On Mon, 5 Jan 2004, Greg Owen wrote: > I tried (you can laugh here) to do it this way, but failed miserably: > > mysql> select class,count(questnum),count(difficulty='0'), > count(difficulty='1'),count(difficulty='2'), > count(in_use='0'),count(in_use='1') from Questions > group by class; >
You can use SELECT class, COUNT(questnum), SUM(IF(difficulty = 0, 1, 0)), SUM(IF(difficulty = 1, 1, 0)), ... cheers, Tobias -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
