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 have two tables
CREATE TABLE `category` (`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8
CREATE TABLE `item` (`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
categoryid` int(10) unsigned DEFAULT NULL,
PRIMARY KEY (`id`), KEY `fk_categoryid_item` (`categoryid`),
CONSTRAINT `fk_categoryid_item` FOREIGN KEY (`categoryid`)
REFERENCES `category` (`id`) ON DELETE CASCADE)
ENGINE=InnoDB DEFAULT CHARSET=utf8
In the table category I have a record with id 2.
In the item I have a record with id = 1, categoryid = 2, with 2 as the foreign key referring to the category table. If I delete the row in the category table with the id 2, the record in the item table that has the categoryid as 2 also gets deleted. This is as expected because of on delete cascade. But If I try to drop the table category, I get the error Error Code:
1217. Cannot delete or update a parent row: a foreign key constraint fails
Why does this happen ? Of course, setting foreign_key_checks = 0 dropping the table becomes possible. But I would like to know why does this happen that we can delete the records, but can not drop the table with on cascade delete option. Does this option only apply for deleting records, but not for dropping tables.
I checked the documentation, I could not find any explanation for this.
Please let me know if there is something fundamental that I am missing or if you point out to the related documentation it would be helpful. I am using MySQL 5.7.
Thanks in advance.
If you delete the table category but do not remove/alter the foreign key, then that will be left pointing to nothing. Internally the database has a management system that reinforces the referential constraints and that prevents you from creating lose ends. See also this, this and this questions.
It has something to do also with the math behind it, it is called relational algebra. I am not at that level either, but I think it breaks the definition of a FK if you delete one of the associated tables.
In database relational modeling and implementation, a unique key is a set of zero or more attributes, the value(s) of which are guaranteed to be unique for each tuple (row) in a relation.
I am trying to insert pseudo data into my db to get going, and in one particular table I have two columns which are FK's and PK's of the table; fk_product_manf_code and fk_content_id. To my understanding, these are considered composite keys in their current state.
So I add data to the table:
fk_product_manf_code fk_content_id
NOV-ABC123 1
I then want to associate another content_id to the same product_manf_code, so I perform the following:
INSERT INTO `mydb`.`package_contents`
(`fk_product_manf_code`, `fk_content_id`)
VALUES
('NOV-ABC123', 2);
However I'm greeted with the following error:
Error Code: 1062. Duplicate entry 'NOV-ABC123' for key 'fk_product_manf_code_UNIQUE'
I don't understand what's going, because I thought a composite key makes 2 columns unique? So why is it kicking up a fuss about just 1 column being unique?
Here is the table CREATE statement
CREATE TABLE `package_contents` (
`fk_product_manf_code` varchar(255) NOT NULL,
`fk_content_id` int(11) NOT NULL,
PRIMARY KEY (`fk_content_id`,`fk_product_manf_code`),
UNIQUE KEY `fk_content_id_UNIQUE` (`fk_content_id`),
UNIQUE KEY `fk_product_manf_code_UNIQUE` (`fk_product_manf_code`),
CONSTRAINT `content_id` FOREIGN KEY (`fk_content_id`) REFERENCES `contents` (`content_id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
CONSTRAINT `product_manf_code` FOREIGN KEY (`fk_product_manf_code`) REFERENCES `products` (`product_manf_code`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
So, you are learning why composite primary keys are a pain, especially for foreign key constraints. Not only are integer keys more efficient, but a single key is easier to work with.
I would suggest changing your table structure to be more like this:
CREATE TABLE package_contents (
package_contents_id int not null auto_increment primary key,
fk_product_manf_id int NOT NULL,
fk_content_id int(11) NOT NULL,
UNIQUE KEY (fk_content_id, fk_product_manf_id),
CONSTRAINT content_id FOREIGN KEY (fk_content_id)
REFERENCES contents(content_id) ON DELETE NO ACTION ON UPDATE NO ACTION,
CONSTRAINT product_manf_code FOREIGN KEY (fk_product_manf_id)
REFERENCES products(product_manf_id) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
Note that I changed the manufacturer code to an id as well. This should also reduce the size of the table, assuming that the "code" is longer than 4 bytes.
If you do this for all your tables, the database will be a bit more efficient, and you won't need superfluous unique constraints. The foreign key constraints should always be to primary keys (unless there is a very good reason for using a different unique key).
In a MySQL database where there are relationships between tables and the primary key of one table is stored as a foreign key in a second table, is there still a need to perform a join?
If there is, what is the point on declaring the relationship? I'd take a stab in the dark and say it's something to do with the indexing or related tables can find related records much faster? I've tried Googleing this, but can't seem to find much. I'm sure there is loads out there on this, but I don't know the keywords to search for.
Here is an example of table 1 and table 2:
------------------- Table 1 ----------------------
CREATE TABLE IF NOT EXISTS `db_hint`.`user` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
`fb_id` INT NOT NULL,
`last_logged_in` DATETIME NULL,
`permissions` INT UNSIGNED NOT NULL,
PRIMARY KEY (`id`),
INDEX `permissions_id_idx` (`permissions` ASC),
CONSTRAINT `permissions_id`
FOREIGN KEY (`permissions`)
REFERENCES `db_hint`.`permissions` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
----------------- Table 2 ----------------------
CREATE TABLE IF NOT EXISTS `db_hint`.`user_stat` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
`user_id` INT UNSIGNED NOT NULL,
PRIMARY KEY (`id`),
INDEX `user_id_idx3` (`user_id` ASC),
CONSTRAINT `user_id`
FOREIGN KEY (`user_id`)
REFERENCES `db_hint`.`user` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
When performing any kind of join, does the InnoDB engine use the relationship in any way? Thanks.
The point of declaring the foreign key is to enforce data consistency.
You will still need the JOIN in order to get desired data.
In MySQL foreign keys will improve performance, but don't expect much comparable to indexes.
To do a query involving two tables, you need JOIN ... ON ... to say how they are related. FOREIGN KEYs are not involved in a SELECT and has zero impact on performance of SELECT. You do not "have to have" FOREIGN KEYs to perform SELECTs.
A FOREIGN KEY is used during INSERTs (and other writes) to verify that a subsequent JOIN will actually find something in the other table. It is an overhead during the write -- the INSERT actively checks (via an index) that the referenced table has the indicated row.
FOREIGN KEYs may also do a cascading operation. For example, a DELETE can cause another DELETE to happen. I prefer to such take control in my application code.
I am creating a database model with Workbench and create the following table:
CREATE TABLE IF NOT EXISTS `Database`.`table1` (
`idtable1` INT NOT NULL ,
`uniquecolumn` INT NOT NULL ,
PRIMARY KEY (`idtable1`) ,
UNIQUE INDEX `UniqueIndex` (`uniquecolumn` ASC) )
ENGINE = InnoDB
It has a primary key, and a unique key on my second column.
When I create foreign key constraints on them, Workbench automatically adds two indexes:
CREATE TABLE IF NOT EXISTS `Database`.`table1` (
`idtable1` INT NOT NULL ,
`uniquecolumn` INT NOT NULL ,
PRIMARY KEY (`idtable1`) ,
UNIQUE INDEX `UniqueIndex` (`uniquecolumn` ASC) ,
INDEX `FKOne` (`idtable1` ASC) , //here
INDEX `FKTwo` (`uniquecolumn` ASC) , //(I don't want this!)
CONSTRAINT `FKOne`
FOREIGN KEY (`idtable1` )
REFERENCES `Database`.`table2` (`idtable2` )
ON DELETE CASCADE
ON UPDATE CASCADE,
CONSTRAINT `FKTwo`
FOREIGN KEY (`uniquecolumn` )
REFERENCES `Database`.`table2` (`idtable2` )
ON DELETE CASCADE
ON UPDATE CASCADE)
ENGINE = InnoDB
(The above is the forward-engineered script after adding the foreign keys to my model)
I have four indexes now.
This is what the MySQL Reference Manual says:
In the referencing table, there must be an index where the foreign key
columns are listed as the first columns in the same order. Such an
index is created on the referencing table automatically if it does not
exist.
So I understand there is no need to create indexes FKOne and FKTwo, since there are already a Primary Key and a Unique index, on the same columns, in the same order. Yet MySQL Workbench doesn't allow me to delete indexes FKOne and FKTwo. And I think I should be able to do this:
CREATE TABLE IF NOT EXISTS `Database`.`table1` (
`idtable1` INT NOT NULL ,
`uniquecolumn` INT NOT NULL ,
PRIMARY KEY (`idtable1`) ,
UNIQUE INDEX `UniqueIndex` (`uniquecolumn` ASC) ,
CONSTRAINT `FKOne`
FOREIGN KEY (`idtable1` )
REFERENCES `Database`.`table2` (`idtable2` )
ON DELETE CASCADE
ON UPDATE CASCADE,
CONSTRAINT `FKTwo`
FOREIGN KEY (`uniquecolumn` )
REFERENCES `Database`.`table2` (`idtable2` )
ON DELETE CASCADE
ON UPDATE CASCADE)
ENGINE = InnoDB
Am I right? Would this code work? Is there some way to do it with Workbench? (Apart from deleting those two lines at the last moment before forward-engineering).
Or maybe MySQL is smart enough to refrain from creating totally redundant indexes and I don't have to worry about it...?
(I'm assuming this is when defining a model.)
See Bug 53277, where I mention the following obscure workaround:
You start with a foreign key and its corresponding generated index that you want to get rid of. Make sure the key is (at least temporarily) on a single non-unique column. In the Indexes tab, change the Type to UNIQUE. Then go to the Columns tab, where UQ is now checked, and uncheck it. The unwanted index is eliminated!