I create a table in mysql using the following script:
CREATE TABLE IF NOT EXISTS users_x_activities(
id int NOT NULL auto_increment,
id_user int unsigned NOT NULL,
id_attivita int unsigned NOT NULL,
PRIMARY KEY (id),
FOREIGN KEY (id_user) REFERENCES utente(id),
FOREIGN KEY (id_attivita) REFERENCES attivita(id)
) ENGINE = INNODB;
When I export the created table from phpMyAdmin, I obtain the following script
CREATE TABLE IF NOT EXISTS `users_x_activities` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`id_user` int(10) unsigned NOT NULL,
`id_attivita` int(10) unsigned NOT NULL,
PRIMARY KEY (`id`),
KEY `id_user` (`id_user`),
KEY `id_attivita` (`id_attivita`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;
So the question are: where is my foreign key constraints? does KEY refer to FK? Seems that the two tables utente and attivita are no longer referenced in the new generated script. where am I doing wrong?
EDIT
In phpMyAdmin, configuring the export of the table I found the option "Display Foreign Key Relationship"
If I flag this option I otain also this code in the script
--
-- RELATIONS FOR TABLE `users_x_activity`:
-- `id_user`
-- `utente` -> `id`
-- `id_attivita`
-- `attivita` -> `id`
--
--
-- Constraints for dumped tables
--
--
-- Constraints for table `users_x_activity`
--
ALTER TABLE `users_x_activity`
ADD CONSTRAINT `users_x_activities_ibfk_1` FOREIGN KEY (`id_user`) REFERENCES `utente` (`id`),
ADD CONSTRAINT `users_x_activities_ibfk_2` FOREIGN KEY (`id_attivita`) REFERENCES `attivita` (`id`);
This means that if I add the option "Display Foreign Key Relationship" I obtain also the FK constrains? in other case not?
So the question are: where is my foreign key constraints?
They are defined in the database. The output from SHOW CREATE TABLE users_x_activities will include the foreign key constraint definitions.
The definitions of the foreign key constraints likely appear in separate ALTER TABLE statements at the end of the generated script.
does KEY refer to FK?
No. KEY id_user (id_user) here refers to an index.
Seems that the two tables utente and attivita are no longer referenced in the new generated script.
Yes, you are correct. The foreign key constraints are not included in the CREATE TABLE statement.
where am I doing wrong?
A MySQL SHOW CREATE TABLE users_x_activities will include the foreign key constraints.
The foreign key constraints are likely included in the script generated by phpMyAdmin, but at the end of the script, in separate ALTER TABLE statements.
There are two type of constraints when you managing your tables with phpmyadmin:
internal: when you set constraints with phpmyadmin designer for example the constraints stored as internal,that will not be included in export.
innoDB: these constraints included in export check out linked video about it
Setting up a foreign key constraint
Follow the following steps :
phpmyadmin configuration
export time customer configuration
Related
I cannot seem to be able to delete primary keys in a table.
All references (FKs) have been removed but it still doesn't let me delete it.
What I'm trying to do is: delete old primary keys to add a new one - but keep the old columns and data (just remove the PK attribute).
What is wrong ?
Table:
CREATE TABLE `employee` (
`User` int(10) unsigned NOT NULL,
`Company` int(10) unsigned NOT NULL,
--unrelated boolean fields
PRIMARY KEY (`User`,`Company`),
KEY `FK_Employee_Company_idx` (`Company`),
CONSTRAINT `FK_Employee_Company` FOREIGN KEY (`Company`) REFERENCES `company` (`ID`) ON DELETE NO ACTION ON UPDATE NO ACTION,
CONSTRAINT `FK_Employee_User` FOREIGN KEY (`User`) REFERENCES `user` (`ID`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
Trying to delete:
alter table Employee
drop primary key;
Issue:
Error 1025: Error on rename of '.\DB_NAME#sql-3640_4' to '.\DB_NAME\employee' (errno: 150 "Foreign key constraint is incorrectly formed") SQL Statement: ALTER TABLE DB_NAME.employee DROP PRIMARY KEY
Nothing references this table anymore. I also checked via statements which select from information_schema.key_column_usage but yields no results.
Wasted the last hours on Google but can't seem to figure it out.
And if that would work, adding a new column:
alter table Employee
add column ID int unsigned not null auto_increment primary key;
The index is still needed for the existing FK constraints.
Adding the following index (first) should satisfy that requirement:
CREATE INDEX xxx ON employee (User, Company);
Test case
I have the following main table:
CREATE TABLE IF NOT EXISTS `table_1` (
`id` BIGINT UNSIGNED NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
And the following table:
CREATE TABLE IF NOT EXISTS `table_2` (
`id_1` BIGINT UNSIGNED NOT NULL,
`id_2` BIGINT UNSIGNED NOT NULL,
PRIMARY KEY (`id_1`,`id_2`),
FOREIGN KEY (`id_1`) REFERENCES table_1(`id`) ON DELETE CASCADE,
FOREIGN KEY (`id_2`) REFERENCES table_2(`id`) ON DELETE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
For some reason when creating the above tables, I get the following index created automatically by MYSQL:
Keyname Type Unique Packed Column Cardinality Collation Null
id_2 BTREE No No id_2 94695 A No
So MYSQL is creating an index on the second column called id_2 in table_2. Strange enough, it's not created on both foreign keys and if I create only 1 foreign key, MYSQL wouldn't create an index like this.
I tried to drop the index and got the following error:
Cannot drop index 'id_2': needed in a foreign key constraint
So why does MYSQL need to create an index like this and why it's created on both keys??
Unlike other databases, MySQL creates indexes for foreign key constraints. As explained in the documentation:
index_name represents a foreign key ID. The index_name value is
ignored if there is already an explicitly defined index on the child
table that can support the foreign key. Otherwise, MySQL implicitly creates a foreign key index that is named according to the following rules:
. . .
In your case, one of the foreign key declarations is handled by the primary key index, because id_1 is the first key in both of them.
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 tried to create a table in MySQL using the CREATE TABLE statement below:
CREATE TABLE `visit` (
`visit_id` int(11) NOT NULL,
`site_id` int(11) DEFAULT NULL,
PRIMARY KEY (`visit_id`),
CONSTRAINT `FK_visit_site` FOREIGN KEY (`site_id`) REFERENCES `site` (`site_id`),
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
I received this error:
ERROR 1005 (HY000): Can't create table 'fooschema.visit' (errno: 121)
I used SHOW ENGINE INNODB STATUS command. This is the error message:
------------------------
LATEST FOREIGN KEY ERROR
------------------------
140222 7:03:17 Error in foreign key constraint creation for table `fooschema`.`visit`.
A foreign key constraint of name `fooschema/FK_visit_site`
already exists. (Note that internally InnoDB adds 'databasename/'
in front of the user-defined constraint name).
Note that InnoDB's FOREIGN KEY system tables store
constraint names as case-insensitive, with the
MySQL standard latin1_swedish_ci collation. If you
create tables or databases whose names differ only in
the character case, then collisions in constraint
names can occur. Workaround: name your constraints
explicitly with unique names.
Then, I used the query below to list all available constraints:
select *
from information_schema.table_constraints
where constraint_schema = 'fooschema'
I didn't see any constraint with name 'FK_visit_site' in the result.
The FK_visit_site constraint was a foreign key constraint of table visit. I dropped the visit table and attempted to recreate it.
Is there a way I can drop this foreign key constraint even when the table it was associated to doesn't exist?
your foreign key already exist , so either drop existed foreign key or rename your second key.
ALTER TABLE `site` DROP FOREIGN KEY `FK_visit_site`;
or rename to other new one.
CREATE TABLE `visit` (
`visit_id` int(11) NOT NULL PRIMARY KEY,
`site_id` int(11) NOT NULL,
CONSTRAINT `FK_visit_site` FOREIGN KEY (`site_id`) REFERENCES `site` (`site_id`),
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
Added PRIMARY KEY to the visit_id line.
Note:
make sure that site_id in the site table have exact same datatype of site_id in visit table.
like that
`site_id` int(11) DEFAULT NULL --//in the `site` table
The two keys you're coupling must have the exact same datatype ( INT NOT NULL), even signedness
AFAIK, you will get this error when you're trying to add a constraint with a name that's already used somewhere else. Means, in your case FK FK_visit_site had already been used before.
If the table you're trying to create includes a foreign key constraint, and you've provided your own name for that constraint, remember that it must be unique within the database.
You can run the below query to find out the same
SELECT
constraint_name,
table_name
FROM
information_schema.table_constraints
WHERE
constraint_type = 'FOREIGN KEY'
AND table_schema = DATABASE()
ORDER BY
constraint_name;
Taken from the post here
http://www.thenoyes.com/littlenoise/?p=81
Try using a different name for your FK like
CREATE TABLE `visit` (
`visit_id` int(11) NOT NULL,
`site_id` int(11) DEFAULT NULL,
PRIMARY KEY (`visit_id`),
CONSTRAINT `FK_visit_site_New` FOREIGN KEY (`site_id`)
REFERENCES `site` (`site_id`),
)
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.