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.
Using MySQL, moving from MyISAM to InnoDB tables. Database design started with dumping the data and re-importing it without foreign keys or constraints. Adding those one at a time to find errors.
I have ParentTable which can either be linked to ChildTableA or ChildTableB, but not both. Should the CREATE syntax be: (using CREATE syntax for simplicity rather than multiple ALTERs)
CREATE TABLE `ParentTable`
`IDParentTable` bigint(20) unsigned NOT NULL auto_increment,
`IDChildTableA` bigint(20) unsigned NOT NULL default '0',
`IDChildTableB` bigint(20) unsigned NOT NULL default '0',
PRIMARY KEY (`IDParentTable`),
KEY `ParentTable_IDChildTableA` (`IDChildTableA`),
KEY `ParentTable_IDChildTableB` (`IDChildTableB`)
ENGINE=InnoDB DEFAULT CHARSET=latin1;
Without thinking about it, I tried including:
CONSTRAINT `ParentTable_IDChildTableA` FOREIGN KEY (`IDChildTableA`) REFERENCES `ChildTableA` (`IDChildTableA`),
CONSTRAINT `ParentTable_IDChildTableB` FOREIGN KEY (`IDChildTableB`) REFERENCES `ChildTableB` (`IDChildTableB`)
Which failed, because many rows have 0 for IDChildTableA, and many rows have 0 for IDChildTableB. But, no rows have 0 for both. It's seeing that no ChildTableA exists with IDChildTableA of 0, and likewise with B.
Is there a proper way to handle this situation while keeping referential integrity? Without splitting ParentTable in two? A way to say it's OK if it's 0 or references a valid related table? Or, does wanting polymorphic tables mean I have to go without constraints?
BTW, I much prefer this route than having a single IDChildTable foreign key and then having another column designating whether it's table A or B... Not how I see that would work either for constraints, just saying I prefer not to go that route...
A column used as a foreign key can be nullable. Use a NULL value in the foreign key column to indicate "no row referenced."
It seems like you have your foreign keys backwards. Usually, the child table has a reference to the parent table.
parent (id int primary key)
childA (id int, parent_id int, ...)
childB (id int, parent_id int, ...)
EDIT
Related to the question regarding a foreign key column referencing two tables (based on a discriminator column)... that's not possible. A foreign key constraint can reference only one table.
To get something like that work, you'd need to add two separate foreign key columns, each referencing one target table. You could make use of the extra discriminator column (A or B) to identify which foreign key column should be used, so one fk column would be populated with a reference, the other fk column would be set to NULL.
However, there is no declarative constraint that would require exactly one of those two fk columns to be populated. That would not be enforced by the database. The extra discriminator column would actually be redundant, because you could derive that based on which foreign key column was populated.
I am in the process of designing the databases for my system. There are a lot of foreign key constraints.
I was wondering whether I could get some advice, whether I should do which of the following:
1) Specify the constraints during table creation itself ie,
CREATE TABLE IF NOT EXISTS abc
(
keyword VARCHAR(20) NOT NULL,
id INT UNSIGNED NOT NULL,
FOREIGN KEY (id) REFERENCES xyz(id) ON DELETE CASCADE ON UPDATE CASCADE
)ENGINE=InnoDB;
2)create the table without FK constraints and 'alter' the table later on ie,
CREATE TABLE IF NOT EXISTS abc
(
keyword VARCHAR(20) NOT NULL,
id INT UNSIGNED NOT NULL,
)ENGINE=InnoDB;
ALTER TABLE abc ADD CONSTRAINT fk_constraint FOREIGN KEY (id) REFERENCES xyz(id)
ON DELETE CASCADE ON UPDATE CASCADE;
Table xyz is simply another table with 'id' as a primary key.
You may create the FK at once. But this is not always possible because they can refer to each other in a circular fashion. Also, you may want to add columns later, with a FK.
It may be slightly faster to add it at once, because MySQL has to validate and rebuild the table structure for some changes (although I'm not sure adding FKs is one of those). But this process will be reasonably fast on empty tables, so it doesn't matter much when you add the FK.
The result will be the same. So, there is no differences.
If I create new database, I'd create table and its foreign key in one statement. The script will look better. But in this case parent tables must be created before the child tables.
If you don't want to take into account dependencies when creating tables, you can create tables in random order in the beginning of the script and then add foreign keys using ALTER TABLE.
I want to add another primary key to a table in mysql.
I use phpmyadmin to communicate with mysql server.
When I click the primary icon for the desired field it gives me this error:
#1075 - Incorrect table definition; there can be only one auto column and it must be defined as a key
Edited:
here's the query:
ALTER TABLE `files` DROP PRIMARY KEY ,ADD PRIMARY KEY ( `file_type` )
How can I do it?
As the name "primary" key says, there may be only one of that (ref: Highlander).
What you might want to try is a UNIQUE KEY, that acts just like a primary for most purpouses. Auto_increment doesn't seem to fulfill any purpouse if used a second time - what'ts the point of two fields carrying exactly the same information?
I believe in your case, what you need is a composite key. I do not know your table structure, but here is a general example taken from here,
CREATE TABLE track(
album CHAR(10),
disk INTEGER,
posn INTEGER,
song VARCHAR(255),
PRIMARY KEY (album, disk, posn)
)
In this case, there is a combination of 3 columns which avoid the duplicate records as you require. Please let me know if I have any mistakes in understanding your scenario.
The error message says it, I think:
the auto_increment column must be key.
So use this query first:
ALTER TABLE 'files' CHANGE 'id' 'id' INT( 10 ) UNSIGNED NOT NULL;
this will remove the auto_increment.
Also, I recommend the Uniqe key as suggested by other answer. I believe there should always (almost) be an Id column in each table.
We can Give Primary Key only once for a table. You can prefer UNIQUE KEY to prevent duplicate records
ALTER TABLE Persons
ADD UNIQUE (P_Id)
You can mark all the fields you want as primary keys, including the existing one. The system internally will drop the existing one and will set all you marked.