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 have three tabs in my Db:
1.pers_info(id(primary), name ,....)
2.contacts(c_id(primary), phone, email, ...)
now 1 person can have multiple rows in contacts tab.
thus to minimize redundancy i made another tab contact_relation(id (foregin key references pers_info(id), c_id (foregin key references contacts(c_id))
i successfully created the relation and also was able to insert apt entries(related id and c_id) to it "using last_insert_id();" to extract id and c_id required.
now the problem..
i dropped tab contacts_rel. and created it this way :
CREATE TABLE contacts_relation (
id INT NOT NULL,
cid INT NOT NULL,
FOREIGN KEY (id)
REFERENCES pers_info(id)
ON UPDATE CASCADE ON DELETE RESTRICT,
FOREIGN KEY (c_id)
REFERENCES contacts(c_id)
);
if now i try to insert into tab cont_rel, it gives error :
cannot add or update a child row foreign key constraint fails.
which makes sense..
i was hoping that adding constraints in the defn of tab cont_rel would save the hassle of inserting entries in it manually using "last_insert_id();"
so is there a way, i could maintain ref integrity with new data
coming.. thanks.
Why the third table? When you many to many relation you need intermediate table. You have one to many relation so two tables are sufficient. Table contacts need FK to table pers_info. Add it if you don't have already.
Here's a simple transaction example with LAST_INSERT_ID(): SQL INSERT INTO multiple tables
Some info about foreign keys and altering table:
Add Foreign Key to existing table
http://dev.mysql.com/doc/refman/5.0/en/alter-table.html
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 have an InnoDB MySQL database with a table that needs to be able to connect to one of 26 other tables via a foreign key. Each record will only connect to one of these 26 at a time. The table will probably consist of no more than 10,000 records. Is there an alternative way to do this?
-- -----------------------------------------------------
-- Table `db_mydb`.`tb_job`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `db_mydb`.`tb_job` (
`job_id` INT(11) NOT NULL AUTO_INCREMENT ,
// Removed 26 other fields that the table requires
`job_foreignkey_a_id` INT(11) NULL DEFAULT NULL ,
`job_foreignkey_b_id` INT(11) NULL DEFAULT NULL ,
`job_foreignkey_c_id` INT(11) NULL DEFAULT NULL ,
// Removed the other 23 foreign keys fields that are the same
PRIMARY KEY (`job_id`) ,
CONSTRAINT `fka_tb_job_tb`
FOREIGN KEY (`job_foreignkey_a_id` )
REFERENCES `db_mydb`.`tb_foreignkey_a` (`foreignkey_a_id` )
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fkb_tb_job_tb`
FOREIGN KEY (`job_foreignkey_b_id` )
REFERENCES `db_mydb`.`tb_foreignkey_b` (`foreignkey_b_id` )
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fkc_tb_job_tb`
FOREIGN KEY (`job_foreignkey_c_id` )
REFERENCES `db_mydb`.`tb_foreignkey_c` (`foreignkey_c_id` )
ON DELETE NO ACTION
ON UPDATE NO ACTION)
// Removed the other 23 foreign keys constraints that are the same
ENGINE = InnoDB
DEFAULT CHARACTER SET = utf8;
CREATE INDEX `fka_tb_job_tb` ON `db_mydb`.`tb_job` (`job_foreignkey_a_id` ASC) ;
CREATE INDEX `fkb_tb_job_tb` ON `db_mydb`.`tb_job` (`job_foreignkey_b_id` ASC) ;
CREATE INDEX `fkc_tb_job_tb` ON `db_mydb`.`tb_job` (`job_foreignkey_c_id` ASC) ;
// Removed the other 23 foreign keys indexes that are the same
This is the problem of generic foreign keys, which MySQL and friends tend not to support. There are two ways you can do this.
The first, as you have done, is nullable foreign keys, one for every type.
The other, as in Django's Content Types, is to have a join table, each row having a row id and a field that specifies the table to look up on. Your code then has to formulate the SQL query depending on the contents of the field. It works well, but has limitations:
The downside of the first one is bloat, but it brings you the upsides of normal FKs, i.e. referential integrity and SQL joins etc, both of which are very valuable. You can't get those with the second method.
Depends if you want to maintain foreign key constraint, you can have one table that references one of the tables by a key or table type. Problem is you will loose the foreign key constraint. Of course, if you can create a function based constraint, then it can work for you. Or you can enforce the relationship using a trigger. Function based constraints are not available in mysql.
Yes, you can do that. These two StackOverflow answers illustrate the underlying principles in a slightly different context.
Same data from different entities in Database - Best Practice - Phone numbers example
Different user types / objects own content in same table - how?
Using MySQL, you'll need to replace critical CHECK() constraints with foreign key references. This doesn't work in the most general case for MySQL, but it does work in this particular application.
If this isn't enough information to get you going, leave me a comment, and I'll try to expand this answer a little more.