I'm having a bit of a strange problem. I'm trying to add a foreign key to one table that references another, but it is failing for some reason. With my limited knowledge of MySQL, the only thing that could possibly be suspect is that there is a foreign key on a different table referencing the one I am trying to reference.
I've done a SHOW CREATE TABLE query on both tables, sourcecodes_tags is the table with the foreign key, sourcecodes is the referenced table.
CREATE TABLE `sourcecodes` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) unsigned NOT NULL,
`language_id` int(11) unsigned NOT NULL,
`category_id` int(11) unsigned NOT NULL,
`title` varchar(40) CHARACTER SET utf8 NOT NULL,
`description` text CHARACTER SET utf8 NOT NULL,
`views` int(11) unsigned NOT NULL,
`downloads` int(11) unsigned NOT NULL,
`time_posted` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`),
KEY `language_id` (`language_id`),
KEY `category_id` (`category_id`),
CONSTRAINT `sourcecodes_ibfk_3` FOREIGN KEY (`language_id`) REFERENCES `languages` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_1` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_2` FOREIGN KEY (`category_id`) REFERENCES `categories` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1
CREATE TABLE `sourcecodes_tags` (
`sourcecode_id` int(11) unsigned NOT NULL,
`tag_id` int(11) unsigned NOT NULL,
KEY `sourcecode_id` (`sourcecode_id`),
KEY `tag_id` (`tag_id`),
CONSTRAINT `sourcecodes_tags_ibfk_1` FOREIGN KEY (`tag_id`) REFERENCES `tags` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1
This is the code that generates the error:
ALTER TABLE sourcecodes_tags ADD FOREIGN KEY (sourcecode_id) REFERENCES sourcecodes (id) ON DELETE CASCADE ON UPDATE CASCADE
Quite likely your sourcecodes_tags table contains sourcecode_id values that no longer exists in your sourcecodes table. You have to get rid of those first.
Here's a query that can find those IDs:
SELECT DISTINCT sourcecode_id FROM
sourcecodes_tags tags LEFT JOIN sourcecodes sc ON tags.sourcecode_id=sc.id
WHERE sc.id IS NULL;
I had the same issue with my MySQL database but finally, I got a solution which worked for me.
Since in my table everything was fine from the mysql point of view(both tables should use InnoDB engine and the datatype of each column should be of the same type which takes part in foreign key constraint).
The only thing that I did was to disable the foreign key check and later on enabled it after performing the foreign key operation.
Steps that I took:
SET foreign_key_checks = 0;
alter table tblUsedDestination add constraint f_operatorId foreign key(iOperatorId) references tblOperators (iOperatorId); Query
OK, 8 rows affected (0.23 sec) Records: 8 Duplicates: 0 Warnings: 0
SET foreign_key_checks = 1;
Use NOT IN to find where constraints are constraining:
SELECT column FROM table WHERE column NOT IN
(SELECT intended_foreign_key FROM another_table)
so, more specifically:
SELECT sourcecode_id FROM sourcecodes_tags WHERE sourcecode_id NOT IN
(SELECT id FROM sourcecodes)
EDIT: IN and NOT IN operators are known to be much faster than the JOIN operators, as well as much easier to construct, and repeat.
Truncate the tables and then try adding the FK Constraint.
I know this solution is a bit awkward but it does work 100%. But I agree that this is not an ideal solution to deal with problem, but I hope it helps.
For me, this problem was a little different and super easy to check and solve.
You must ensure BOTH of your tables are InnoDB. If one of the tables, namely the reference table is a MyISAM, the constraint will fail.
SHOW TABLE STATUS WHERE Name = 't1';
ALTER TABLE t1 ENGINE=InnoDB;
This also happens when setting a foreign key to parent.id to child.column if the child.column has a value of 0 already and no parent.id value is 0
You would need to ensure that each child.column is NULL or has value that exists in parent.id
And now that I read the statement nos wrote, that's what he is validating.
I had the same problem today. I tested for four things, some of them already mentioned here:
Are there any values in your child column that don't exist in the parent column (besides NULL, if the child column is nullable)
Do child and parent columns have the same datatype?
Is there an index on the parent column you are referencing? MySQL seems to require this for performance reasons (http://dev.mysql.com/doc/refman/5.5/en/create-table-foreign-keys.html)
And this one solved it for me: Do both tables have identical collation?
I had one table in UTF-8 and the other in iso-something. That didn't work. After changing the iso-table to UTF-8 collation the constraints could be added without problems. In my case, phpMyAdmin didn't even show the child table in iso-encoding in the dropdown for creating the foreign key constraint.
It seems there is some invalid value for the column line 0 that is not a valid foreign key so MySQL cannot set a foreign key constraint for it.
You can follow these steps:
Drop the column which you have tried to set FK constraint for.
Add it again and set its default value as NULL.
Try to set a foreign key constraint for it again.
I'd the same problem, I checked rows of my tables and found there was some incompatibility with the value of fields that I wanted to define a foreign key. I corrected those value, tried again and the problem was solved.
I end up delete all the data in my table, and run alter again. It works. Not the brilliant one, but it save a lot time, especially your application is still in development stage without any customer data.
try this
SET foreign_key_checks = 0;
ALTER TABLE sourcecodes_tags ADD FOREIGN KEY (sourcecode_id) REFERENCES sourcecodes (id) ON DELETE CASCADE ON UPDATE CASCADE
SET foreign_key_checks = 1;
I had this exact same problem about three different times. In each instance it was because one (or more) of my records did not conform to the new foreign key. You may want to update your existing records to follow the syntax constraints of the foreign key before trying to add the key itself. The following example should generally isolate the problem records:
SELECT * FROM (tablename)
WHERE (candidate key) <> (proposed foreign key value)
AND (candidate key) <> (next proposed foreign key value)
repeat AND (candidate key) <> (next proposed foreign key value) within your query for each value in the foreign key.
If you have a ton of records this can be difficult, but if your table is reasonably small it shouldn't take too long. I'm not super amazing in SQL syntax, but this has always isolated the issue for me.
Empty both your tables' data and run the command. It will work.
I was getting this error when using Laravel and eloquent, trying to make a foreign key link would cause a 1452. The problem was lack of data in the linked table.
Please see here for an example: http://mstd.eu/index.php/2016/12/02/laravel-eloquent-integrity-constraint-violation-1452-foreign-key-constraint/
You just need to answer one question:
Is your table already storing data? (Especially the table included foreign key.)
If the answer is yes, then the only thing you need to do is to delete all the records, then you are free to add any foreign key to your table.
Delete instruction: From child(which include foreign key table) to parent table.
The reason you cannot add in foreign key after data entries is due to the table inconsistency, how are you going to deal with a new foreign key on the former data-filled the table?
If the answer is no, then follow other instructions.
I was readying this solutions and this example may help.
My database have two tables (email and credit_card) with primary keys for their IDs. Another table (client) refers to this tables IDs as foreign keys. I have a reason to have the email apart from the client data.
First I insert the row data for the referenced tables (email, credit_card) then you get the ID for each, those IDs are needed in the third table (client).
If you don't insert first the rows in the referenced tables, MySQL wont be able to make the correspondences when you insert a new row in the third table that reference the foreign keys.
If you first insert the referenced rows for the referenced tables, then the row that refers to foreign keys, no error occurs.
Hope this helps.
Make sure the value is in the other table otherwise you will get this error, in the assigned corresponding column.
So if it is assigned column is assigned to a row id of another table , make sure there is a row that is in the table otherwise this error will appear.
you can try this exapmple
START TRANSACTION;
SET foreign_key_checks = 0;
ALTER TABLE `job_definers` ADD CONSTRAINT `job_cities_foreign` FOREIGN KEY
(`job_cities`) REFERENCES `drop_down_lists`(`id`) ON DELETE CASCADE ON UPDATE CASCADE;
SET foreign_key_checks = 1;
COMMIT;
Note : if you are using phpmyadmin just uncheck Enable foreign key checks
as example
hope this soloution fix your problem :)
UPDATE sourcecodes_tags
SET sourcecode_id = NULL
WHERE sourcecode_id NOT IN (
SELECT id FROM sourcecodes);
should help to get rid of those IDs. Or if null is not allowed in sourcecode_id, then remove those rows or add those missing values to the sourcecodes table.
I had the same problem and found solution, placing NULL instead of NOT NULL on foreign key column. Here is a query:
ALTER TABLE `db`.`table1`
ADD COLUMN `col_table2_fk` INT UNSIGNED NULL,
ADD INDEX `col_table2_fk_idx` (`col_table2_fk` ASC),
ADD CONSTRAINT `col_table2_fk1`
FOREIGN KEY (`col_table2_fk`)
REFERENCES `db`.`table2` (`table2_id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION;
MySQL has executed this query!
In my case, I created a new table with the same structure, created the relationships with the other tables, then extracted the data in CSV from the old table that has the problem, then imported the CSV to the new table and disabled foreign key checking and disabled import interruption, all my data are inserted to the new table that has no problem successfully, then deleted the old table.
It worked for me.
I'm trying to constraint two different tables fields through mySql.
table(field) : comments(item_id) with item(item_ID) with this code:
ALTER TABLE 'comments' ADD CONSTRAINT comment_item
FOREIGN KEY(item_id) REFERENCES items(item_ID)
ON UPDATE CASCADE
ON DELETE CASCADE;
but I'm getting this error message :
1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''comments' ADD CONSTRAINT comment_item
FOREIGN KEY(item_id) REFERENCES items(it' at line 1
what is the problem ?
I'm using InnoDB as a Storage Engine, and utf_general_ci as a Collation .
Actually you dont need single quotes, according to MySQL documentation you could write your statement like this:
ALTER TABLE comments ADD CONSTRAINT comment_item
FOREIGN KEY(item_id) REFERENCES items(item_ID)
ON UPDATE CASCADE
ON DELETE CASCADE;
You can also check here for an example
change
ALTER TABLE 'comments' ADD CONSTRAINT comment_item
to
ALTER TABLE `comments` ADD CONSTRAINT comment_item
Use ` instead of '.
or just write without ' like this:
ALTER TABLE comments
I want to add foreign key Constraint on my existing tables(type InnoDB)
,I have two tables listed below:
tbl_users : [ user_id(INT,AUTO_INCREMENT), user_name,--- etc]
tbl_users_meta : [ user_meta_id(INT,AUTO_INCREMENT), user_id(INT),--- etc]
I have already created index 'user_id' on 'tbl_users_meta' listed below :
Here is my Query but i am getting (MySQL Error Code 1215: ) each time.what is the problem i can getting it ?
ALTER TABLE `tbl_users_meta`
ADD CONSTRAINT `fk users user_id`
FOREIGN KEY (`user_id`) REFERENCES
`sanskrut`.`tbl_users`(`user_id`)
ON DELETE NO ACTION ON UPDATE CASCADE;
TRY with the same query you have writen with a small modification:
ALTER IGNORE TABLE `tbl_users_meta`
ADD CONSTRAINT `fk users user_id`
FOREIGN KEY (`user_id`) REFERENCES
`sanskrut`.`tbl_users`(`user_id`)
ON DELETE NO ACTION ON UPDATE CASCADE;
Hope it'll work
I can't see anything wrong with what you've written, so there must be something you haven't posted affecting the problem. For example, on MySQL version 5.5.44, I can successfully execute the following:
CREATE DATABASE sanskrut;
USE sanskrut;
CREATE TABLE tbl_users (user_id INT AUTO_INCREMENT PRIMARY KEY, user_name VARCHAR(50));
CREATE TABLE tbl_users_meta (user_meta_id INT AUTO_INCREMENT PRIMARY KEY, user_id INT);
CREATE INDEX user_id ON tbl_users_meta (user_id);
ALTER TABLE `tbl_users_meta`
ADD CONSTRAINT `fk users user_id`
FOREIGN KEY (`user_id`) REFERENCES
`sanskrut`.`tbl_users`(`user_id`)
ON DELETE NO ACTION ON UPDATE CASCADE;
...which finishes with your exact problem statement, copied and pasted, and works just fine.
I'd suspect, as I mentioned in the comments, that there's an index missing or possibly a datatype mismatch. You may find it helpful to add the output of:
DESC tbl_users;
DESC tbl_users_meta;
SHOW indexes FROM tbl_users;
SHOW indexes FROM tbl_users_meta;
...to your question.
So -- I have two tables that I'm trying to relate: tournament and match. They can be outlined as followed:
tournament
-id (int)
-league (varchar)
-status (varchar)
-create_time (datetime)
-update_time (datetime)
match
-id (int)
-tournament_id (int)
-status (varchar)
-create_time (datetime)
-update_time (datetime)
I'm attempting to add a foreign key constraint on the match table with the following SQL:
ALTER TABLE 'match' ADD CONSTRAINT
('FK_match_tournament') FOREIGN KEY
('tournament_id') REFERENCES
'tournament' ('id') ON DELETE CASCADE
ON UPDATE RESTRICT;
however, I am getting the following error message from MySQL:
#1064 - You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version for the right syntax to use near
''match' ADD CONSTRAINT ('FK_match_tournament') FOREIGN KEY ('tournament_id') REF'
at line 1
I looked at the syntax for adding FK constraints on the MySQL website, and everything looks right to me. Any ideas?
First Suggestion (manuelpedrera):
ALTER TABLE `match` ADD CONSTRAINT ('FK_match_tournament') FOREIGN KEY ('tournament_id') REFERENCES `tournament` ('id') ON DELETE CASCADE ON UPDATE RESTRICT;
Results:
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '('FK_match_tournament') FOREIGN KEY ('tournament_id') REFERENCES `tournament` ('' at line 1
Turns out 'match' is a reserved word.
You cannot use regular quotes when referring to a table.
Instead of ALTER TABLE 'match'
use
ALTER TABLE match
or
ALTER TABLE `match`
Edit 1:
Try with this
ALTER TABLE `match` ADD FOREIGN KEY (`tournament_id`) REFERENCES `tournament`(`id`) ON DELETE CASCADE ON UPDATE RESTRICT;
I am trying to create a table in MySql using php. My code looks like this:
$sql = "CREATE TABLE qotwQuestion1111
(
QuestionId int PRIMARY KEY AUTOINCREMENT,
Question varchar(5000),
MemberId varchar(255) FOREIGN KEY REFERENCES qotwMember(MemberId),
PostDate date,
Vote int
)";
mysql_query($sql,$con);
i am unable to create this table, the error is in the "AUTOINCREMENT" and also the "FOREIGN KEY" ..
Can someone tell me what am i doing wrong here. and what should i do to solve this problem please
Best
Zeeshan
http://dev.mysql.com/doc/refman/5.0/en/example-auto-increment.html
The keyword is AUTO INCREMENT
http://dev.mysql.com/doc/refman/5.1/en/innodb-foreign-key-constraints.html
foreign keys are specified as constraints or by FOREIGN KEY fk_name REFERENCES table(key) .
When you have a problem with your sql syntax, it tells you to check your sql manual near where the error occurred. I suggest you take its advice.