MySQL Error Number 1005
Can’t create table ‘.\mydb\#sql-328_45.frm’ (errno: 150)
If you get this error while trying to create a foreign key, it can be pretty frustrating. The error about not being able to create a .frm file seems like it would be some kind of OS file permission error or something but this is not the case. This error has been reported as a bug on the MySQL developer list for ages, but it is actually just a misleading error message.
In every case this is due to something about the relationship that MySQL doesn’t like. Unfortunately it doesn’t specify what the exact issue is. Here is a running list of causes that people have reported for the dreaded errno 150. I’ve tried to put them in order based on the frequency that I hear about a particular cause.
You may want to start by running the MySQL command “SHOW INNODB STATUS” immediately after receiving the error. This command displays log info and error details. (Thanks Jonathan for the tip)
Note: If your script runs fine on one server, but gives an error when you try to run it on a different server, then there is a good chance that #6 is the problem. Different versions of MySQL have different default charset setting.
Known Causes:
- 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(10) as well and not INT(11) or TINYINT. You may want to confirm the field size using SHOW CREATE TABLE because Query Browser will sometimes visually show just INTEGER for both INT(10) and INT(11). You should also check that one is not SIGNED and the other is UNSIGNED. They both need to be exactly the same. (More about signed vs unsigned here).
- 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. (thanks to Venkatesh and Erichero and Terminally Incoherent for this tip)
- 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. (Thanks to Niels for this tip)
- 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. (Thanks to Sammy and J Jammin)
- 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. (Thanks to FRR for this tip)
- You have a default value (ie default=0) on your foreign key column (Thanks to Omar for the tip)
- One of the fields in the relationship is part of a combination (composite) key and does not have it’s 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. (Thanks to Alex for this tip)
- You have a syntax error in your ALTER statement or you have mistyped one of the field names in the relationship (Thanks to Christian & Mateo for the tip)
- The name of your foreign key exceeds the max length of 64 chars. (Thanks to Nyleta for the tip)
The MySQL documentation includes a page explaining requirements for foreign keys. Though they don’t specifically indicate it, these are all potential causes of errno 150. If you still haven’t solved your problem you may want to check there for deeper technical explainations.If you run into this error and find that it’s caused by something else, please leave a comment and I’ll add it to the list.
Stephane
November 8, 2006 at 2:26 pm
No no no no no and No!!! I’ve been looking closer and closer and closer again. I have 2 tables, the first one with 2 fields and the second one with only one. The PK I’m trying to build is for 2 Unsigned INT(10) fields. It just doesn’t work. I’m fed up of this MySQL-Administrator.
I’m not happy
((((((
Jason
November 10, 2006 at 4:28 pm
Hey Stephane, I know this error sucks. MySQL doesn’t tell you what the problem is. But, something about the fields are not matching or else there may be a conflicting key or something. I don’t know I’ve never tried a table with only 1 column before – maybe MySQL doesn’t like that? If you find the answer, post here and maybe it will help someone at some point.
Niels
November 18, 2006 at 12:59 pm
Hi, I had this error also. My problem was that I was trying to create a foreign key with a name, which I had already given another foreign key.
I would try checking that the foreign key name is unique.
I am no MySQL expert, but it just a suggestion.
Jason
November 19, 2006 at 9:29 pm
Thanks Niels – I updated the post to include that info as well. Hopefully will prevent one of us from pulling his/her hair out one day!
Stephane
December 13, 2006 at 9:21 am
Okay, I reinstalled mysql-server-5.1.6_2, mysql-administrator and mysql-query-browser. Now mysql-administrator let me enter the FK and displays a message that the command completed successfully. However, the FK never shows up, nor after I refreshed the schema, shutdown the box, etc. It’s not there at all… I tried it on a W2K box and everything goes smooth. I had this setup working on a FreeBSD 5.4 in the past, I don’t know what’s buggy with 6.1. I’ll try to update all my ports and reinstall the damn thing again. Mmmmm…Ubuntu? Maybe
Jason
December 13, 2006 at 1:52 pm
Hey Stephane, it sounds to me like you may be trying to create foreign keys on a “MyISAM” table? If you need to use foreign keys, you have to use “InnoDB” tables. You can change the table type with the MySQL Query Browser utility.
Stephane
December 18, 2006 at 6:49 am
Jason,
I am using InnoDB as well. I solved my problem: I returned to the command line on the FreeBSD 6.1 box. I don’t know what’s going on since I installed FreeBSD 6.1 but MySql-Administrator goes nuts. And just to be sure it wasn’t something weird with my setup, I set up another FreBSD 6.1 box with the same stuff and still, no luck, same problems. The reason why I’m sure there’s a problem with this mix (FreeBSD 6.1, mysql-server 5.1.6_2 and mysql-administrator 1.1.6) is because everything works fine with mysql-administrator from a remote machine. I’ve been successfully doing my foreign keys remotely from a Win32 box as well as from a Fedora box.
I’ll dig that out eventually. Thanks to all for your input.
BTW, I also tried Ubuntu as a curiosity but naaahh… Too bulky
Stephane
December 18, 2006 at 2:14 pm
Guys, I’d like to insert some revealing snapshots here. I’ll setup a blog and provide you with a link for it. MySql-Administrator 1.1.6 and FreeBSD 6.1 do not cooperate well.
Paul
December 19, 2006 at 1:07 am
Thanks for the tip.
I was getting fustrated anf finally just did a search for MySQL Error 1005 and came across your blog. Had it fixed in a minute.
Srinath
January 24, 2007 at 1:47 am
Very useful tip.. Was a bit frustrated and got help from this article!!
FRR
March 7, 2007 at 3:12 pm
Hi, I had the same problem but in my case the cause of the problem was that the “Colum Charset†and “Colum Collate†was different between the key fields. It’s interesting although the both tables have already configured the same “Charset†and “Collateâ€, the fields could have different values in these parameters.
Jason
March 24, 2007 at 1:50 pm
Thanks FRR – I’ll add that to the list as well. That one would definitely be difficult to troubleshoot
Venkatesh Naicker
April 10, 2007 at 5:33 pm
Index missing on the column that is being tried to have foreign key. Creating index on that column fixed it.
andy
April 11, 2007 at 9:48 am
Hi folks,
I’ve just discover another interesting case. You can get also errno: 150 when you try to do the following: create table A, create table B with the relation many to 1 (respectively) and then create table C which is in relation one to many with table B and it is also in relation one to many with table A. I though this relations should work fine but I figure out if I take one of them I can create all the tables A, B and C if you know what I mean
Erichero
April 19, 2007 at 2:21 am
You might want to mention for newbies like me, that your foreign key must point to a primary key. I assumed that you could just point to anything.
Jason
April 19, 2007 at 11:09 am
Thanks everyone for posting your causes – I’ll continue to update the article with your ideas.
Christian
April 19, 2007 at 1:31 pm
Jesus, this is really a crap error message. It’s also thrown if you have the wrong syntax:
ALTER TABLE ADD CONSTRAINT fk_foo FOREIGN KEY (foo_id) references foo
throws this error. It should of course throw a syntax exception… This works:
ALTER TABLE ADD CONSTRAINT fk_foo FOREIGN KEY (foo_id) references foo (foo_id)
sime
May 14, 2007 at 8:54 am
Bravo! In my case, problem was about MyISAM and INNODB table types.
Justin
May 18, 2007 at 12:05 pm
Thanks for this post. Was beating my head against the wall on this one until I found this.
In my case, it turned out to be a problem with the jack ass trying to create the table. I was a little overzealous in my copy/pasting writing the SQL to create the tables and forgot to change the name of one of the columns. When I tried to create another table later with a foreign key that referenced the correct name, it failed. Took me a little while to realize that the problem wasn’t with the table that was erroring out, but with the one I was referencing.
Oh, and I’m dumb. I’m sure that contributed.
Thanks again.
Ezequiel from Buenos Aires
May 29, 2007 at 2:12 pm
My problem is that you have to create an INDEX on the table where you have the referencing column. Example:
CREATE TABLE `gcom_dbo`.`PERFILES` (
`IDPerfil` INTEGER NOT NULL AUTO_INCREMENT,
`Nombre` VARCHAR(20) NULL,
PRIMARY KEY (`IDPerfil`)
)
ENGINE= INNODB;
CREATE TABLE `gcom_dbo`.`USUARIOS` (
`IDUsuario` INTEGER NOT NULL AUTO_INCREMENT,
`IDPerfil` INTEGER NOT NULL,
`UserName` VARCHAR(50) NULL,
`Password` VARCHAR(50) NULL,
PRIMARY KEY (`IDUsuario`),
INDEX (`IDPerfil`),
CONSTRAINT `FK_USUARIOS_PERFILES` FOREIGN KEY `FK_USUARIOS_PERFILES` (`IDPerfil`)
REFERENCES `gcom_dbo`.`perfiles` (`IDPerfil`)
ON DELETE NO ACTION
ON UPDATE NO ACTION
)
ENGINE= INNODB;
Made me lost like 3 hours to work it out. It’s strange that nobody else said this… we’re migrating from MSSQL Server 2000 to MySQL 4.0.20 (pretty old stuff, I still have to figure out what to do with triggers!)
Good Luck!
Omar
May 31, 2007 at 3:36 am
Thanks for the tricks, but after hours of testing and re testing we found another issue : There should be no ‘defaut 0′ close on the foreign key column in the child table.
Greetings,
Sammy
July 5, 2007 at 6:43 am
Make sure when use of ON DELETE SET NULL with a foreign key, it is not setted to NOT NULL
Terminally Incoherent
July 9, 2007 at 2:00 pm
[...] is not the case – it is just a silly, misleading message. I started googling for it and found some good tips on how to avoid this error at VerySimple Dev Blog. I spent over two hours testing all possible solutions listed in that post. [...]
Richard Kennard
July 30, 2007 at 10:39 pm
THANK YOU! As many others have commented above, this blog entry saved me a great deal of time and hair pulling.
(my 150 was, for your statistics, caused by differing collations)
Vamsikrishna Nadella
August 2, 2007 at 11:11 am
This Article helped me a lot. I was really frustrated with the error and the InnoDB/MyISAM combination was killing me. One of the tables was MyISAM. Thanks you very much for putting up this article. It saved me a lot of time.
– Vamsi
Correa Rodrigo
August 2, 2007 at 1:02 pm
I can’t alter tables, drop key, nothig. I can’t do nothing with my database. And always show this message:
#1005 – Can’t create table ‘#sql-a2f_2e680′ (errno: 13)
ILVC
August 3, 2007 at 10:01 am
Thank you for your blog! My problem was not on the list but at least this article gave me light to hunt what was happening. In my case the problem was simply that I had TWO primary keys in the referenced table by accident! I simply dropped the other primary key and presto.
Jonathan Kohler
August 3, 2007 at 3:44 pm
Sometimes when you get this error you can get more information by checking the InnoDB log with ‘SHOW INNODB STATUS;’ It helped me pinpoint the problem.
freak
September 3, 2007 at 4:56 am
In my case datatypes of both tables were not same, one was int with unsigned and other was just int. I put the unsigned in the other one and it worked fine. Both tables should be innoDB
Mateo
September 26, 2007 at 10:18 pm
If you think you’ve tried it all, check that you are trying to reference an EXISTING field. So, perhaps you have “id_table” instead of “idtable”. The “_” can be a baddie sometimes
Sergio
October 23, 2007 at 7:20 pm
This post really helped me. My problem was that the referenced table was MyISAM. I changed to InnoDB and everything works nice now. Thank you Jason.
Edd
November 7, 2007 at 5:12 pm
My problem of migratina an MS Access db to MySQL was solved by rewriting the ALTER TABLE table1 ADD FOREIGN KEY… statement. There was a syntax error, I guess, from the fact that the two FK were a composite PK in table1. Excellent blog, btw.
J Jammin
November 17, 2007 at 9:27 am
I had this problem, but it was the famous user error. I defined the foreign key column as ‘NOT NULL’ and then specified that on delete set NULL. Makes perfect sense that MYSQL refused to create the table.
John
November 23, 2007 at 5:00 pm
Thanks – mine was #4
crisu
December 3, 2007 at 6:47 pm
ammm…can i allso add that none of all that is here helped me
(
crisu
December 3, 2007 at 6:48 pm
a good solution would be to use the plain old SQL
B
December 8, 2007 at 10:34 am
THANK YOU SO MUCH! This has been driving me crazy for the last couple days and I finally got it working. I didn’t know every foreign key has to have a unique name withing the database. What a pain in the ass!
Simon
December 9, 2007 at 3:13 pm
Thanks, this was a big help.
Gustavo
December 21, 2007 at 9:43 am
Thanks, you saved me a lot of time and headache. Mine was #1.
George Calm
January 3, 2008 at 1:23 am
Thank you, Jason. I was struggling with this error for hours now. Excellent posting!
Alex Grim
January 6, 2008 at 2:16 am
Great post. I happened across this while researching this error. One thing that you mentioned what that “one or the other must be indexed”.
But i wish to add that if one or the other (PK or FK) is part of a combined primary key/index, you will STILL get this error, you MUST create a separate index for the field you wish to reference with your FK.
Thanx
Jason
January 9, 2008 at 11:30 pm
Thanks Alex. I didn’t know that was the case but I don’t use combo keys too often. I can see that one stumping somebody so I added your tip to the article.
Victoria
January 12, 2008 at 2:59 pm
Ty #3 !!
Mat
January 14, 2008 at 2:59 pm
Also, UNSIGNED and ZEROFILL properties for the foreign key columns must match.
Jim
January 19, 2008 at 7:13 pm
Here is my SQL code: NO CLUE what is going on with this error…
CREATE TABLE PCP
(
PCPID MEDIUMINT NOT NULL,
PCPName VARCHAR(75),
PCPMD MEDIUMINT NOT NULL,
PRIMARY KEY (PCPID),
FOREIGN KEY (PCPMD) REFERENCES PCP
);
Dougie
February 4, 2008 at 11:39 am
I placed the primary key field when using the foreign and that work, just as Christian did further above.
Jim try this code:
CREATE TABLE PCP
(
PCPID MEDIUMINT NOT NULL,
PCPName VARCHAR(75),
PCPMD MEDIUMINT NOT NULL,
PRIMARY KEY (PCPID),
FOREIGN KEY (PCPMD) REFERENCES PCP (PCPMD)
);
Hope this works as this did for me.
BTW i noticed in you code your pulling data using the (foreign key) from itself as the primary key is in the same table, have you tried pulling data from a primary key of another table.
Example
FOREIGN KEY (PCPMD) REFERENCES PCP2 (PCPMD)
shob
February 17, 2008 at 3:12 pm
Thanks a lot.. i was going nuts on this bizzare err.. this aricle saved a lot … mine was INNODB == MYISAM … [:)]
Dave Miller
February 27, 2008 at 10:10 am
Mine was number 5, thank you!!!
Arturo
February 29, 2008 at 1:03 am
I had a problem where the primary key of the referenced table was not the first column on the table and I couldn’t use it as a foreign key. Droping the table and having the primary key as the first column fixed the issue. Both tables were Innodb
Cheers,
Arturo
Ivan
March 26, 2008 at 6:33 pm
Thanks!
Number 3 worked for me. Actually it is very obvious that foreign key name must be unique
Ivan
varma
March 29, 2008 at 1:24 am
i nead to know very soon.. is there any possible to use char 0r varchar to set auto_incriment
Jason
March 29, 2008 at 2:44 pm
varma – i don’t think it is possible to use varchar as auto-increment. i’m not sure the reason to do that unless it is to support some legacy schema?
Matt and Cody
April 3, 2008 at 1:06 pm
Thank you so much. This helped us find the solution to our problem. You are the best
Mamta
April 4, 2008 at 5:06 pm
Dear Jason,
Thanks a lot.. your blog helped me to solve my problem. thanks again.
Andreas
April 8, 2008 at 7:12 am
Thank you verry much! Just didn’t know there has to be uniq foreign key names.
Pete
April 24, 2008 at 3:39 pm
Thank you for this article. Just printed it out. It just saved the remaining hair I am left with!
Vinod Pillai
May 7, 2008 at 2:42 am
It is true that MySQL is having such a problem. This is the solution.
1) First of all make sure both parent and child table should be InnoDB.
Now when you create the foreign key in child table it creates the key_Name by default now you have to change that key_Name.
2) Now make sure that Foreign Column name should exactly match with the field name of the parent table.
That’s all I hope this will solve the problem.
Dele Agagu
May 9, 2008 at 8:49 am
Whao!! couldnt help but post.i tried all sorts but nothing worked until i search for the error and whola!! your blog blew my mind!! thanks for this
Dele Agagu
May 9, 2008 at 12:03 pm
Discovered another reason for this error:
Make sure that if your referenced table has the primary key column as UNSIGNED, then the referencing table needs to have the referencing column as UNSIGNED too.
Shift Instinct » Blog Archive » MySQL Error Number 1005 Can’t create table ‘.mydb#sql-328_45.frm’ (errno: 150)
May 9, 2008 at 12:12 pm
[...] VerySimpleDevBlog [...]
Tor Andre
May 15, 2008 at 4:24 pm
Very nice article.
I was using ActiveRecord/NHibernate to keep my db up to date. Since this was before release 1.0 I had ActiveRecord to automatically update my database (using ‘hibernate.hbm2ddl.auto’). This results in dropping every table and generate them again every time I rebuild my project.
I got stuck with this error, and there was one table I was unable to create. I tried everything, manually without foreign keys as well.
What solved it was to create the table WITH a foreign key but not naming it. Once doing that, I edited the table in Query Browser and found two (2) foreign keys with same reference. It seemed as the table was dropped as it should be, but the previous foreign key was stuck somewhere (couldn’t find it in the system tables either).
I dropped both the foreign keys and the table, then clean build the project (resulting in drop and create statements for all tables), and it WORKED!
Happy debugging!
Rahul Master
May 18, 2008 at 1:09 am
Thanks, the 4th cause worked for me
Aswin Anand
May 31, 2008 at 8:54 pm
If the table for which the foreign key has to be applied contains data, this could happen. I took a backup of the existing data, truncated the table, ran the alter table query again. This time it went on smoothly.
Find an Appraiser
June 12, 2008 at 6:12 pm
I checked everything in this list and still got the error. Then I just dropped and recreated the database, ran new create script with only difference being my new table having PK varchar(255) instead of int(3), and it all worked fine. Strange, but there you have it.
Amin Abbasopour
June 15, 2008 at 7:37 am
You’re great!
Spenner
July 2, 2008 at 10:20 am
Just a little extra info with MySQL
I had quite a few problems creating FK’s and found out that it’s good practice to use different FK names for each table.
Apparently MySQL does not like multiple instances of the same FK name in the same DB for different tables which I also think is quite logical.
Thanks to everyone else on this post for all the info provided as well
pavan
July 7, 2008 at 6:23 am
thanks a lot
Marco Agurto
July 18, 2008 at 3:48 pm
I don’t write in english very well so a try to do my best work. I try to fix my alter with the nine steps but I cant fix my problem So I could fix my problem using the alter in the same order. In table one DE_PROC, CO_PROC in the table two CO_PROC, DE_PROC change the order of table two, I am happy again lucky for everybody
jegreat
September 10, 2008 at 11:25 am
make sure that the foreign key u r trying to add is already present in the reference table!! if this value is not a primary key in the reference table; then u wil get this error
Nyleta
October 7, 2008 at 2:28 am
After a lot of swearing and reading through your suggestions I tried something from left field and shortened the name of my foreign key, and bingo! So if you are still adding to your list up the top, can I suggest shortening of the foreign key name.
werutzb
October 7, 2008 at 9:43 pm
Hi!
I would like improve my SQL experience.
I red that many SQL resources and would like to
get more about SQL for my work as mysql database manager.
What would you recommend?
Thanks,
Werutz
Jason
October 8, 2008 at 1:46 pm
Thanks Nyleta, I found the max length is 64 chars, so I’ll add this to the list
King`Babylon` » Blog Archive » Mysql : Foreign Key
October 23, 2008 at 9:40 am
[...] Un lien d’aide en cas d’erreur 1005 [...]
anonymous
October 24, 2008 at 5:25 pm
It the definition of the local columns that is different to the ForeignKey-Column (such like “UNSIGNED”, (), etc.)
Paresh
November 11, 2008 at 7:20 pm
Thanks for the information. I had the problem of one is SIGNED and the other is UNSIGNED.
Your blog helped a lot!!!
Thanks again
Nayyereh
November 12, 2008 at 7:34 am
Hi,
Thank you Niels, I had this problem and with your suggestion solved.
Deepak
November 13, 2008 at 8:19 am
The solution that worked for me (having ensured all the above were verified was the length of the name of FK. I gave it a shorter name and it worked.
Brian
November 15, 2008 at 12:55 am
“You should also check that one is not SIGNED and the other is UNSIGNED.” Ahh..HAH!
Very good! Thank you sir.
-Brian
Ignace
November 16, 2008 at 5:25 am
Some good advice: make sure all your tables are dropped in your database before running your sql installation script! I personally use:
DROP TABLE IF EXISTS ..
before i place my table definition, however when i ran the script when their were already tables in the db i got the #1005 error, when i used the same script on an empty db it worked just fine.
the.janitor
November 18, 2008 at 8:47 am
thanks a lot, very helpful information
Shan
December 3, 2008 at 5:28 pm
This was incredibly helpful. I usually avoid these kind of sites, but this helped what was proving to be an extremely frustrating issue. Brilliant.
MySQL Error Number 1005 Can’t create table ‘.mydb#sql-328_45.frm’ (errno: 150) « Fabio Cepeda
December 4, 2008 at 4:53 pm
[...] Reasons for this error 2. Tips of importing to a [...]
chicco
December 15, 2008 at 7:42 am
It’s a dirty way to solve it but if you enclose the alter table statement between
SET FOREIGN_KEY_CHECKS=0;
and
SET FOREIGN_KEY_CHECKS=1;
it works
HKASLCA
January 4, 2009 at 12:51 pm
Thanks.
I’ve solved the problem.
Lainey
February 3, 2009 at 7:56 pm
Can I say I love you? I have spent hours trying to figure out this issue and it turned out that both tables were MyISAM tables! Gotta love the internet and people like you who post solutions to problems!
CJ
February 12, 2009 at 12:06 pm
ONE MORE:
Make sure your tables are written in the definition file BEFORE you try and make a foreign key on that table.
Stupid rubbish mysql parser.
kirov
March 12, 2009 at 4:19 am
yeah, one more time THANK YOU, I was ready unninstal MySQL because of that, and the problem was stupid index on column. Why they couldn’t do a error message about this, I don’t understand.
MySQL Error Number 1005 Can’t create table ‘.mydb#sql-328_45.frm’ (errno: 150)
March 15, 2009 at 10:14 am
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ynabid
April 10, 2009 at 11:36 am
when I change my tables engine to innoDB and I want to create a foreign key for a attribut that has data type=varchar, that not work but when I set flag to binary it work.
Plat
April 15, 2009 at 7:55 pm
Thanks for the troubleshootin’ list. This saved me an incredible amount of time. I caught an UNSIGNED/SIGNED mismatch prior to this page, but didn’t think to check InnoDB vs MyISAM. My new editor must be defaulting to InnoDB for some reason.
Gerhard Kratz
April 21, 2009 at 1:23 pm
Gerhard Kratz
April 22, 2009 at 12:19 am
If one wants to
- CREATE TABLE A with FOREIGN KEY ( ab ) REFERENCES B
- CREATE TABLE B with FOREIGN kEY ( ba ) REFERENCES A
then one of the foreign keys has to be declared without the table it references already existing.
The problem is remedied proceeding as follows:
- CREATE TABLE A
- CREATE TABLE B with FOREIGN kEY ( ba ) REFERENCES A
- ALTER TABLE A ADD FOREIGN KEY ( ab ) REFERENCES B
I always give the name of the primary key of the table referenced to the foreign key constraint.
Lauro Valente
April 22, 2009 at 8:49 pm
Make sure the tables are clean, with no data in them….
Jason
April 24, 2009 at 12:39 am
that’s a good point Lauro. You can actually have data in the tables, but you have to make sure that every row works with the new key. a lot of times when you try to add a key you’ll have to clean up the data first.
Heti
April 29, 2009 at 11:01 am
thaaaaanks a ton … got the prblem fixed in lesser time than expected.
))))) i am so happppyyyy
Radek
May 22, 2009 at 5:55 pm
Thanks for that checklist.
Here’s what was wrong in my case: mySQL 5.0 under windows lowercased all column names when I created the tables. Under UNIX the tables were created with orignal cases. Then I tried copy pasting the ALTER TABLE statement from Windows DEV DB with the lowercase version. That is where erno 150 came from.
Dave
June 24, 2009 at 3:45 am
Thanks heaps!!!!!!!!!!!!!!!
My problem was point 3 – duplicate fk name
thao
June 27, 2009 at 9:46 am
thankyou!
Can i translate this Posts to vietnamese and post it in our public site?
Jason
June 27, 2009 at 4:03 pm
hi thao, that would be ok – it would be appreciated if you link back to my site as well.
zachariah
June 29, 2009 at 7:51 am
Hi………. Thanks a lot. Great work. Helped me a lot.
che
July 6, 2009 at 7:32 am
ciao, thank you for this. my problem was #8
Sax
July 14, 2009 at 6:02 am
Issue number 6 was my problem today. Thanks for the solution.
Rohit
July 22, 2009 at 9:07 am
Really helpful post mate
alk
July 22, 2009 at 9:19 am
Dude ur the best! i solved my pproblem using #6! Thanks alot.
ann_an
July 23, 2009 at 5:16 am
nice this trick saved my hell amount of time, thanks a ton
What I like! » Blog Archive » Error #1005 – Can’t create table… en MySQL al intentar crear una relacion 1-M
July 25, 2009 at 6:58 pm
[...] Pueden conseguir mas informacion en este blog http://www.verysimple.com/blog/2006/10/22/mysql-error-number-1005-cant-create-table-mydbsql-328_45fr... [...]
Xavier Montero
August 13, 2009 at 4:56 pm
In the case it serves to somebody: Do not skip step6.
I’ve been messing with the same problem. The DATA TYPE was identical. In both cases a VARCHAR(36) which I use to store as primary keys standard UUID values as “string”.
Nevertheless, being them text, they have a “charset”, and one table was “UTF” and the other “LATIN”.
They did not link for that reason.
Changing the charset to be coherent (both to utf8 or both to latin) I was able to create the FOREIGN KEY without the error of “can’t create table…”
Chris
October 19, 2009 at 3:25 pm
Also:
If the primary key you want to connect to your foreign key is UNSIGNED make it G** D*** SURE that your foreign key is also UNSIGNED.
wirawit
November 12, 2009 at 5:14 am
Hey, Fixed my Bug! Thx!
Firas Abd Alrahman
November 19, 2009 at 3:53 pm
I encountered this problem in this case
I created a foreign key
On Delete = Set Null
On Update = Set Null
but the foreign key field does Have “Not null ” flag
Just removed the check by mysql GUI administrator and Everything was just fine
Amal Kaluarachchi
November 30, 2009 at 10:14 pm
I had the same problem and spent many hours on troubleshooting. This article helped me to figure it out the issue. It was I have set the ondelete to set null where the table field is set not null.
Thanks
Brian Weiss
December 17, 2009 at 12:02 pm
My issue came up because I was trying to import/build a new table that had foreign keys in tables that didn’t exist. The export from the old MYSQL server exported the tables in alpha numeric order, but on the new server, my CONSTRAINT statement was referencing foreign keys in tables that didn’t yet exist as they were further down in the file. So I just cut and pasted the Create table statements into a new order, where all the ones that needed foreign keys were built after the tables they referenced. And it worked! Hope this helps others…
AJ
January 8, 2010 at 3:24 pm
If you use the SQL command,
SHOW ENGINE INNODB STATUS;
it will give you are more detailed error message.
Erika
January 13, 2010 at 9:53 am
Thanks, It solved my problem. One table had UNSIGNED checked and the other not. Pretty simple but very hard to find.
Daniel Cairol
January 20, 2010 at 5:26 pm
THANKS BABY! My solution was number 4.
Kai-Jin
February 3, 2010 at 5:41 am
thy! uft8 != latin1
Amit Singh
February 10, 2010 at 7:32 am
Thanks! it helped me I solved my permanent problem
Kulish kushwah
March 18, 2010 at 3:32 am
The error may occur, if both the fields having different storage engine.
sim4000
May 19, 2010 at 7:02 am
Many thanks.
I remove “NOT NULL” from the FK column and it works!
_Jon
May 21, 2010 at 1:53 pm
Thanks! #1 was my match.
For some reason, this table was created with INT(5) for the PK, where all previous tables were SMALLINT(5). I have no idea why the GUI did that.
But now I know.
Thanks again.
VISHNU
May 25, 2010 at 12:19 am
unable to create partition on a parent a child table.
query executed:
ALTER TABLE table_name1 PARTITION BY KEY (pk_column_name1,pk_column_name2);
ALTER TABLE table_name2 PARTITION BY KEY (pk_column_name1,pk_column_name2);
parent table name:table_name1
child table name: table_name2
need to alter both the tables.
pkak
June 22, 2010 at 7:38 am
dude, thanks. it was number 6
csen
July 9, 2010 at 5:21 pm
Good!As I have try the first one ,my problem was solved!
haha,thanks a lot!
Lloyd
July 13, 2010 at 4:11 am
You’re a saviour! My error was number 5, I had FK set to “on delete set null” which was referencing a column that was “not null” in hind sight my mistake was so blindingly obvious but at the time I just couldn’t see the wood for the trees.
This is fantastic list and a great resource
Cheers
Nikki
July 23, 2010 at 12:10 am
The thing that I noticed is the Character set of both the columns should be same….for e.g. A table has primary key named ID,Now the table which want to use this key as foreign key ,the column to which it is relating to be the foreign key should have same character set as that of the A table as well as the size should also be same.
N3r1
August 6, 2010 at 2:23 pm
hi just run into another one
let me show you
it doesn’t work
constraint `fk_res_num_occ` foreign key(`occ_res_num`) references`hotel`.`reservation`(`res_num`),
it works
constraint `fk_res_num_occ` foreign key(`occ_res_num`) references `hotel`.`reservation`(`res_num`),
there was no space after references
Joe Devon
August 7, 2010 at 10:16 pm
I had a really weird one I just fixed… Nothing on this list made a difference…but I noticed that a second column (NOT USED AT ALL IN ANY WAY BY THE FK) in the original table had a unique index…so on a whim, I deleted it…was able to add a foreign key, then reinstate the unique index…
No idea why that should matter…but if none of the above helps you, try it…
Rolf
August 15, 2010 at 12:07 pm
If MySQL is acting like the usual piece of crap that it is, you might want to do this:
SET FOREIGN_KEY_CHECKS = 0;
To disable checks. It worked for me so far, as I couldn’t find the problem, despite your wonderfully compiled list, and don’t intend so spend the rest of my day on this.
Once done, there is still the option to re-enable the sadistic masochistic checks:
SET FOREIGN_KEY_CHECKS = 1;
Hope this info will be added to your post