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.
Related
I am trying to create a table in MySQL with two foreign keys, which reference the primary keys in 2 other tables, but I am getting an errno: 150 error and it will not create the table.
Here is the SQL for all 3 tables:
CREATE TABLE role_groups (
`role_group_id` int(11) NOT NULL `AUTO_INCREMENT`,
`name` varchar(20),
`description` varchar(200),
PRIMARY KEY (`role_group_id`)
) ENGINE=InnoDB;
CREATE TABLE IF NOT EXISTS `roles` (
`role_id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50),
`description` varchar(200),
PRIMARY KEY (`role_id`)
) ENGINE=InnoDB;
create table role_map (
`role_map_id` int not null `auto_increment`,
`role_id` int not null,
`role_group_id` int not null,
primary key(`role_map_id`),
foreign key(`role_id`) references roles(`role_id`),
foreign key(`role_group_id`) references role_groups(`role_group_id`)
) engine=InnoDB;
These conditions must be satisfied to not get error 150 re ALTER TABLE ADD FOREIGN KEY:
The Parent table must exist before you define a foreign key to reference it. You must define the tables in the right order: Parent table first, then the Child table. If both tables references each other, you must create one table without FK constraints, then create the second table, then add the FK constraint to the first table with ALTER TABLE.
The two tables must both support foreign key constraints, i.e. ENGINE=InnoDB. Other storage engines silently ignore foreign key definitions, so they return no error or warning, but the FK constraint is not saved.
The referenced columns in the Parent table must be the left-most columns of a key. Best if the key in the Parent is PRIMARY KEY or UNIQUE KEY.
The FK definition must reference the PK column(s) in the same order as the PK definition. For example, if the FK REFERENCES Parent(a,b,c) then the Parent's PK must not be defined on columns in order (a,c,b).
The PK column(s) in the Parent table must be the same data type as the FK column(s) in the Child table. For example, if a PK column in the Parent table is UNSIGNED, be sure to define UNSIGNED for the corresponding column in the Child table field.
Exception: length of strings may be different. For example, VARCHAR(10) can reference VARCHAR(20) or vice versa.
Any string-type FK column(s) must have the same character set and collation as the corresponding PK column(s).
If there is data already in the Child table, every value in the FK column(s) must match a value in the Parent table PK column(s). Check this with a query like:
SELECT COUNT(*) FROM Child LEFT OUTER JOIN Parent ON Child.FK = Parent.PK
WHERE Parent.PK IS NULL;
This must return zero (0) unmatched values. Obviously, this query is an generic example; you must substitute your table names and column names.
Neither the Parent table nor the Child table can be a TEMPORARY table.
Neither the Parent table nor the Child table can be a PARTITIONED table.
If you declare a FK with the ON DELETE SET NULL option, then the FK column(s) must be nullable.
If you declare a constraint name for a foreign key, the constraint name must be unique in the whole schema, not only in the table in which the constraint is defined. Two tables may not have their own constraint with the same name.
If there are any other FK's in other tables pointing at the same field you are attempting to create the new FK for, and they are malformed (i.e. different collation), they will need to be made consistent first. This may be a result of past changes where SET FOREIGN_KEY_CHECKS = 0; was utilized with an inconsistent relationship defined by mistake. See #andrewdotn's answer below for instructions on how to identify these problem FK's.
MySQL’s generic “errno 150” message “means that a foreign key constraint was not correctly formed.” As you probably already know if you are reading this page, the generic “errno: 150” error message is really unhelpful. However:
You can get the actual error message by running SHOW ENGINE INNODB STATUS; and then looking for LATEST FOREIGN KEY ERROR in the output.
For example, this attempt to create a foreign key constraint:
CREATE TABLE t1
(id INTEGER);
CREATE TABLE t2
(t1_id INTEGER,
CONSTRAINT FOREIGN KEY (t1_id) REFERENCES t1 (id));
fails with the error Can't create table 'test.t2' (errno: 150). That doesn’t tell anyone anything useful other than that it’s a foreign key problem. But run SHOW ENGINE INNODB STATUS; and it will say:
------------------------
LATEST FOREIGN KEY ERROR
------------------------
130811 23:36:38 Error in foreign key constraint of table test/t2:
FOREIGN KEY (t1_id) REFERENCES t1 (id)):
Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
It says that the problem is it can’t find an index. SHOW INDEX FROM t1 shows that there aren’t any indexes at all for table t1. Fix that by, say, defining a primary key on t1, and the foreign key constraint will be created successfully.
Make sure that the properties of the two fields you are trying to link with a constraint are exactly the same.
Often, the 'unsigned' property on an ID column will catch you out.
ALTER TABLE `dbname`.`tablename` CHANGE `fieldname` `fieldname` int(10) UNSIGNED NULL;
What's the current state of your database when you run this script? Is it completely empty? Your SQL runs fine for me when creating a database from scratch, but errno 150 usually has to do with dropping & recreating tables that are part of a foreign key. I'm getting the feeling you're not working with a 100% fresh and new database.
If you're erroring out when "source"-ing your SQL file, you should be able to run the command "SHOW ENGINE INNODB STATUS" from the MySQL prompt immediately after the "source" command to see more detailed error info.
You may want to check out the manual entry too:
If you re-create a table that was dropped, it must have a definition that conforms to the foreign key constraints referencing it. It must have the right column names and types, and it must have indexes on the referenced keys, as stated earlier. If these are not satisfied, MySQL returns error number 1005 and refers to error 150 in the error message. If MySQL reports an error number 1005 from a CREATE TABLE statement, and the error message refers to error 150, table creation failed because a foreign key constraint was not correctly formed.
— MySQL 5.1 reference manual.
For people who are viewing this thread with the same problem:
There are a lot of reasons for getting errors like this. For a fairly complete list of causes and solutions of foreign key errors in MySQL (including those discussed here), check out this link:
MySQL Foreign Key Errors and Errno 150
For others that find this SO entry via Google: Be sure that you aren't trying to do a SET NULL action on a foreign key (to be) column defined as "NOT NULL." That caused great frustration until I remembered to do a CHECK ENGINE INNODB STATUS.
Definitely it is not the case but I found this mistake pretty common and unobvious. The target of a FOREIGN KEY could be not PRIMARY KEY. Te answer which become useful for me is:
A FOREIGN KEY always must be pointed to a PRIMARY KEY true field of other table.
CREATE TABLE users(
id INT AUTO_INCREMENT PRIMARY KEY,
username VARCHAR(40));
CREATE TABLE userroles(
id INT AUTO_INCREMENT PRIMARY KEY,
user_id INT NOT NULL,
FOREIGN KEY(user_id) REFERENCES users(id));
As pointed by #andrewdotn the best way is to see the detailed error(SHOW ENGINE INNODB STATUS;) instead of just an error code.
One of the reasons could be that an index already exists with the same name, may be in another table. As a practice, I recommend prefixing table name before the index name to avoid such collisions. e.g. instead of idx_userId use idx_userActionMapping_userId.
Please make sure at first that
you are using InnoDB tables.
field for FOREIGN KEY has the same type and length (!) as source field.
I had the same trouble and I've fixed it. I had unsigned INT for one field and just integer for other field.
Helpful tip, use SHOW WARNINGS; after trying your CREATE query and you will receive the error as well as the more detailed warning:
---------------------------------------------------------------------------------------------------------+
| Level | Code | Message |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
| Warning | 150 | Create table 'fakeDatabase/exampleTable' with foreign key constraint failed. There is no index in the referenced table where the referenced columns appear as the first columns.
|
| Error | 1005 | Can't create table 'exampleTable' (errno:150) |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
So in this case, time to re-create my table!
This is usually happening when you try to source file into existing database.
Drop all the tables first (or the DB itself).
And then source file with SET foreign_key_checks = 0; at the beginning and SET foreign_key_checks = 1; at the end.
I've found another reason this fails... case sensitive table names.
For this table definition
CREATE TABLE user (
userId int PRIMARY KEY AUTO_INCREMENT,
username varchar(30) NOT NULL
) ENGINE=InnoDB;
This table definition works
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES **u**ser(userId)
) ENGINE=InnoDB;
whereas this one fails
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES User(userId)
) ENGINE=InnoDB;
The fact that it worked on Windows and failed on Unix took me a couple of hours to figure out. Hope that helps someone else.
MySQL Workbench 6.3 for Mac OS.
Problem: errno 150 on table X when trying to do Forward Engineering on a DB diagram, 20 out of 21 succeeded, 1 failed. If FKs on table X were deleted, the error moved to a different table that wasn't failing before.
Changed all tables engine to myISAM and it worked just fine.
Also worth checking that you aren't accidentally operating on the wrong database. This error will occur if the foreign table does not exist. Why does MySQL have to be so cryptic?
Make sure that the foreign keys are not listed as unique in the parent. I had this same problem and I solved it by demarcating it as not unique.
In my case it was due to the fact that the field that was a foreign key field had a too long name, ie. foreign key (some_other_table_with_long_name_id). Try sth shorter. Error message is a bit misleading in that case.
Also, as #Jon mentioned earlier - field definitions have to be the same (watch out for unsigned subtype).
(Side notes too big for a Comment)
There is no need for an AUTO_INCREMENT id in a mapping table; get rid of it.
Change the PRIMARY KEY to (role_id, role_group_id) (in either order). This will make accesses faster.
Since you probably want to map both directions, also add an INDEX with those two columns in the opposite order. (There is no need to make it UNIQUE.)
More tips: http://mysql.rjweb.org/doc.php/index_cookbook_mysql#speeding_up_wp_postmeta
When the foreign key constraint is based on varchar type, then in addition to the list provided by marv-el the target column must have an unique constraint.
execute below line before creating table :
SET FOREIGN_KEY_CHECKS = 0;
FOREIGN_KEY_CHECKS option specifies whether or not to check foreign key constraints for InnoDB tables.
-- Specify to check foreign key constraints (this is the default)
SET FOREIGN_KEY_CHECKS = 1;
-- Do not check foreign key constraints
SET FOREIGN_KEY_CHECKS = 0;
When to Use :
Temporarily disabling referential constraints (set FOREIGN_KEY_CHECKS to 0) is useful when you need to re-create the tables and load data in any parent-child order
I encountered the same problem, but I check find that I hadn't the parent table. So I just edit the parent migration in front of the child migration. Just do it.
I am trying to create a table in MySQL with two foreign keys, which reference the primary keys in 2 other tables, but I am getting an errno: 150 error and it will not create the table.
Here is the SQL for all 3 tables:
CREATE TABLE role_groups (
`role_group_id` int(11) NOT NULL `AUTO_INCREMENT`,
`name` varchar(20),
`description` varchar(200),
PRIMARY KEY (`role_group_id`)
) ENGINE=InnoDB;
CREATE TABLE IF NOT EXISTS `roles` (
`role_id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50),
`description` varchar(200),
PRIMARY KEY (`role_id`)
) ENGINE=InnoDB;
create table role_map (
`role_map_id` int not null `auto_increment`,
`role_id` int not null,
`role_group_id` int not null,
primary key(`role_map_id`),
foreign key(`role_id`) references roles(`role_id`),
foreign key(`role_group_id`) references role_groups(`role_group_id`)
) engine=InnoDB;
These conditions must be satisfied to not get error 150 re ALTER TABLE ADD FOREIGN KEY:
The Parent table must exist before you define a foreign key to reference it. You must define the tables in the right order: Parent table first, then the Child table. If both tables references each other, you must create one table without FK constraints, then create the second table, then add the FK constraint to the first table with ALTER TABLE.
The two tables must both support foreign key constraints, i.e. ENGINE=InnoDB. Other storage engines silently ignore foreign key definitions, so they return no error or warning, but the FK constraint is not saved.
The referenced columns in the Parent table must be the left-most columns of a key. Best if the key in the Parent is PRIMARY KEY or UNIQUE KEY.
The FK definition must reference the PK column(s) in the same order as the PK definition. For example, if the FK REFERENCES Parent(a,b,c) then the Parent's PK must not be defined on columns in order (a,c,b).
The PK column(s) in the Parent table must be the same data type as the FK column(s) in the Child table. For example, if a PK column in the Parent table is UNSIGNED, be sure to define UNSIGNED for the corresponding column in the Child table field.
Exception: length of strings may be different. For example, VARCHAR(10) can reference VARCHAR(20) or vice versa.
Any string-type FK column(s) must have the same character set and collation as the corresponding PK column(s).
If there is data already in the Child table, every value in the FK column(s) must match a value in the Parent table PK column(s). Check this with a query like:
SELECT COUNT(*) FROM Child LEFT OUTER JOIN Parent ON Child.FK = Parent.PK
WHERE Parent.PK IS NULL;
This must return zero (0) unmatched values. Obviously, this query is an generic example; you must substitute your table names and column names.
Neither the Parent table nor the Child table can be a TEMPORARY table.
Neither the Parent table nor the Child table can be a PARTITIONED table.
If you declare a FK with the ON DELETE SET NULL option, then the FK column(s) must be nullable.
If you declare a constraint name for a foreign key, the constraint name must be unique in the whole schema, not only in the table in which the constraint is defined. Two tables may not have their own constraint with the same name.
If there are any other FK's in other tables pointing at the same field you are attempting to create the new FK for, and they are malformed (i.e. different collation), they will need to be made consistent first. This may be a result of past changes where SET FOREIGN_KEY_CHECKS = 0; was utilized with an inconsistent relationship defined by mistake. See #andrewdotn's answer below for instructions on how to identify these problem FK's.
MySQL’s generic “errno 150” message “means that a foreign key constraint was not correctly formed.” As you probably already know if you are reading this page, the generic “errno: 150” error message is really unhelpful. However:
You can get the actual error message by running SHOW ENGINE INNODB STATUS; and then looking for LATEST FOREIGN KEY ERROR in the output.
For example, this attempt to create a foreign key constraint:
CREATE TABLE t1
(id INTEGER);
CREATE TABLE t2
(t1_id INTEGER,
CONSTRAINT FOREIGN KEY (t1_id) REFERENCES t1 (id));
fails with the error Can't create table 'test.t2' (errno: 150). That doesn’t tell anyone anything useful other than that it’s a foreign key problem. But run SHOW ENGINE INNODB STATUS; and it will say:
------------------------
LATEST FOREIGN KEY ERROR
------------------------
130811 23:36:38 Error in foreign key constraint of table test/t2:
FOREIGN KEY (t1_id) REFERENCES t1 (id)):
Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
It says that the problem is it can’t find an index. SHOW INDEX FROM t1 shows that there aren’t any indexes at all for table t1. Fix that by, say, defining a primary key on t1, and the foreign key constraint will be created successfully.
Make sure that the properties of the two fields you are trying to link with a constraint are exactly the same.
Often, the 'unsigned' property on an ID column will catch you out.
ALTER TABLE `dbname`.`tablename` CHANGE `fieldname` `fieldname` int(10) UNSIGNED NULL;
What's the current state of your database when you run this script? Is it completely empty? Your SQL runs fine for me when creating a database from scratch, but errno 150 usually has to do with dropping & recreating tables that are part of a foreign key. I'm getting the feeling you're not working with a 100% fresh and new database.
If you're erroring out when "source"-ing your SQL file, you should be able to run the command "SHOW ENGINE INNODB STATUS" from the MySQL prompt immediately after the "source" command to see more detailed error info.
You may want to check out the manual entry too:
If you re-create a table that was dropped, it must have a definition that conforms to the foreign key constraints referencing it. It must have the right column names and types, and it must have indexes on the referenced keys, as stated earlier. If these are not satisfied, MySQL returns error number 1005 and refers to error 150 in the error message. If MySQL reports an error number 1005 from a CREATE TABLE statement, and the error message refers to error 150, table creation failed because a foreign key constraint was not correctly formed.
— MySQL 5.1 reference manual.
For people who are viewing this thread with the same problem:
There are a lot of reasons for getting errors like this. For a fairly complete list of causes and solutions of foreign key errors in MySQL (including those discussed here), check out this link:
MySQL Foreign Key Errors and Errno 150
For others that find this SO entry via Google: Be sure that you aren't trying to do a SET NULL action on a foreign key (to be) column defined as "NOT NULL." That caused great frustration until I remembered to do a CHECK ENGINE INNODB STATUS.
Definitely it is not the case but I found this mistake pretty common and unobvious. The target of a FOREIGN KEY could be not PRIMARY KEY. Te answer which become useful for me is:
A FOREIGN KEY always must be pointed to a PRIMARY KEY true field of other table.
CREATE TABLE users(
id INT AUTO_INCREMENT PRIMARY KEY,
username VARCHAR(40));
CREATE TABLE userroles(
id INT AUTO_INCREMENT PRIMARY KEY,
user_id INT NOT NULL,
FOREIGN KEY(user_id) REFERENCES users(id));
As pointed by #andrewdotn the best way is to see the detailed error(SHOW ENGINE INNODB STATUS;) instead of just an error code.
One of the reasons could be that an index already exists with the same name, may be in another table. As a practice, I recommend prefixing table name before the index name to avoid such collisions. e.g. instead of idx_userId use idx_userActionMapping_userId.
Please make sure at first that
you are using InnoDB tables.
field for FOREIGN KEY has the same type and length (!) as source field.
I had the same trouble and I've fixed it. I had unsigned INT for one field and just integer for other field.
Helpful tip, use SHOW WARNINGS; after trying your CREATE query and you will receive the error as well as the more detailed warning:
---------------------------------------------------------------------------------------------------------+
| Level | Code | Message |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
| Warning | 150 | Create table 'fakeDatabase/exampleTable' with foreign key constraint failed. There is no index in the referenced table where the referenced columns appear as the first columns.
|
| Error | 1005 | Can't create table 'exampleTable' (errno:150) |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
So in this case, time to re-create my table!
This is usually happening when you try to source file into existing database.
Drop all the tables first (or the DB itself).
And then source file with SET foreign_key_checks = 0; at the beginning and SET foreign_key_checks = 1; at the end.
I've found another reason this fails... case sensitive table names.
For this table definition
CREATE TABLE user (
userId int PRIMARY KEY AUTO_INCREMENT,
username varchar(30) NOT NULL
) ENGINE=InnoDB;
This table definition works
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES **u**ser(userId)
) ENGINE=InnoDB;
whereas this one fails
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES User(userId)
) ENGINE=InnoDB;
The fact that it worked on Windows and failed on Unix took me a couple of hours to figure out. Hope that helps someone else.
MySQL Workbench 6.3 for Mac OS.
Problem: errno 150 on table X when trying to do Forward Engineering on a DB diagram, 20 out of 21 succeeded, 1 failed. If FKs on table X were deleted, the error moved to a different table that wasn't failing before.
Changed all tables engine to myISAM and it worked just fine.
Also worth checking that you aren't accidentally operating on the wrong database. This error will occur if the foreign table does not exist. Why does MySQL have to be so cryptic?
Make sure that the foreign keys are not listed as unique in the parent. I had this same problem and I solved it by demarcating it as not unique.
In my case it was due to the fact that the field that was a foreign key field had a too long name, ie. foreign key (some_other_table_with_long_name_id). Try sth shorter. Error message is a bit misleading in that case.
Also, as #Jon mentioned earlier - field definitions have to be the same (watch out for unsigned subtype).
(Side notes too big for a Comment)
There is no need for an AUTO_INCREMENT id in a mapping table; get rid of it.
Change the PRIMARY KEY to (role_id, role_group_id) (in either order). This will make accesses faster.
Since you probably want to map both directions, also add an INDEX with those two columns in the opposite order. (There is no need to make it UNIQUE.)
More tips: http://mysql.rjweb.org/doc.php/index_cookbook_mysql#speeding_up_wp_postmeta
When the foreign key constraint is based on varchar type, then in addition to the list provided by marv-el the target column must have an unique constraint.
execute below line before creating table :
SET FOREIGN_KEY_CHECKS = 0;
FOREIGN_KEY_CHECKS option specifies whether or not to check foreign key constraints for InnoDB tables.
-- Specify to check foreign key constraints (this is the default)
SET FOREIGN_KEY_CHECKS = 1;
-- Do not check foreign key constraints
SET FOREIGN_KEY_CHECKS = 0;
When to Use :
Temporarily disabling referential constraints (set FOREIGN_KEY_CHECKS to 0) is useful when you need to re-create the tables and load data in any parent-child order
I encountered the same problem, but I check find that I hadn't the parent table. So I just edit the parent migration in front of the child migration. Just do it.
I am trying to figure out relationships and deletion options.
I have two tables, User and UserStaff, with a 1:n relationship from User to UserStaff (a user can have multiple staff members).
When my User is deleted, I want to delete all of the UserStaff tables associated with that User. When my UserStaff is deleted, I don't want anything to happen to User. I understand that this is a cascading relationship, but I'm not sure which way.
i.e. Do I select the existing foreign key in my UserStaff table and make it cascading, or do I create a new foreign key in User and set that to cascading?
Yes, it's possible. You should make the FK in UserStaff table. In this way:
User Table
CREATE TABLE `User` (
`Id` int(11) NOT NULL AUTO_INCREMENT,
`Name` varchar(255) DEFAULT NULL,
PRIMARY KEY (`Id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
UserStaff Table
CREATE TABLE `UserStaff` (
`Id` int(11) NOT NULL AUTO_INCREMENT,
`UserId` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`Id`),
KEY `UserId` (`UserId`),
CONSTRAINT `UserStaff_ibfk_1`
FOREIGN KEY (`UserId`)
REFERENCES `User` (`Id`)
ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
From Wikipedia:
CASCADE
Whenever rows in the master (referenced) table are deleted (resp. updated), the respective rows of the child (referencing) table with a matching foreign key column will get deleted (resp. updated) as well. This is called a cascade delete (resp. update[2]).
Here, User is the master table, and UserStaff is the child table. So, yes, you'll want to create the foreign key in UserStaff, with ON DELETE CASCADE
It's been a while since I've used this, but here goes (btw, I use Toad for MySql - a great IDE, and it's free too - http://www.toadworld.com/Freeware/ToadforMySQLFreeware/tabid/561/Default.aspx!)
You need to add a Constraint to the User table. If you have an id column (and the corresponding foreign userid key in UserStaff) then the SouceColumn should be id, the destination table UserStaff and the destination column userid. You can then set the OnDelete action to be 'Cascade'
The other options are pretty self-explanatory - Restrict limits values to the values in the source column, Set Null sets the foreign key matches to Null and No Action does, er, nothing.
This stuff is very easy to do via the Toad IDE. I used MySqlAdmin tools for ages but recently discovered Toad (and it has diff and compare tools too!).
The ON DELETE CASCADE is specified on the foreign key in the UserStaff table. For additional info on foreign keys the MySQL documentation has a number of examples. The User table does not have a foreign key pointing to UserStaff, so it will not be affected by changes to the UserStaff table.
The easiest way might be to make two quick tables and try it out. But since you didn't I can tell you that the outcome will be that it work the way that you want to.
When you have a table User and a table UserStaff were a field in UserStaff uses a foreign key to reference a field in User; then if you delete a record from UserStaff that will be removed wihtout having any affect on the User table. The other way around will delete all records related to that record.
Short version: A field in UserStaff should reference a field in User with CASCADE
I want to add foreign key Constraint on my existing tables(type InnoDB)
,I have two tables listed below:
tbl_users : [ user_id(INT,AUTO_INCREMENT), user_name,--- etc]
tbl_users_meta : [ user_meta_id(INT,AUTO_INCREMENT), user_id(INT),--- etc]
I have already created index 'user_id' on 'tbl_users_meta' listed below :
Here is my Query but i am getting (MySQL Error Code 1215: ) each time.what is the problem i can getting it ?
ALTER TABLE `tbl_users_meta`
ADD CONSTRAINT `fk users user_id`
FOREIGN KEY (`user_id`) REFERENCES
`sanskrut`.`tbl_users`(`user_id`)
ON DELETE NO ACTION ON UPDATE CASCADE;
TRY with the same query you have writen with a small modification:
ALTER IGNORE TABLE `tbl_users_meta`
ADD CONSTRAINT `fk users user_id`
FOREIGN KEY (`user_id`) REFERENCES
`sanskrut`.`tbl_users`(`user_id`)
ON DELETE NO ACTION ON UPDATE CASCADE;
Hope it'll work
I can't see anything wrong with what you've written, so there must be something you haven't posted affecting the problem. For example, on MySQL version 5.5.44, I can successfully execute the following:
CREATE DATABASE sanskrut;
USE sanskrut;
CREATE TABLE tbl_users (user_id INT AUTO_INCREMENT PRIMARY KEY, user_name VARCHAR(50));
CREATE TABLE tbl_users_meta (user_meta_id INT AUTO_INCREMENT PRIMARY KEY, user_id INT);
CREATE INDEX user_id ON tbl_users_meta (user_id);
ALTER TABLE `tbl_users_meta`
ADD CONSTRAINT `fk users user_id`
FOREIGN KEY (`user_id`) REFERENCES
`sanskrut`.`tbl_users`(`user_id`)
ON DELETE NO ACTION ON UPDATE CASCADE;
...which finishes with your exact problem statement, copied and pasted, and works just fine.
I'd suspect, as I mentioned in the comments, that there's an index missing or possibly a datatype mismatch. You may find it helpful to add the output of:
DESC tbl_users;
DESC tbl_users_meta;
SHOW indexes FROM tbl_users;
SHOW indexes FROM tbl_users_meta;
...to your question.
I am trying to create a table in MySQL with two foreign keys, which reference the primary keys in 2 other tables, but I am getting an errno: 150 error and it will not create the table.
Here is the SQL for all 3 tables:
CREATE TABLE role_groups (
`role_group_id` int(11) NOT NULL `AUTO_INCREMENT`,
`name` varchar(20),
`description` varchar(200),
PRIMARY KEY (`role_group_id`)
) ENGINE=InnoDB;
CREATE TABLE IF NOT EXISTS `roles` (
`role_id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50),
`description` varchar(200),
PRIMARY KEY (`role_id`)
) ENGINE=InnoDB;
create table role_map (
`role_map_id` int not null `auto_increment`,
`role_id` int not null,
`role_group_id` int not null,
primary key(`role_map_id`),
foreign key(`role_id`) references roles(`role_id`),
foreign key(`role_group_id`) references role_groups(`role_group_id`)
) engine=InnoDB;
These conditions must be satisfied to not get error 150 re ALTER TABLE ADD FOREIGN KEY:
The Parent table must exist before you define a foreign key to reference it. You must define the tables in the right order: Parent table first, then the Child table. If both tables references each other, you must create one table without FK constraints, then create the second table, then add the FK constraint to the first table with ALTER TABLE.
The two tables must both support foreign key constraints, i.e. ENGINE=InnoDB. Other storage engines silently ignore foreign key definitions, so they return no error or warning, but the FK constraint is not saved.
The referenced columns in the Parent table must be the left-most columns of a key. Best if the key in the Parent is PRIMARY KEY or UNIQUE KEY.
The FK definition must reference the PK column(s) in the same order as the PK definition. For example, if the FK REFERENCES Parent(a,b,c) then the Parent's PK must not be defined on columns in order (a,c,b).
The PK column(s) in the Parent table must be the same data type as the FK column(s) in the Child table. For example, if a PK column in the Parent table is UNSIGNED, be sure to define UNSIGNED for the corresponding column in the Child table field.
Exception: length of strings may be different. For example, VARCHAR(10) can reference VARCHAR(20) or vice versa.
Any string-type FK column(s) must have the same character set and collation as the corresponding PK column(s).
If there is data already in the Child table, every value in the FK column(s) must match a value in the Parent table PK column(s). Check this with a query like:
SELECT COUNT(*) FROM Child LEFT OUTER JOIN Parent ON Child.FK = Parent.PK
WHERE Parent.PK IS NULL;
This must return zero (0) unmatched values. Obviously, this query is an generic example; you must substitute your table names and column names.
Neither the Parent table nor the Child table can be a TEMPORARY table.
Neither the Parent table nor the Child table can be a PARTITIONED table.
If you declare a FK with the ON DELETE SET NULL option, then the FK column(s) must be nullable.
If you declare a constraint name for a foreign key, the constraint name must be unique in the whole schema, not only in the table in which the constraint is defined. Two tables may not have their own constraint with the same name.
If there are any other FK's in other tables pointing at the same field you are attempting to create the new FK for, and they are malformed (i.e. different collation), they will need to be made consistent first. This may be a result of past changes where SET FOREIGN_KEY_CHECKS = 0; was utilized with an inconsistent relationship defined by mistake. See #andrewdotn's answer below for instructions on how to identify these problem FK's.
MySQL’s generic “errno 150” message “means that a foreign key constraint was not correctly formed.” As you probably already know if you are reading this page, the generic “errno: 150” error message is really unhelpful. However:
You can get the actual error message by running SHOW ENGINE INNODB STATUS; and then looking for LATEST FOREIGN KEY ERROR in the output.
For example, this attempt to create a foreign key constraint:
CREATE TABLE t1
(id INTEGER);
CREATE TABLE t2
(t1_id INTEGER,
CONSTRAINT FOREIGN KEY (t1_id) REFERENCES t1 (id));
fails with the error Can't create table 'test.t2' (errno: 150). That doesn’t tell anyone anything useful other than that it’s a foreign key problem. But run SHOW ENGINE INNODB STATUS; and it will say:
------------------------
LATEST FOREIGN KEY ERROR
------------------------
130811 23:36:38 Error in foreign key constraint of table test/t2:
FOREIGN KEY (t1_id) REFERENCES t1 (id)):
Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
It says that the problem is it can’t find an index. SHOW INDEX FROM t1 shows that there aren’t any indexes at all for table t1. Fix that by, say, defining a primary key on t1, and the foreign key constraint will be created successfully.
Make sure that the properties of the two fields you are trying to link with a constraint are exactly the same.
Often, the 'unsigned' property on an ID column will catch you out.
ALTER TABLE `dbname`.`tablename` CHANGE `fieldname` `fieldname` int(10) UNSIGNED NULL;
What's the current state of your database when you run this script? Is it completely empty? Your SQL runs fine for me when creating a database from scratch, but errno 150 usually has to do with dropping & recreating tables that are part of a foreign key. I'm getting the feeling you're not working with a 100% fresh and new database.
If you're erroring out when "source"-ing your SQL file, you should be able to run the command "SHOW ENGINE INNODB STATUS" from the MySQL prompt immediately after the "source" command to see more detailed error info.
You may want to check out the manual entry too:
If you re-create a table that was dropped, it must have a definition that conforms to the foreign key constraints referencing it. It must have the right column names and types, and it must have indexes on the referenced keys, as stated earlier. If these are not satisfied, MySQL returns error number 1005 and refers to error 150 in the error message. If MySQL reports an error number 1005 from a CREATE TABLE statement, and the error message refers to error 150, table creation failed because a foreign key constraint was not correctly formed.
— MySQL 5.1 reference manual.
For people who are viewing this thread with the same problem:
There are a lot of reasons for getting errors like this. For a fairly complete list of causes and solutions of foreign key errors in MySQL (including those discussed here), check out this link:
MySQL Foreign Key Errors and Errno 150
For others that find this SO entry via Google: Be sure that you aren't trying to do a SET NULL action on a foreign key (to be) column defined as "NOT NULL." That caused great frustration until I remembered to do a CHECK ENGINE INNODB STATUS.
Definitely it is not the case but I found this mistake pretty common and unobvious. The target of a FOREIGN KEY could be not PRIMARY KEY. Te answer which become useful for me is:
A FOREIGN KEY always must be pointed to a PRIMARY KEY true field of other table.
CREATE TABLE users(
id INT AUTO_INCREMENT PRIMARY KEY,
username VARCHAR(40));
CREATE TABLE userroles(
id INT AUTO_INCREMENT PRIMARY KEY,
user_id INT NOT NULL,
FOREIGN KEY(user_id) REFERENCES users(id));
As pointed by #andrewdotn the best way is to see the detailed error(SHOW ENGINE INNODB STATUS;) instead of just an error code.
One of the reasons could be that an index already exists with the same name, may be in another table. As a practice, I recommend prefixing table name before the index name to avoid such collisions. e.g. instead of idx_userId use idx_userActionMapping_userId.
Please make sure at first that
you are using InnoDB tables.
field for FOREIGN KEY has the same type and length (!) as source field.
I had the same trouble and I've fixed it. I had unsigned INT for one field and just integer for other field.
Helpful tip, use SHOW WARNINGS; after trying your CREATE query and you will receive the error as well as the more detailed warning:
---------------------------------------------------------------------------------------------------------+
| Level | Code | Message |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
| Warning | 150 | Create table 'fakeDatabase/exampleTable' with foreign key constraint failed. There is no index in the referenced table where the referenced columns appear as the first columns.
|
| Error | 1005 | Can't create table 'exampleTable' (errno:150) |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
So in this case, time to re-create my table!
This is usually happening when you try to source file into existing database.
Drop all the tables first (or the DB itself).
And then source file with SET foreign_key_checks = 0; at the beginning and SET foreign_key_checks = 1; at the end.
I've found another reason this fails... case sensitive table names.
For this table definition
CREATE TABLE user (
userId int PRIMARY KEY AUTO_INCREMENT,
username varchar(30) NOT NULL
) ENGINE=InnoDB;
This table definition works
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES **u**ser(userId)
) ENGINE=InnoDB;
whereas this one fails
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES User(userId)
) ENGINE=InnoDB;
The fact that it worked on Windows and failed on Unix took me a couple of hours to figure out. Hope that helps someone else.
MySQL Workbench 6.3 for Mac OS.
Problem: errno 150 on table X when trying to do Forward Engineering on a DB diagram, 20 out of 21 succeeded, 1 failed. If FKs on table X were deleted, the error moved to a different table that wasn't failing before.
Changed all tables engine to myISAM and it worked just fine.
Also worth checking that you aren't accidentally operating on the wrong database. This error will occur if the foreign table does not exist. Why does MySQL have to be so cryptic?
Make sure that the foreign keys are not listed as unique in the parent. I had this same problem and I solved it by demarcating it as not unique.
In my case it was due to the fact that the field that was a foreign key field had a too long name, ie. foreign key (some_other_table_with_long_name_id). Try sth shorter. Error message is a bit misleading in that case.
Also, as #Jon mentioned earlier - field definitions have to be the same (watch out for unsigned subtype).
(Side notes too big for a Comment)
There is no need for an AUTO_INCREMENT id in a mapping table; get rid of it.
Change the PRIMARY KEY to (role_id, role_group_id) (in either order). This will make accesses faster.
Since you probably want to map both directions, also add an INDEX with those two columns in the opposite order. (There is no need to make it UNIQUE.)
More tips: http://mysql.rjweb.org/doc.php/index_cookbook_mysql#speeding_up_wp_postmeta
When the foreign key constraint is based on varchar type, then in addition to the list provided by marv-el the target column must have an unique constraint.
execute below line before creating table :
SET FOREIGN_KEY_CHECKS = 0;
FOREIGN_KEY_CHECKS option specifies whether or not to check foreign key constraints for InnoDB tables.
-- Specify to check foreign key constraints (this is the default)
SET FOREIGN_KEY_CHECKS = 1;
-- Do not check foreign key constraints
SET FOREIGN_KEY_CHECKS = 0;
When to Use :
Temporarily disabling referential constraints (set FOREIGN_KEY_CHECKS to 0) is useful when you need to re-create the tables and load data in any parent-child order
I encountered the same problem, but I check find that I hadn't the parent table. So I just edit the parent migration in front of the child migration. Just do it.