Just a note...
> Thanks - this did not work for me as I am on 4.0.17 - presumably this works > on 4.1 (seems to need the SubQuery feature)? If so I will upgrade > immediately! This isn't a subquery -- this is a Derived Table. With regards, Martijn Tonies Database Workbench - developer tool for InterBase, Firebird, MySQL & MS SQL Server. Upscene Productions http://www.upscene.com > >Following query does what you want: > > > >SELECT COUNT(*) from (SELECT COUNT(*) as c FROM pet GROUP BY owner HAVING > >c>1) as temp > > > >-Yayati -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
