What does mysql error 1025 (HY000): Error on rename of './foo' (errorno: 150) mean?

Question

I tried this in mysql:

mysql> alter table region drop column country_id;

And got this:

ERROR 1025 (HY000): Error on rename of './product/#sql-14ae_81' to
'./product/region' (errno: 150)

Any ideas? Foreign key stuff?

Ingrained

@skiphoppy - Are you trying to give a bounty to an already-given-answer? Is that even allowed? Or is your case different, in which case you should start another thread? – Biophysics

@RickJames Yes it is. However, skiphoppy should add her comment under the answer she elected, as the bouty message will disappear when the bounty is over. – Performance

Answer 1

You usually get this error if your tables use the InnoDB engine. In that case you would have to drop the foreign key, and then do the alter table and drop the column.

But the tricky part is that you can't drop the foreign key using the column name, but instead you would have to find the name used to index it. To find that, issue the following select:

SHOW CREATE TABLE region;

This should show you the name of the index, something like this:

CONSTRAINT region_ibfk_1 FOREIGN KEY (country_id) REFERENCES country (id) ON DELETE NO ACTION ON UPDATE NO ACTION

Now simply issue an:

alter table region drop foreign key region_ibfk_1;

And finally an:

alter table region drop column country_id;

And you are good to go!

Ronrona

Note that you can also drop the foreign key and the column in one ALTER TABLE query; – Strobilaceous

nice! we should drop foreign key , all thing will be okay! Thanks! – Malina

Answer 2

It is indeed a foreign key error, you can find out using perror:

shell$ perror 150
MySQL error code 150: Foreign key constraint is incorrectly formed

To find out more details about what failed, you can use SHOW ENGINE INNODB STATUS and look for the LATEST FOREIGN KEY ERROR section it contains details about what is wrong.

In your case, it is most likely cause something is referencing the country_id column.

Gony

Again, another misleading MySQL displayed error... Thanks. – Mide

In phpMyAdmin, you can find the status of the engine in Storage Engines, InnoDB, InnoDB Status – Mahmud

Answer 3

You can get also get this error trying to drop a non-existing foreign key. So when dropping foreign keys, always make sure they actually exist.

If the foreign key does exist, and you are still getting this error try the following:

SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0;
SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0;
SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='TRADITIONAL';

// Drop the foreign key here!

SET SQL_MODE=@OLD_SQL_MODE;
SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS;
SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS;

This always does the trick for me :)

Oppilate

I needed this while switching to a different character set of a table, because of a mysql error: ERROR 1025 (HY000): Error on rename of './table/#sql-14ae_81' to (errno: 150) – Margie

Answer 4

Simply run the alter table query using 'KEY' instead of 'FOREIGN KEY' in the drop statement. I hope it will help to solve the issue, and will drop the foreign key constraint and you can change the table columns and drop the table.

ALTER TABLE slide_image_sub DROP  KEY  FK_slide_image_sub;

here in DROP KEY instead of DROP FOREIGN KEY,

hope it will help.

Thanks

Gayelord

Answer 5

I know, this is an old post, but it's the first hit on everyone's favorite search engine if you are looking for error 1025.

However, there is an easy "hack" for fixing this issue:

Before you execute your command(s) you first have to disable the foreign key constraints check using this command:

SET FOREIGN_KEY_CHECKS = 0;

Then you are able to execute your command(s).

After you are done, don't forget to enable the foreign key constraints check again, using this command:

SET FOREIGN_KEY_CHECKS = 1;

Good luck with your endeavor.

Untaught

Was having trouble installing octobercms on xampp, finally this helped. – Augsburg

Answer 6

I had a similar issues once. I deleted the primary key from TABLE A but when I was trying to delete the foreign key column from table B I was shown the above same error.

You can't drop the foreign key using the column name and to bypass this in PHPMyAdmin or with MySQL, first remove the foreign key constraint before renaming or deleting the attribute.

Gur

Answer 7

Take a look in error file for your mysql database. According to Bug #26305 my sql do not give you the cause. This bug exists since MySQL 4.1 ;-)

Laliberte

I found some of your answer difficult to follow, but +1 for linking to a bug report. Apparently, it's now been fixed in MySQL v5.6.6. :) – Descombes

Answer 8

If you are using a client like MySQL Workbench, right click the desired table from where a foreign key is to be deleted, then select the foreign key tab and delete the indexes.

Then you can run the query like this:

alter table table_name drop foreign_key_col_name;

Hailstone

Answer 9

There is probably another table with a foreign key referencing the primary key you are trying to change.

To find out which table caused the error you can run SHOW ENGINE INNODB STATUS and then look at the LATEST FOREIGN KEY ERROR section

Use SHOW CREATE TABLE categories to show the name of constraint.

Most probably it will be categories_ibfk_1

Use the name to drop the foreign key first and the column then:

ALTER TABLE categories DROP FOREIGN KEY categories_ibfk_1;
ALTER TABLE categories DROP COLUMN assets_id;

Tolentino

Answer 10

I got this error with MySQL 5.6 but it had nothing to do with Foreign keys. This was on a Windows 7 Professional machine acting as a server on a small LAN.

The client application was doing a batch operation that creates a table fills it with some external data then runs a query joining with permanent tables then dropping the "temporary" table. This batch does this approximately 300 times and this particular routine had been running week in week out for several years when suddenly we get the Error 1025 Unable to rename problem at a random point in the batch.

In my case the application was using 4 DDL statements a CREATE TABLE followed by 3 CREATE INDEX, there is no foreign key. However only 2 of the indexes actually get created and the actual table .frm file was renamed, at the point of failure.

My solution was to get rid of the separate CREATE INDEX statements and create them using the CREATE TABLE statement. This at the time of writing has solved the issue for me and my help someone else scratching their head when they find this thread.

Martimartial

Answer 11

I'd guess foreign key constraint problem. Is country_id used as a foreign key in another table?

I'm not DB guru but I think I solved a problem like this (where there was a fk constraint) by removing the fk, doing my alter table stuff and then redoing the fk stuff.

I'll be interested to hear what the outcome is - sometime mysql is pretty cryptic.

Aglaia

Mysql can be awful. Good usually, but those error messages, sheesh. If we get a concise explanation from someone in the office, I'll throw it up here. – Ingrained

Answer 12

In my case, I was using MySQL workbench and I faced the same issue while dropping one of my columns in a table. I could not find the name of the foreign key. I followed the following steps to resolve the issue:

1. Rt. click on your schema and select 'schema inspector'. This gives you various tables, columns, indexes, ect.

2. Go to the tab named 'Indexes' and search the name of the column under the column named 'Column'. Once found check the name of the table for this record under the column name 'Table'. If it matches the name of the table you want, then note down the name of the foreign key from the column named 'Name'.

3. Now execute the query : ALTER table tableNamexx DROP KEY foreignKeyName;

4. Now you can execute the drop statement which shall execute successfully.

Minaminabe

Answer 13

Doing

SET FOREIGN_KEY_CHECKS=0;

before the Operation can also do the trick.

Regulation

Answer 14

averageRatings= FOREACH groupedRatings GENERATE group AS movieID, AVG(ratings.rating) AS avgRating, COUNT(ratings.rating) AS numRatings;

If you are using any command like above you must use group in small letters. This may solve your problem it solved mine. At least in PIG script.

Organo

I don't think this is problem related, he or she is hitting mysql problem – Paphian