I have got 3 different tables:
review table-> this has mov_id as the primary key.
users_table -> this has username as the primary key.
review_table -> can I make a composite primary key for this table from movie_id and username?
users table
reviews table
comments table
I can't connect these 3 tables and I can't figure out why.
This is the error I'm getting while creating the comments table
You can create composite primary key with foreign keys (primary keys from others tables), you have nothing special to do. Just constrain the foreign keys and declare them as primary.
With the names of tables and columns you provided, here's a query to create the table you want :
CREATE TABLE review_table (
movie_id INT,
username VARCHAR(255),
-- Change INT and VARCHAR(255) by what you used on the others tables
PRIMARY KEY (movie_id, username),
FOREIGN KEY (movie_id) REFERENCES movies_table(movie_id),
FOREIGN KEY (username) REFERENCES users_table(username)
)
I am trying to create a table in MySQL with two foreign keys, which reference the primary keys in 2 other tables, but I am getting an errno: 150 error and it will not create the table.
Here is the SQL for all 3 tables:
CREATE TABLE role_groups (
`role_group_id` int(11) NOT NULL `AUTO_INCREMENT`,
`name` varchar(20),
`description` varchar(200),
PRIMARY KEY (`role_group_id`)
) ENGINE=InnoDB;
CREATE TABLE IF NOT EXISTS `roles` (
`role_id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50),
`description` varchar(200),
PRIMARY KEY (`role_id`)
) ENGINE=InnoDB;
create table role_map (
`role_map_id` int not null `auto_increment`,
`role_id` int not null,
`role_group_id` int not null,
primary key(`role_map_id`),
foreign key(`role_id`) references roles(`role_id`),
foreign key(`role_group_id`) references role_groups(`role_group_id`)
) engine=InnoDB;
These conditions must be satisfied to not get error 150 re ALTER TABLE ADD FOREIGN KEY:
The Parent table must exist before you define a foreign key to reference it. You must define the tables in the right order: Parent table first, then the Child table. If both tables references each other, you must create one table without FK constraints, then create the second table, then add the FK constraint to the first table with ALTER TABLE.
The two tables must both support foreign key constraints, i.e. ENGINE=InnoDB. Other storage engines silently ignore foreign key definitions, so they return no error or warning, but the FK constraint is not saved.
The referenced columns in the Parent table must be the left-most columns of a key. Best if the key in the Parent is PRIMARY KEY or UNIQUE KEY.
The FK definition must reference the PK column(s) in the same order as the PK definition. For example, if the FK REFERENCES Parent(a,b,c) then the Parent's PK must not be defined on columns in order (a,c,b).
The PK column(s) in the Parent table must be the same data type as the FK column(s) in the Child table. For example, if a PK column in the Parent table is UNSIGNED, be sure to define UNSIGNED for the corresponding column in the Child table field.
Exception: length of strings may be different. For example, VARCHAR(10) can reference VARCHAR(20) or vice versa.
Any string-type FK column(s) must have the same character set and collation as the corresponding PK column(s).
If there is data already in the Child table, every value in the FK column(s) must match a value in the Parent table PK column(s). Check this with a query like:
SELECT COUNT(*) FROM Child LEFT OUTER JOIN Parent ON Child.FK = Parent.PK
WHERE Parent.PK IS NULL;
This must return zero (0) unmatched values. Obviously, this query is an generic example; you must substitute your table names and column names.
Neither the Parent table nor the Child table can be a TEMPORARY table.
Neither the Parent table nor the Child table can be a PARTITIONED table.
If you declare a FK with the ON DELETE SET NULL option, then the FK column(s) must be nullable.
If you declare a constraint name for a foreign key, the constraint name must be unique in the whole schema, not only in the table in which the constraint is defined. Two tables may not have their own constraint with the same name.
If there are any other FK's in other tables pointing at the same field you are attempting to create the new FK for, and they are malformed (i.e. different collation), they will need to be made consistent first. This may be a result of past changes where SET FOREIGN_KEY_CHECKS = 0; was utilized with an inconsistent relationship defined by mistake. See #andrewdotn's answer below for instructions on how to identify these problem FK's.
MySQL’s generic “errno 150” message “means that a foreign key constraint was not correctly formed.” As you probably already know if you are reading this page, the generic “errno: 150” error message is really unhelpful. However:
You can get the actual error message by running SHOW ENGINE INNODB STATUS; and then looking for LATEST FOREIGN KEY ERROR in the output.
For example, this attempt to create a foreign key constraint:
CREATE TABLE t1
(id INTEGER);
CREATE TABLE t2
(t1_id INTEGER,
CONSTRAINT FOREIGN KEY (t1_id) REFERENCES t1 (id));
fails with the error Can't create table 'test.t2' (errno: 150). That doesn’t tell anyone anything useful other than that it’s a foreign key problem. But run SHOW ENGINE INNODB STATUS; and it will say:
------------------------
LATEST FOREIGN KEY ERROR
------------------------
130811 23:36:38 Error in foreign key constraint of table test/t2:
FOREIGN KEY (t1_id) REFERENCES t1 (id)):
Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
It says that the problem is it can’t find an index. SHOW INDEX FROM t1 shows that there aren’t any indexes at all for table t1. Fix that by, say, defining a primary key on t1, and the foreign key constraint will be created successfully.
Make sure that the properties of the two fields you are trying to link with a constraint are exactly the same.
Often, the 'unsigned' property on an ID column will catch you out.
ALTER TABLE `dbname`.`tablename` CHANGE `fieldname` `fieldname` int(10) UNSIGNED NULL;
What's the current state of your database when you run this script? Is it completely empty? Your SQL runs fine for me when creating a database from scratch, but errno 150 usually has to do with dropping & recreating tables that are part of a foreign key. I'm getting the feeling you're not working with a 100% fresh and new database.
If you're erroring out when "source"-ing your SQL file, you should be able to run the command "SHOW ENGINE INNODB STATUS" from the MySQL prompt immediately after the "source" command to see more detailed error info.
You may want to check out the manual entry too:
If you re-create a table that was dropped, it must have a definition that conforms to the foreign key constraints referencing it. It must have the right column names and types, and it must have indexes on the referenced keys, as stated earlier. If these are not satisfied, MySQL returns error number 1005 and refers to error 150 in the error message. If MySQL reports an error number 1005 from a CREATE TABLE statement, and the error message refers to error 150, table creation failed because a foreign key constraint was not correctly formed.
— MySQL 5.1 reference manual.
For people who are viewing this thread with the same problem:
There are a lot of reasons for getting errors like this. For a fairly complete list of causes and solutions of foreign key errors in MySQL (including those discussed here), check out this link:
MySQL Foreign Key Errors and Errno 150
For others that find this SO entry via Google: Be sure that you aren't trying to do a SET NULL action on a foreign key (to be) column defined as "NOT NULL." That caused great frustration until I remembered to do a CHECK ENGINE INNODB STATUS.
Definitely it is not the case but I found this mistake pretty common and unobvious. The target of a FOREIGN KEY could be not PRIMARY KEY. Te answer which become useful for me is:
A FOREIGN KEY always must be pointed to a PRIMARY KEY true field of other table.
CREATE TABLE users(
id INT AUTO_INCREMENT PRIMARY KEY,
username VARCHAR(40));
CREATE TABLE userroles(
id INT AUTO_INCREMENT PRIMARY KEY,
user_id INT NOT NULL,
FOREIGN KEY(user_id) REFERENCES users(id));
As pointed by #andrewdotn the best way is to see the detailed error(SHOW ENGINE INNODB STATUS;) instead of just an error code.
One of the reasons could be that an index already exists with the same name, may be in another table. As a practice, I recommend prefixing table name before the index name to avoid such collisions. e.g. instead of idx_userId use idx_userActionMapping_userId.
Please make sure at first that
you are using InnoDB tables.
field for FOREIGN KEY has the same type and length (!) as source field.
I had the same trouble and I've fixed it. I had unsigned INT for one field and just integer for other field.
Helpful tip, use SHOW WARNINGS; after trying your CREATE query and you will receive the error as well as the more detailed warning:
---------------------------------------------------------------------------------------------------------+
| Level | Code | Message |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
| Warning | 150 | Create table 'fakeDatabase/exampleTable' with foreign key constraint failed. There is no index in the referenced table where the referenced columns appear as the first columns.
|
| Error | 1005 | Can't create table 'exampleTable' (errno:150) |
+---------+------+-------------------------------------------------------------------------- -------------------------------------------------------------------------------------------- ---------------+
So in this case, time to re-create my table!
This is usually happening when you try to source file into existing database.
Drop all the tables first (or the DB itself).
And then source file with SET foreign_key_checks = 0; at the beginning and SET foreign_key_checks = 1; at the end.
I've found another reason this fails... case sensitive table names.
For this table definition
CREATE TABLE user (
userId int PRIMARY KEY AUTO_INCREMENT,
username varchar(30) NOT NULL
) ENGINE=InnoDB;
This table definition works
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES **u**ser(userId)
) ENGINE=InnoDB;
whereas this one fails
CREATE TABLE product (
id int PRIMARY KEY AUTO_INCREMENT,
userId int,
FOREIGN KEY fkProductUser1(userId) REFERENCES User(userId)
) ENGINE=InnoDB;
The fact that it worked on Windows and failed on Unix took me a couple of hours to figure out. Hope that helps someone else.
MySQL Workbench 6.3 for Mac OS.
Problem: errno 150 on table X when trying to do Forward Engineering on a DB diagram, 20 out of 21 succeeded, 1 failed. If FKs on table X were deleted, the error moved to a different table that wasn't failing before.
Changed all tables engine to myISAM and it worked just fine.
Also worth checking that you aren't accidentally operating on the wrong database. This error will occur if the foreign table does not exist. Why does MySQL have to be so cryptic?
Make sure that the foreign keys are not listed as unique in the parent. I had this same problem and I solved it by demarcating it as not unique.
In my case it was due to the fact that the field that was a foreign key field had a too long name, ie. foreign key (some_other_table_with_long_name_id). Try sth shorter. Error message is a bit misleading in that case.
Also, as #Jon mentioned earlier - field definitions have to be the same (watch out for unsigned subtype).
(Side notes too big for a Comment)
There is no need for an AUTO_INCREMENT id in a mapping table; get rid of it.
Change the PRIMARY KEY to (role_id, role_group_id) (in either order). This will make accesses faster.
Since you probably want to map both directions, also add an INDEX with those two columns in the opposite order. (There is no need to make it UNIQUE.)
More tips: http://mysql.rjweb.org/doc.php/index_cookbook_mysql#speeding_up_wp_postmeta
When the foreign key constraint is based on varchar type, then in addition to the list provided by marv-el the target column must have an unique constraint.
execute below line before creating table :
SET FOREIGN_KEY_CHECKS = 0;
FOREIGN_KEY_CHECKS option specifies whether or not to check foreign key constraints for InnoDB tables.
-- Specify to check foreign key constraints (this is the default)
SET FOREIGN_KEY_CHECKS = 1;
-- Do not check foreign key constraints
SET FOREIGN_KEY_CHECKS = 0;
When to Use :
Temporarily disabling referential constraints (set FOREIGN_KEY_CHECKS to 0) is useful when you need to re-create the tables and load data in any parent-child order
I encountered the same problem, but I check find that I hadn't the parent table. So I just edit the parent migration in front of the child migration. Just do it.
I want to make doctorid a foreign key in my patient table.
So I have all of my tables created - the main problem is that when I go to the table > structure > relation view only the primary key comes up that I can create a foreign key (and it is already the primary key of the certain table that I want to keep - i.e Patient table patient is enabled to be changed but the doctor Id -I have a doctor table also- is not enabled).
I have another table with two composite keys (medicineid and patientid) in relation view it enables me to change both
Do I have to chance the index of doctor ID in patient table to something else? both cannot be primary keys as patient ID is the primary for the patient table - doctor is the foreign.
I hope anyone can help
Kind regards
You can do it the old fashioned way... with an SQL statement that looks something like this
ALTER TABLE table_1_name
ADD CONSTRAINT fk_foreign_key_name
FOREIGN KEY (table_1_column_name)
REFERENCES target_table(target_table_column_name);
For example:
If you have books table with column created_by which refers to column id in users table:
ALTER TABLE books
ADD CONSTRAINT books_FK_1
FOREIGN KEY (created_by)
REFERENCES users(id);
This assumes the keys already exist in the relevant table
The key must be indexed to apply foreign key constraint. To do that follow the steps.
Open table structure. (2nd tab)
See the last column action where multiples action options are there. Click on Index, this will make the column indexed.
Open relation view and add foreign key constraint.
You will be able to assign DOCTOR_ID as foreign now.
To be able to create a relation, the table Storage Engine must be InnoDB. You can edit in Operations tab.
Then, you need to be sure that the id column in your main table has been indexed. It should appear at Index section in Structure tab.
Finally, you could see the option Relations View in Structure tab. When edditing, you will be able to select the parent column in foreign table to create the relation.
See attachments. I hope this could be useful for anyone.
Create a categories table:
CREATE TABLE categories(
cat_id int not null auto_increment primary key,
cat_name varchar(255) not null,
cat_description text
) ENGINE=InnoDB;
Create a products table and reference categories table:
CREATE TABLE products(
prd_id int not null auto_increment primary key,
prd_name varchar(355) not null,
prd_price decimal,
cat_id int not null,
FOREIGN KEY fk_cat(cat_id)
REFERENCES categories(cat_id)
ON UPDATE CASCADE
ON DELETE RESTRICT
)ENGINE=InnoDB;
Create a vendors table and modify products table:
CREATE TABLE vendors(
vdr_id int not null auto_increment primary key,
vdr_name varchar(255)
)ENGINE=InnoDB;
ALTER TABLE products
ADD COLUMN vdr_id int not null AFTER cat_id;
To add a foreign key (referencing vendors table) to the products table, you use the following statement:
ALTER TABLE products
ADD FOREIGN KEY fk_vendor(vdr_id)
REFERENCES vendors(vdr_id)
ON DELETE NO ACTION
ON UPDATE CASCADE;
If you wish to drop that key then:
ALTER TABLE table_name
DROP FOREIGN KEY constraint_name;
In phpmyadmin, Go to Structure tab, select Relation view as shown in image below.
Here you will find the tabular form to add foreign key constrain name, current table column, foreign key database, table and column
I'm trying to create a database in MySQL on phpMyAdmin. I am able to create the tables without any trouble, but I also want to add some foreign keys. In this case I want to link the BIDS and CLIENTS tables via the CLIENTID attribute.
CREATE TABLE BIDS (
BIDID NUMERIC(3) NOT NULL PRIMARY KEY,
CLIENTID NUMERIC(3) NOT NULL
);
CREATE TABLE CLIENTS (
CLIENTID NUMERIC(3) NOT NULL,
EMAILADDRESSES VARCHAR(100) NOT NULL,
PHONENUMBERS VARCHAR(11) NOT NULL,
FOREIGN KEY (CLIENTID) REFERENCES BIDS (CLIENTID),
PRIMARY KEY (CLIENTID,EMAILADDRESSES,PHONENUMBERS)
);
Research has told me that the syntax is correct, but this code returns the following error.
1005 - Can't create table 'CLIENTS' (errno: 150)
Apparently, a solution might be involved with something called 'InnoDB'. How can I use it to fix my problem?
Syntax is fine but problem is with FORIEGN KEY statement as below. You can't create FK on a non-key column. In BIDS table it's BIDID which is defined as Primary Key and not CLIENTID
FOREIGN KEY (CLIENTID) REFERENCES BIDS (CLIENTID)
So, your FORIEGN KEY definition should actually be
FOREIGN KEY (CLIENTID) REFERENCES BIDS (BIDID)
See a demo here http://sqlfiddle.com/#!2/f1c9ec
i've tried to set the primary and foreign key using the method that i learn at http://fellowtuts.com/php/setting-up-foreign-key-in-phpmyadmin/ but an error came up stating that
#1025 - Error on rename of '.\sistem_akaun\#sql-1b70_7d' to '.\sistem_akaun\detail_akaun' (errno: 150 - Foreign key constraint is incorrectly formed)
can i know what's the problem here?sorry if this question sounds stupid,just a newbie
Check to make sure that the Primary Key you are referencing exists. If, in your main table, you have id_main = 0 on your main table, where id_main is a foreign key referencing id_ref (which is the primary key of the other table) but you have reference ref_id = 1 and no 0 value, you will get an error.
Check to make sure your foreign key is the primary key of the other table.
Check to make sure they are the same data type, length, unsigned status. Sometimes these matter sometimes not.
Sometimes I've had trouble where both Foreign Key and Primary Key are both named "id". This can be a problem depending on what software/methods you are using.
You may find it easier to specify a foreign key manually, in SQL.
ALTER TABLE Table
ADD FOREIGN KEY (Column)
REFERENCES TableToReference (ColumnToReference)
(Where Table is the table you wish to add a foreign key to)
#itsfawwaz, You can also do this by below way. Check below example.
Example : (Table Orders)
CREATE TABLE Orders
(
O_Id int NOT NULL,
OrderNo int NOT NULL,
P_Id int,
PRIMARY KEY (O_Id),
CONSTRAINT fk_PerOrders FOREIGN KEY (P_Id)
REFERENCES Persons(P_Id)
)
Above is sample example, you can use your own table fields !
Let me know if still you have any issues.