Hi there,
here a simple sample (basicall taken from the documents) with 3 tables.
2 points to watch out for:
1) first declare the primary keys
here (p1.id and p2.id)
2) index the foreign key column
here (child.id1 and child.id2)
mysql> CREATE TABLE p1(id INT PRIMARY KEY)TYPE=INNODB;
Query OK, 0 rows affected (0.03 sec)
mysql> CREATE TABLE p2(id INT PRIMARY KEY)TYPE=INNODB;
Query OK, 0 rows affected (0.01 sec)
mysql> CREATE TABLE child(
-> id1 INT,
-> id2 INT,
-> INDEX(id1),
-> INDEX(id2),
-> FOREIGN KEY (id1) REFERENCES p1(id),
-> FOREIGN KEY (id2) REFERENCES p2(id))TYPE=InnoDB;
Query OK, 0 rows affected (0.01 sec)
In you case you haven't indexed IDTr
Create table y (
IDTr INT NOT NULL,
INDEX (IDTr)
constraint FOREIGN KEY IDTr REFERENCES x(ID)
)Type=Innodb;
That should work. Best regards
Nils Valentin
Tokyo/Japan
2003年 8月 12日 火曜日 04:31、b b さんは書きました:
> I am using MYSQL 4. I understand that it allows for
> foreign keys. Could someone show me an example of how
> to declare a foriegn key. I tried a combination of
> statements but I always got a syntax error.
>
> Here is what I am trying for example ...
>
> Create table x (
> ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
> whatever int
>
> )Type=Innodb;
>
>
> Create table y (
> IDTr INT NOT NULL,
> constraint FOREIGN KEY IDTr REFERENCES x(ID)
> )Type=Innodb;
>
> How would I create a foreign key linking IDTr to
> x(ID)?
>
> Cheers.
>
> Cheers.
>
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Internet Technology
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