Error Code: 1005. Can't create table '...' (errno: 150)

Question

I searched for a solution to this problem on the Internet and checked the Stack Overflow questions, but none of the solutions worked for my case.

I want to create a foreign key from table sira_no to metal_kod.

ALTER TABLE sira_no
    ADD CONSTRAINT METAL_KODU FOREIGN KEY(METAL_KODU)
    REFERENCES metal_kod(METAL_KODU)
    ON DELETE SET NULL
    ON UPDATE SET NULL ;

This script returns:

Error Code: 1005. Can't create table 'ebs.#sql-f48_1a3' (errno: 150)

I tried adding an index to the referenced table:

CREATE INDEX METAL_KODU_INDEX ON metal_kod (METAL_KODU);

I checked METAL_KODU on both tables (charset and collation), but I couldn't find a solution to this problem. How can I fix this problem?

Here is the metal_kod table:

METAL_KODU    varchar(4)    NO    PRI
DURUM    bit(1)    NO
METAL_ISMI    varchar(30)    NO
AYAR_YOGUNLUK    smallint(6)    YES        100

Can you show the schema for the `metal_kod` table ... which field in that table should the Foreign Key reference ? — Manse, Jan 26 '12 at 13:28

score 281 · Accepted Answer · edited Dec 03 '21 at 00:53

281

Error Code: 1005 -- there is a wrong primary key reference in your code

Usually it's due to a referenced foreign key field that does not exist. It might be you have a typo mistake, or check case it should be same, or there's a field-type mismatch. Foreign key-linked fields must match definitions exactly.

Some known causes may be:

The two key fields type and/or size doesn’t match exactly. For example, if one is INT(10) the key field needs to be INT as well and not BIGINT or SMALLINT or TINYINT. You should also check that one is not SIGNED and the other is UNSIGNED. They both need to be exactly the same.
One of the key field that you are trying to reference does not have an index and/or is not a primary key. If one of the fields in the relationship is not a primary key, you must create an index for that field.
The foreign key name is a duplicate of an already existing key. Check that the name of your foreign key is unique within your database. Just add a few random characters to the end of your key name to test for this.
One or both of your tables is a MyISAM table. In order to use foreign keys, the tables must both be InnoDB. (Actually, if both tables are MyISAM then you won’t get an error message - it just won’t create the key.) In Query Browser, you can specify the table type.
You have specified a cascade ON DELETE SET NULL, but the relevant key field is set to NOT NULL. You can fix this by either changing your cascade or setting the field to allow NULL values.
Make sure that the Charset and Collate options are the same both at the table level as well as individual field level for the key columns.
You have a default value (that is, default=0) on your foreign key column
One of the fields in the relationship is part of a combination (composite) key and does not have its own individual index. Even though the field has an index as part of the composite key, you must create a separate index for only that key field in order to use it in a constraint.
You have a syntax error in your ALTER statement or you have mistyped one of the field names in the relationship
The name of your foreign key exceeds the maximum length of 64 characters.

For more details, refer to: MySQL Error Number 1005 Can’t create table

edited Dec 03 '21 at 00:53

Bill Karwin

538,548
86
673
828

answered Jan 26 '12 at 13:26

4

the problem was that the foreign key charsets were not matching. Thanks for the answer. – lamostreta Jan 31 '12 at 09:53
4

`SHOW ENGINE INNODB STATUS` as mentioned in [this question](http://stackoverflow.com/questions/2799021/mysql-error-1005-hy000-cant-create-table-foo-sql-12c-4-errno-150) helped me diagnose my particular problem (PEBCAK, in my case...) – Hobo Jun 30 '12 at 12:37
1

damn, even the it is a primary key. You need to create individual index for that key. thanks this solved my problem. – RSB Aug 15 '12 at 06:28
3

#4 was my problem - one of the tables was MyISAM and the script tried to create an InnoDB table. I ran into this problem when I was trying to deploy an old system that was initially running MySQL 5.0 or similar version, where the default storage engine was MyISAM and the scripts were running OK. My current environment is 5.5 and the default storage is InnoDB. Adding `set names 'utf8', storage_engine=MYISAM;` at the beginning of the script solved the issue for me. Thank you @user319198 and @Stefano for the elaborate answer! :o) – Boris Chervenkov Jul 17 '13 at 20:01
After hours and hours trying to figure this out #2 was my problem. The column I was referencing was a primary key, but it was a composite primary key. I solved this by adding a generic index on top of it being part of the primary key and that solved the issue. – Nick Nov 28 '13 at 07:10
I had a FK to columns with the same charset, but different collation. That's also a no-no. – Mopper Jan 02 '14 at 13:47
1

Mine was missing the 'unsigned' attribute as mentioned in #1, thanks! – helvete Jun 18 '15 at 13:45
thanks.. :) my issue was unassigned attribute and it's resolved – Venu Feb 27 '17 at 07:52
#8 is not correct. A composite index is fine as long as the referenced field is the leftmost field in the index. – Shadow Jul 12 '17 at 20:50
Thanks for the answer, the column size issue was the exact solution to my problem – Aadil Nov 17 '17 at 06:08
#6 fixed my problem, one is utf8, another is utf8mb4 – Mitoxys Oct 22 '18 at 03:41
Had the same problem, it was field-type mismatch on FK, thanks for the help! – MNN Jul 31 '19 at 18:48
I edited the first cause about INT types. The number argument to an integer type (e.g. `INT(10)` vs. `INT(11)`) makes no difference. You can mix and match these, since that argument doesn't affect the size of the data type anyway. But you cannot mix other types of INT, like BIGINT, SMALLINT, TINYINT. – Bill Karwin Dec 03 '21 at 00:54

score 11 · Answer 2 · answered May 07 '15 at 13:27

This could also happen when exporting your database from one server to another and the tables are listed in alphabetical order by default.
So, your first table could have a foreign key of another table that is yet to be created. In such cases, disable foreign_key_checks and create the database.

Just add the following to your script:

SET FOREIGN_KEY_CHECKS=0;

and it shall work.

score 5 · Answer 3 · edited Jun 12 '19 at 11:54

5

Sometimes it is due to the master table is dropped (maybe by disabling foreign_key_checks), but the foreign key CONSTRAINT still exists in other tables. In my case I had dropped the table and tried to recreate it, but it was throwing the same error for me.

So try dropping all the foreign key CONSTRAINTs from all the tables if there are any and then update or create the table.

edited Jun 12 '19 at 11:54

Peter Mortensen

30,738
21
105
131

answered Feb 01 '15 at 11:21

rajug

77
1
5

score 4 · Answer 4 · edited Jun 12 '19 at 11:53

4

Very often it happens when the foreign key and the reference key don't have the same type or same length.

edited Jun 12 '19 at 11:53

Peter Mortensen

30,738
21
105
131

answered Aug 25 '14 at 12:29

zahid9i

596
1
8
17

score 2 · Answer 5 · edited Jun 12 '19 at 12:00

Error Code: 1005

I had a similar issue, so here are few things that I did try (not in any order, except for the solution :) )

Changed the foreign key names (it didn't work)
Reduced the foreign key length
Verified the datatypes (darn nothing wrong)
Check indexes
Check the collations (everything fine, darn again)
Truncated the table, of no good use
Dropped the table and re-created
Tried to see if any circular reference is being created --- all fine
Finally, I saw that I had two editors open. One that in PhpStorm (JetBrains) and the other MySQL workbench. It seems that the PhpStorm / MySQL Workbench creates some kind of edit lock.

I closed PhpStorm just to check if locking was the case (it could have been the other way around). This solved my problem.

score 2 · Answer 6 · answered Jul 14 '16 at 08:34

I had a similar error. The problem had to do with the child and parent table not having the same charset and collation. This can be fixed by appending ENGINE = InnoDB DEFAULT CHARACTER SET = utf8;

CREATE TABLE IF NOT EXISTS `country` (`id` INT(11) NOT NULL AUTO_INCREMENT,...) ENGINE = InnoDB DEFAULT CHARACTER SET = utf8;

... on the SQL statement means that there is some missing code.

score 2 · Answer 7 · edited Jun 12 '19 at 12:01

2

I had the very same error message. Finally I figured out I misspelled the name of the table in the command:

ALTER TABLE `users` ADD FOREIGN KEY (country_id) REFERENCES country (id);

versus

ALTER TABLE `users` ADD FOREIGN KEY (country_id) REFERENCES countries (id);

I wonder why on earth MySQL cannot tell such a table does not exist...

edited Jun 12 '19 at 12:01

Peter Mortensen

30,738
21
105
131

answered Feb 07 '17 at 03:04

Csongor Halmai

3,239
29
30

score 2 · Answer 8 · answered Feb 15 '17 at 13:10

The foreign key has to have the exact same type as the primary key that it references. For the example has the type “INT UNSIGNED NOT NULL” the foreing key also have to “INT UNSIGNED NOT NULL”

CREATE TABLE employees(
id_empl INT UNSIGNED NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id)
);
CREATE TABLE offices(
id_office INT UNSIGNED NOT NULL AUTO_INCREMENT,
id_empl INT UNSIGNED NOT NULL,
PRIMARY KEY(id),
CONSTRAINT `constraint1` FOREIGN KEY (`id_empl`) REFERENCES `employees` (`id_empl`) ON DELETE CASCADE ON UPDATE CASCADE ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='my offices';

unsigned was the problem for me. Thanks! – Gabo Dec 04 '17 at 05:36 — Gabo, Dec 04 '17 at 05:36

score 1 · Answer 9 · edited Jun 12 '19 at 11:55

1

It happened in my case, because the name of the table being referenced in the constraint declaration wasn't correct (I forgot the upper case in the table name):

ALTER TABLE `Window` ADD CONSTRAINT `Windows_ibfk_1` FOREIGN KEY (`WallId`) REFERENCES `Wall` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;

edited Jun 12 '19 at 11:55

Peter Mortensen

30,738
21
105
131

answered Mar 10 '15 at 15:45

Abdellah Stands with Palestine

5,032
1
26
36

score 1 · Answer 10 · answered Jan 20 '16 at 23:29

MyISAM has been just mentioned. Simply try adding ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; at the end of a statement, assuming that your other tables were created with MyISAM.

CREATE TABLE IF NOT EXISTS `tablename` (
  `key` bigint(20) NOT NULL AUTO_INCREMENT,
  FOREIGN KEY `key` (`key`) REFERENCES `othertable`(`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

score 1 · Answer 11 · answered Mar 28 '16 at 02:01

1

In my case, it happened when one table is InnoB and other is MyISAM. Changing engine of one table, through MySQL Workbench, solves for me.

answered Mar 28 '16 at 02:01

rkawano

2,443
22
22

score 1 · Answer 12 · answered Sep 21 '19 at 02:35

1

check both tables has same schema InnoDB MyISAM. I made them all the same in my case InnoDB and worked

answered Sep 21 '19 at 02:35

Brian Sanchez

832
1
13
11

score 1 · Answer 13 · edited Jun 08 '20 at 09:05

1

My problem was not listed, it was something so silly ..... The table that has the FK as PK was a composite PK that was declared like this: primary key (CNPJ, CEP) I wanted the CEP field to be FK in another table and I was stuck in this error, the moral of the story just inverted the code above for Primary key (CEP, CNPJ) and it worked. Get tip their friends.

edited Jun 08 '20 at 09:05

Karthick Nagarajan

1,327
2
15
27

answered Jun 08 '20 at 04:49

fagner willys

11
1

Error Code: 1005. Can't create table '...' (errno: 150)

13 Answers13

Linked

Related