New to MySQL / PHPMyAdmin although I've done some SQL before. I've tried to create a foreign key constraint on a table and was successful in creating this constraint with one table (Account - Standard) but not for (Account - Premium). This link contains screenshots and further explanations of the errors I am receiving. https://imgur.com/a/0s9VUA5
Constraint name should be unique. For example:
fk_username_standard, fk_username_premium ...
Here's the script
ALTER TABLE `candycorn`.`bb_users`
ADD CONSTRAINT `pf_minecraftusername`
FOREIGN KEY (`pf_minecraftusername`)
REFERENCES `candycorn`.`bb_profile_fields_data` (`pf_minecraftusername`)
ON DELETE NO ACTION
ON UPDATE NO ACTION;
and the error description
ERROR 1452: Cannot add or update a child row: a foreign key constraint fails
(`candycorn`.<result 2 when explaining filename '#sql-4e4_1785d'>, CONSTRAINT
`pf_minecraftusername` FOREIGN KEY (`pf_minecraftusername`) REFERENCES
`bb_profile_fields_data` (`pf_minecraftusern)
if somebody could, would you explain what I'm missing out here?
The constraint that you are trying to add isn't satisfied by some data already in the tables. This can be because a value in bb_users table isn't found in the corresponding column (pf_minecraftusername) in bb_profile_fields_data.
If you have to add the constraint to tables with data already in them, you have to clean up the tables by hand first. Alternatively, you can empty the tables (with truncate, or by making the tables afresh after doing a "drop database"), then add the constraint, and then run whatever scripts you have to put data in the tables.
Of course, if this is a production system, you'll need to do more complex data fixing before you can successfully add the constraint.
Finally, I suggest that you make a dummy copy of the database with no data at all and add the constraint there, just to check that the constraint is properly specified. You don't want to be barking up the wrong tree.
ALTER TABLE tablename with NOCHECK
ADD CONSTRAINT [FK_1] FOREIGN KEY ([Column name])
REFERENCES restaurants([column name])
It seems data is already present in table.. So
You need to modify table using above query.. No need to recreate table nd data
There are several other questions about this topic that I have gone through, but I can't seem to figure out how their solutions apply to my tables. Check out the sqlfiddle. You can see it builds the schema just fine.
Basically, one table is a table of contacts/people. The second table is a table of countries. I am attempting to create a foreign key reference between contacts.country_id and countries.id.
Now, add the following to the panel on the left side:
ALTER TABLE `ultra_contacts`
ADD INDEX `fk_test` (`country_id`),
ADD CONSTRAINT `fk_test` FOREIGN KEY (`country_id`) REFERENCES `ultra_countries` (`id`) ON UPDATE CASCADE ON DELETE CASCADE`
The alter table code is not working for some reason. Any help would be appreciated.
The error is: Schema Creation Failed: Can't create table 'db_e342e.#sql-7711_1a4d2' (errno: 150): Using a 3rd party program (HeidiSQL) the error is a bit more detailed:
Foreign key constraint is incorrectly formed
You're trying to use foreign keys on a MyISAM table, which is not allowed (they only work with InnoDB). Take a look here: http://sqlfiddle.com/#!2/64951 All I've changed from your original is the table type (from MyISAM to InnoDB) and then I added the constraint. Worked fine.
Full disclosure - I'm the author of SQL Fiddle :)
I am giving following error while adding the object into database.
Foreign key 'fk4150' references object 'TESTPOLY2' which is not a user
table. Could not create constraint. See previous errors. in SQL:
ALTER TABLE [WORKLOG2SETUP] ADD CONSTRAINT fk4150 FOREIGN KEY
(related_id) REFERENCES TESTPOLY2
Please let me know if anyone knows solution.
Cheers,
Pravin
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?
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!
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.
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 :)
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
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.
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.
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 ;-)
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;
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;
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.
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.
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:
Rt. click on your schema and select 'schema inspector'. This gives you various tables, columns, indexes, ect.
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'.
Now execute the query : ALTER table tableNamexx DROP KEY foreignKeyName;
Now you can execute the drop statement which shall execute successfully.
Doing
SET FOREIGN_KEY_CHECKS=0;
before the Operation can also do the trick.
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.