza.png
Moreover, it has following properties set: https://dl.dropboxusercontent.com/u/3843729/indexes.png
SQL generated:
CREATE TABLE IF NOT EXISTS `test`.`User_has_Menu` (
`User_id` BIGINT UNIQUE NOT NULL,
`Menu_id` BIGINT UNIQUE NOT NULL,
PRIMARY KEY (`User_id`, `Menu_id`),
INDEX `fk_User_has_Menu_Menu1_idx` (`Menu_id` ASC),
INDEX `fk_User_has_Menu_User1_idx` (`User_id` ASC),
CONSTRAINT `fk_User_has_Menu_User1`
FOREIGN KEY (`User_id`)
REFERENCES `test`.`User` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fk_User_has_Menu_Menu1`
FOREIGN KEY (`Menu_id`)
REFERENCES `test`.`Menu` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
As you can see, 'User_id' and 'Menu_id' have UNIQUE keyword, so as a result I cannot insert single user_id (or menu_id) in two rows, for example:
insert into user_has_menu (user_id, menu_id) values (1,1);
insert into user_has_menu (user_id, menu_id) values (1,3);
Second insert statement is not executed as user_id must be unique.
Note: If I create new diagram in MySQL Workbench, everything is working fine and the SQL generated is without problematic UNIQUE constraints.
How can I remove those constraints without creating new diagram?
first find the unique constraint names:
select *
from
information_schema.key_column_usage
where
and table_schema = 'my_database'
and table_name = 'my_table'
then drop them
alter table my_table drop index my_contraint
And because unique constraints are handled as indexes on MySQL you can try make sure you don't have these indexes on the MySQL graphical interface.
I want to add a Foreign Key to a table called "katalog".
ALTER TABLE katalog
ADD CONSTRAINT `fk_katalog_sprache`
FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL;
When I try to do this, I get this error message:
Error Code: 1005. Can't create table 'mytable.#sql-7fb1_7d3a' (errno: 150)
Error in INNODB Status:
120405 14:02:57 Error in foreign key constraint of table
mytable.#sql-7fb1_7d3a:
FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL:
Cannot resolve table name close to:
(`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL
When i use this query it works, but with wrong "on delete" action:
ALTER TABLE `katalog`
ADD FOREIGN KEY (`Sprache` ) REFERENCES `sprache` (`ID` )
Both tables are InnoDB and both fields are "INT(11) not null". I'm using MySQL 5.1.61. Trying to fire this ALTER Query with MySQL Workbench (newest) on a MacBook Pro.
Table Create Statements:
CREATE TABLE `katalog` (
`ID` int(11) unsigned NOT NULL AUTO_INCREMENT,
`Name` varchar(50) COLLATE utf8_unicode_ci NOT NULL,
`AnzahlSeiten` int(4) unsigned NOT NULL,
`Sprache` int(11) NOT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `katalogname_uq` (`Name`)
) ENGINE=InnoDB AUTO_INCREMENT=12 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci ROW_FORMAT=DYNAMIC$$
CREATE TABLE `sprache` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`Bezeichnung` varchar(45) NOT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `Bezeichnung_UNIQUE` (`Bezeichnung`),
KEY `ix_sprache_id` (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8
To add a foreign key (grade_id) to an existing table (users), follow the following steps:
ALTER TABLE users ADD grade_id SMALLINT UNSIGNED NOT NULL DEFAULT 0;
ALTER TABLE users ADD CONSTRAINT fk_grade_id FOREIGN KEY (grade_id) REFERENCES grades(id);
Simply use this query, I have tried it as per my scenario and it works well
ALTER TABLE katalog ADD FOREIGN KEY (`Sprache`) REFERENCES Sprache(`ID`);
Simple Steps...
ALTER TABLE t_name1 ADD FOREIGN KEY (column_name) REFERENCES t_name2(column_name)
FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL;
But your table has:
CREATE TABLE `katalog` (
`Sprache` int(11) NOT NULL,
It cant set the column Sprache to NULL because it is defined as NOT NULL.
check this link. It has helped me with errno 150:
http://verysimple.com/2006/10/22/mysql-error-number-1005-cant-create-table-mydbsql-328_45frm-errno-150/
On the top of my head two things come to mind.
Is your foreign key index a unique name in the whole database (#3 in the list)?
Are you trying to set the table PK to NULL on update (#5 in the list)?
I'm guessing the problem is with the set NULL on update (if my brains aren't on backwards today as they so often are...).
Edit: I missed the comments on your original post. Unsigned/not unsigned int columns maybe resolved your case. Hope my link helps someone in the future thought.
How to fix Error Code: 1005. Can't create table 'mytable.#sql-7fb1_7d3a' (errno: 150) in mysql.
alter your table and add an index to it..
ALTER TABLE users ADD INDEX index_name (index_column)
Now add the constraint
ALTER TABLE foreign_key_table
ADD CONSTRAINT foreign_key_name FOREIGN KEY (foreign_key_column)
REFERENCES primary_key_table (primary_key_column) ON DELETE NO ACTION
ON UPDATE CASCADE;
Note if you don't add an index it wont work.
After battling with it for about 6 hours I came up with the solution
I hope this save a soul.
MySQL will execute this 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;
Cheers!
When you add a foreign key constraint to a table using ALTER TABLE, remember to create the required indexes first.
Create index
Alter table
try all in one query
ALTER TABLE users ADD grade_id SMALLINT UNSIGNED NOT NULL DEFAULT 0,
ADD CONSTRAINT fk_grade_id FOREIGN KEY (grade_id) REFERENCES grades(id);
step 1: run this script
SET FOREIGN_KEY_CHECKS=0;
step 2: add column
ALTER TABLE mileage_unit ADD COLUMN COMPANY_ID BIGINT(20) NOT NULL
step 3: add foreign key to the added column
ALTER TABLE mileage_unit
ADD FOREIGN KEY (COMPANY_ID) REFERENCES company_mst(COMPANY_ID);
step 4: run this script
SET FOREIGN_KEY_CHECKS=1;
ALTER TABLE child_table_name ADD FOREIGN KEY (child_table_column) REFERENCES parent_table_name(parent_table_column);
child_table_name is that table in which we want to add constraint.
child_table_column is that table column in which we want to add foreign key.
parent table is that table from which we want to take reference.
parent_table_column is column name of the parent table from which we take reference
this is basically happens because your tables are in two different charsets. as a example one table created in charset=utf-8 and other tables is created in CHARSET=latin1 so you want be able add foriegn key to these tables. use same charset in both tables then you will be able to add foriegn keys. error 1005 foriegn key constraint incorrectly formed can resolve from this
The foreign key constraint must be the same data type as the primary key in the reference table and column
ALTER TABLE TABLENAME ADD FOREIGN KEY (Column Name) REFERENCES TableName(column name)
Example:-
ALTER TABLE Department ADD FOREIGN KEY (EmployeeId) REFERENCES Employee(EmployeeId)
i geted through the same problem. I my case the table already have data and there were key in this table that was not present in the reference table. So i had to delete this rows that disrespect the constraints and everything worked.
Double check if the engine and charset of the both tables are the same.
If not, it will show this error.
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`);
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
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.
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.
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
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;
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
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`)
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
You need to delete it by order
There are dependency in the tables
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!
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.
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');
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.
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 ;
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.
Use MySQL workbench
Click on Database -> Reverse Engineering
Select a correct connection
Next till the end, remember to select database & tables that need examine
Now you have the ER diagram, you can see which table have FK conflict
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 work
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")
*/
Maybe you should try ON DELETE CASCADE
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.