Re: Query help...

2001-10-24 Thread David Hugh-Jones
Try SELECT lastip,username FROM users HAVING count(*) > 1 GROUP BY lastip; (or possibly SELECT lastip,username FROM users GROUP BY lastip HAVING count(*) > 1; ) david On Wednesday 24 October 2001 2:21 pm, David Wolf wrote: > I'm trying to come up with a query to do the following... I'm no

RE: selecting rows where all rows meet a criterion

2001-10-23 Thread David Hugh-Jones
gt; [EMAIL PROTECTED] > > > -Original Message- > From: David Hugh-Jones [mailto:[EMAIL PROTECTED]] > Sent: Tuesday, 23 October, 2001 13:10 > To: [EMAIL PROTECTED] > Subject: selecting rows where all rows meet a criterion > > > > OK, I have a tricky SQL syntax pro

RE: selecting rows where all rows meet a criterion

2001-10-23 Thread David Hugh-Jones
ere students.id=grades.id > and min(grades.grade - students.average_grade) > 0 > group by students.id > > > Tom Haapanen > [EMAIL PROTECTED] > > > -Original Message- > From: David Hugh-Jones [mailto:[EMAIL PROTECTED]] > Sent: Tuesday, 23 October, 2001 13:1

selecting rows where all rows meet a criterion

2001-10-23 Thread David Hugh-Jones
OK, I have a tricky SQL syntax problem I would like to share. For simplicity's sake, I'll use an analogy. I have two tables, of students and grades, each with a "ID" column for use in joins, corresponding to a student's ID. The student table has an "average grade" column which shows that stud

dynamic SET and ENUM

2001-04-30 Thread David Hugh-Jones
, but just wanted to check. sql sql sql. dumb email filter. -- David Hugh-Jones iDomainCart - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive