I need to ALTER my existing database to add a column. Consequently I also want to update the UNIQUE field to encompass that new column. I'm trying to remove the current index but keep getting the error MySQL Cannot drop index needed in a foreign key constraint
CREATE TABLE mytable_a (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
CREATE TABLE mytable_b (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
CREATE TABLE mytable_c (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
CREATE TABLE `mytable` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`AID` tinyint(5) NOT NULL,
`BID` tinyint(5) NOT NULL,
`CID` tinyint(5) NOT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `AID` (`AID`,`BID`,`CID`),
KEY `BID` (`BID`),
KEY `CID` (`CID`),
CONSTRAINT `mytable_ibfk_1` FOREIGN KEY (`AID`) REFERENCES `mytable_a` (`ID`) ON DELETE CASCADE,
CONSTRAINT `mytable_ibfk_2` FOREIGN KEY (`BID`) REFERENCES `mytable_b` (`ID`) ON DELETE CASCADE,
CONSTRAINT `mytable_ibfk_3` FOREIGN KEY (`CID`) REFERENCES `mytable_c` (`ID`) ON DELETE CASCADE
) ENGINE=InnoDB;
mysql> ALTER TABLE mytable DROP INDEX AID;
ERROR 1553 (HY000): Cannot drop index 'AID': needed in a foreign key constraint
You have to drop the foreign key. Foreign keys in MySQL automatically create an index on the table (There was a SO Question on the topic).
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_1 ;
Step 1
List foreign key ( NOTE that its different from index name )
SHOW CREATE TABLE <Table Name>
The result will show you the foreign key name.
Format:
CONSTRAINT `FOREIGN_KEY_NAME` FOREIGN KEY (`FOREIGN_KEY_COLUMN`) REFERENCES `FOREIGN_KEY_TABLE` (`id`),
Step 2
Drop (Foreign/primary/key) Key
ALTER TABLE <Table Name> DROP FOREIGN KEY <Foreign key name>
Step 3
Drop the index.
If you mean that you can do this:
CREATE TABLE mytable_d (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
ALTER TABLE mytable
ADD COLUMN DID tinyint(5) NOT NULL,
ADD CONSTRAINT mytable_ibfk_4
FOREIGN KEY (DID)
REFERENCES mytable_d (ID) ON DELETE CASCADE;
> OK.
But then:
ALTER TABLE mytable
DROP KEY AID ;
gives error.
You can drop the index and create a new one in one ALTER TABLE statement:
ALTER TABLE mytable
DROP KEY AID ,
ADD UNIQUE KEY AID (AID, BID, CID, DID);
A foreign key always requires an index. Without an index enforcing the constraint would require a full table scan on the referenced table for every inserted or updated key in the referencing table. And that would have an unacceptable performance impact.
This has the following 2 consequences:
When creating a foreign key, the database checks if an index exists. If not an index will be created. By default, it will have the same name as the constraint.
When there is only one index that can be used for the foreign key, it can't be dropped. If you really wan't to drop it, you either have to drop the foreign key constraint or to create another index for it first.
Because you have to have an index on a foreign key field you can just create a simple index on the field 'AID'
CREATE INDEX aid_index ON mytable (AID);
and only then drop the unique index 'AID'
ALTER TABLE mytable DROP INDEX AID;
I think this is easy way to drop the index.
set FOREIGN_KEY_CHECKS=0; //disable checks
ALTER TABLE mytable DROP INDEX AID;
set FOREIGN_KEY_CHECKS=1; //enable checks
drop the index and the foreign_key in the same query like below
ALTER TABLE `your_table_name` DROP FOREIGN KEY `your_index`;
ALTER TABLE `your_table_name` DROP COLUMN `your_foreign_key_id`;
Dropping FK is tedious and risky. Simply create the new index with new columns and new index name, such as AID2. After the new Unique Index is created, you can drop the old one with no issue. Or you can use the solution given above to incorporate the "drop index, add unique index" in the same alter table command. Both solutions will work
In my case I dropped the foreign key and I still could not drop the index. That was because there was yet another table that had a foreign key to this table on the same fields. After I dropped the foreign key on the other table I could drop the indexes on this table.
If you are using PhpMyAdmin sometimes it don't show the foreign key to delete.
The error code gives us the name of the foreign key and the table where it was defined, so the code is:
ALTER TABLE your_table DROP FOREIGN KEY foreign_key_name;
You can show Relation view in phpMyAdmin and first delete foreign key. After this you can remove index.
You can easily check it with DBeaver. Example:
As you can see there are 3 FKs but only 2 FK indexes. There is no index for FK_benefCompanyNumber_beneficiaries_benefId as UK index provide uniqueness for that FK.
To drop that UK you need to:
DROP FK_benefCompanyNumber_beneficiaries_benefId
DROP UK
CREATE FK_benefCompanyNumber_beneficiaries_benefId
The current most upvoted answer is not complete.
One needs to remove all the foreign keys whose "source" column is also present in the UNIQUE KEY declaration.
So in this case, it is not enough to remove mytable_ibfk_1 for the error to go away, mytable_ibfk_2 and mytable_ibfk_3 must be deleted as well.
This is the complete answer:
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_1;
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_2;
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_3;
Its late now but I found a solution which might help somebody in future.
Just go to table's structure and drop foreign key from foreign keys list. Now you will be able to delete that column.
Related
Some how, my database has gotten into a bad state. I previously had a table named live_stream. When I tried to drop a foreign key constraint, I got an error that mariadb could not rename #sql-26_e7a to live_stream. Now when I try to run the following statement, I get this error.
Can't create table live_stream (errno: 150 "Foreign key constraint is incorrectly formed")
CREATE TABLE live_stream
(idbigint(20) NOT NULL PRIMARY KEY);
As you can see I don't have any foreign key constraints in the definition. If I try the exact same definition with a different table name, it works. If I try to drop the table, mariadb complains that live_stream doesn't exist. Its like the table or foreign key are stuck in a transaction or something like that.
I am using galara with maria db 10.3.
UPDATE
I believe the problem was introduced when a foreign key and unique index were given the same name. I recreated the scenario, and when I try to drop the index, mariadb prevents it.
* UPDATE 2 *
Here is the output of SHOW ENGINE INNODB STATUS;
* UPDATE3 *
Here are the steps to reproduce.
create table tb1
(
id bigint null,
constraint tb1_pk
primary key (id)
);
create table tb2
(
id bigint null,
tb1_id bigint null,
constraint tb2_pk
primary key (id),
constraint tb2_tb1_id_fk
foreign key (tb1_id) references tb1 (id)
);
ALTER TABLE tb2 ADD CONSTRAINT tb2_tb1_id_fk UNIQUE (tb1_id, tb1_id);
drop index tb2_tb1_id_fk on tb2;
The problem is that the unique constraint has the same name as the foreign key and references the same column twice.
MySQL 5.6.16
Two tables. Altering Table 1 to have a foreign key to Table 2's primary key. SQL Error 1215.
If I drop Table 1 and incorporate the foreign key constraint into the build, it accepts the constraint just fine. Only altering the tables after creation causes a problem.
Any ideas? Below are two attempts at writing the alter statement, followed by the creation script.
ALTER TABLE c_users ADD FOREIGN KEY fk_user_prof_position_tid(professional_position_tid) REFERENCES d_taxonomy(tid);
ALTER TABLE c_users ADD CONSTRAINT fk_user_prof_position_tid FOREIGN KEY (professional_position_tid) REFERENCES d_taxonomy(tid);
CREATE TABLE c_users (
user_id INT(11) NOT NULL AUTO_INCREMENT COMMENT 'Primary, auto-generated key',
professional_position_tid INT(11),
...
PRIMARY KEY (user_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE INDEX i_user_id ON c_users (user_id) USING BTREE;
CREATE TABLE d_taxonimy (
tid INT(11) NOT NULL COMMENT '',
...
PRIMARY KEY (tid)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE INDEX i_tid ON d_taxonimy (tid) USING BTREE;
Typos:
ALTER TABLE [...snip...] REFERENCES d_taxonomy(tid);
^----
CREATE TABLE d_taxonimy (
^----
Plus, if you're running the statements in that order, you can't alter a table which doesn't exist yet, or create a foreign key when the foreign field/table don't exist yet either.
I am trying to delete a multi-column unique key from a table that also has a foreign key. I keep getting 'errno 150', unless I delete the foreign key first.
For example, if I create the table:
CREATE TABLE `testtable` (
`testtable_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`testtable_value` char(255) DEFAULT NULL,
`othertable_id` int(10) unsigned NOT NULL,
PRIMARY KEY (`testtable_id`),
UNIQUE KEY `tt_unique_key` (`othertable_id`,`testtable_value`),
CONSTRAINT `tt_foreign_key` FOREIGN KEY (`othertable_id`) REFERENCES `othertable` (`othertable_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
and I try to remove the unique key like this:
ALTER TABLE `testtable` DROP KEY `tt_unique_key`;
It generates the error:
Error Code: 1025
Error on rename of './testdb/#sql-374_27' to './testdb/testtable' (errno: 150)
I tried setting FOREIGN_KEY_CHECKS = 0, but I get the same error:
SET FOREIGN_KEY_CHECKS = 0;
ALTER TABLE `testtable` DROP KEY `tt_unique_key`;
SET FOREIGN_KEY_CHECKS = 1;
This generates the same error message as above.
However, if I first delete the foreign key, then delete the unique key, then recreate the foreign key, everything works:
ALTER TABLE `testtable` DROP FOREIGN KEY `tt_foreign_key`;
ALTER TABLE `testtable` DROP KEY `tt_unique_key`;
ALTER TABLE `testtable` ADD CONSTRAINT `tt_foreign_key` FOREIGN KEY (`othertable_id`) REFERENCES `othertable` (`othertable_id`);
This seems really inefficient. Can anyone explain what is going on? Is there a way to drop the unique key without dropping the foreign key first?
a FOREIGN KEY REFERENCES, require a key,
the only key that can be used is t_unique_key,
thats why you can't remove it.
so add another matching key first, and then remove the old key, in your case the othertable_id field
ALTER TABLE `testtable`
ADD KEY (othertable_id),
DROP KEY `tt_unique_key`;
I want to add a Foreign Key to a table called "katalog".
ALTER TABLE katalog
ADD CONSTRAINT `fk_katalog_sprache`
FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL;
When I try to do this, I get this error message:
Error Code: 1005. Can't create table 'mytable.#sql-7fb1_7d3a' (errno: 150)
Error in INNODB Status:
120405 14:02:57 Error in foreign key constraint of table
mytable.#sql-7fb1_7d3a:
FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL:
Cannot resolve table name close to:
(`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL
When i use this query it works, but with wrong "on delete" action:
ALTER TABLE `katalog`
ADD FOREIGN KEY (`Sprache` ) REFERENCES `sprache` (`ID` )
Both tables are InnoDB and both fields are "INT(11) not null". I'm using MySQL 5.1.61. Trying to fire this ALTER Query with MySQL Workbench (newest) on a MacBook Pro.
Table Create Statements:
CREATE TABLE `katalog` (
`ID` int(11) unsigned NOT NULL AUTO_INCREMENT,
`Name` varchar(50) COLLATE utf8_unicode_ci NOT NULL,
`AnzahlSeiten` int(4) unsigned NOT NULL,
`Sprache` int(11) NOT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `katalogname_uq` (`Name`)
) ENGINE=InnoDB AUTO_INCREMENT=12 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci ROW_FORMAT=DYNAMIC$$
CREATE TABLE `sprache` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`Bezeichnung` varchar(45) NOT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `Bezeichnung_UNIQUE` (`Bezeichnung`),
KEY `ix_sprache_id` (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8
To add a foreign key (grade_id) to an existing table (users), follow the following steps:
ALTER TABLE users ADD grade_id SMALLINT UNSIGNED NOT NULL DEFAULT 0;
ALTER TABLE users ADD CONSTRAINT fk_grade_id FOREIGN KEY (grade_id) REFERENCES grades(id);
Simply use this query, I have tried it as per my scenario and it works well
ALTER TABLE katalog ADD FOREIGN KEY (`Sprache`) REFERENCES Sprache(`ID`);
Simple Steps...
ALTER TABLE t_name1 ADD FOREIGN KEY (column_name) REFERENCES t_name2(column_name)
FOREIGN KEY (`Sprache`)
REFERENCES `Sprache` (`ID`)
ON DELETE SET NULL
ON UPDATE SET NULL;
But your table has:
CREATE TABLE `katalog` (
`Sprache` int(11) NOT NULL,
It cant set the column Sprache to NULL because it is defined as NOT NULL.
check this link. It has helped me with errno 150:
http://verysimple.com/2006/10/22/mysql-error-number-1005-cant-create-table-mydbsql-328_45frm-errno-150/
On the top of my head two things come to mind.
Is your foreign key index a unique name in the whole database (#3 in the list)?
Are you trying to set the table PK to NULL on update (#5 in the list)?
I'm guessing the problem is with the set NULL on update (if my brains aren't on backwards today as they so often are...).
Edit: I missed the comments on your original post. Unsigned/not unsigned int columns maybe resolved your case. Hope my link helps someone in the future thought.
How to fix Error Code: 1005. Can't create table 'mytable.#sql-7fb1_7d3a' (errno: 150) in mysql.
alter your table and add an index to it..
ALTER TABLE users ADD INDEX index_name (index_column)
Now add the constraint
ALTER TABLE foreign_key_table
ADD CONSTRAINT foreign_key_name FOREIGN KEY (foreign_key_column)
REFERENCES primary_key_table (primary_key_column) ON DELETE NO ACTION
ON UPDATE CASCADE;
Note if you don't add an index it wont work.
After battling with it for about 6 hours I came up with the solution
I hope this save a soul.
MySQL will execute this 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;
Cheers!
When you add a foreign key constraint to a table using ALTER TABLE, remember to create the required indexes first.
Create index
Alter table
try all in one query
ALTER TABLE users ADD grade_id SMALLINT UNSIGNED NOT NULL DEFAULT 0,
ADD CONSTRAINT fk_grade_id FOREIGN KEY (grade_id) REFERENCES grades(id);
step 1: run this script
SET FOREIGN_KEY_CHECKS=0;
step 2: add column
ALTER TABLE mileage_unit ADD COLUMN COMPANY_ID BIGINT(20) NOT NULL
step 3: add foreign key to the added column
ALTER TABLE mileage_unit
ADD FOREIGN KEY (COMPANY_ID) REFERENCES company_mst(COMPANY_ID);
step 4: run this script
SET FOREIGN_KEY_CHECKS=1;
ALTER TABLE child_table_name ADD FOREIGN KEY (child_table_column) REFERENCES parent_table_name(parent_table_column);
child_table_name is that table in which we want to add constraint.
child_table_column is that table column in which we want to add foreign key.
parent table is that table from which we want to take reference.
parent_table_column is column name of the parent table from which we take reference
this is basically happens because your tables are in two different charsets. as a example one table created in charset=utf-8 and other tables is created in CHARSET=latin1 so you want be able add foriegn key to these tables. use same charset in both tables then you will be able to add foriegn keys. error 1005 foriegn key constraint incorrectly formed can resolve from this
The foreign key constraint must be the same data type as the primary key in the reference table and column
ALTER TABLE TABLENAME ADD FOREIGN KEY (Column Name) REFERENCES TableName(column name)
Example:-
ALTER TABLE Department ADD FOREIGN KEY (EmployeeId) REFERENCES Employee(EmployeeId)
i geted through the same problem. I my case the table already have data and there were key in this table that was not present in the reference table. So i had to delete this rows that disrespect the constraints and everything worked.
Double check if the engine and charset of the both tables are the same.
If not, it will show this error.
Here's my table:
CREATE TABLE `alums_alumphoto` (
`id` int(11) NOT NULL auto_increment,
`alum_id` int(11) NOT NULL,
`photo_id` int(11) default NULL,
`media_id` int(11) default NULL,
`updated` datetime NOT NULL,
PRIMARY KEY (`id`),
KEY `alums_alumphoto_alum_id` (`alum_id`),
KEY `alums_alumphoto_photo_id` (`photo_id`),
KEY `alums_alumphoto_media_id` (`media_id`),
CONSTRAINT `alums_alumphoto_ibfk_1` FOREIGN KEY (`media_id`) REFERENCES `media_mediaitem` (`id`),
CONSTRAINT `alum_id_refs_id_706915ea` FOREIGN KEY (`alum_id`) REFERENCES `alums_alum` (`id`),
CONSTRAINT `photo_id_refs_id_63282119` FOREIGN KEY (`photo_id`) REFERENCES `media_mediaitem` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=63 DEFAULT CHARSET=utf8
I want to delete the column photo_id, which presumably will also require deleting the foreign key constraint and the index.
The problem is that I get errors when I try to drop the column:
ERROR 1025 (HY000): Error on rename of '.\dbname\#sql-670_c5c' to '.\dbname\alums_alumphoto' (errno: 150)
... when I try to drop the index (same as above), and when I try to drop the foreign key constraint:
ERROR 1091 (42000): Can't DROP 'photo_id_refs_id_63282119'; check that column/key exists)
What order should I be doing all of this in? What precise commands should I be using?
Precisely, try this :
First drop the Foreign Key or Constraint :
ALTER TABLE `alums_alumphoto` DROP FOREIGN KEY `photo_id_refs_id_63282119`;
The previous command removes the Foreign Key Constraint on the column. Now you can drop the column photo_id (the index is removed by MySQL on dropping the column) :
ALTER TABLE `alums_alumphoto` DROP COLUMN `photo_id`;
Aternatively, you could combine these 2 operations into one :
ALTER TABLE `alums_alumphoto`
DROP FOREIGN KEY `photo_id_refs_id_63282119` ,
DROP COLUMN `photo_id`;
The sure thing is to make a duplicate table.
> CREATE TABLE alums_alumphoto_new LIKE alums_alumphoto;
> ALTER TABLE .... // Drop constraint
> ALTER TABLE .... // Drop KEY
> ALTER TABLE .... // Drop the column
> INSERT INTO alums_alumphoto_new (SELECT id, alum_id, photo_id, media_id, updated FROM alums_alumphoto);
> RENAME TABLE alums_alumphoto TO alums_alumphoto_old, alums_alumphoto_new TO alums_alumphoto;
If there's an error executing RENAME TABLE, some other tables might have foreign key constraints referencing this table, in which case this whole approach is stupid. :)
Try combining the DROP KEY and DROP FOREIGN KEY statements.
ALTER TABLE `alums_alumphoto`
DROP KEY KEY `alums_alumphoto_photo_id`,
DROP FOREIGN KEY `photo_id_refs_id_63282119`;
ALTER TABLE `alums_alumphoto`
DROP COLUMN `photo_id`;