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. > > __________________________________ > Do you Yahoo!? > Yahoo! SiteBuilder - Free, easy-to-use web site design software > http://sitebuilder.yahoo.com -- --- Valentin Nils Internet Technology E-Mail: [EMAIL PROTECTED] URL: http://www.knowd.co.jp Personal URL: http://www.knowd.co.jp/staff/nils -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]