Re: Advanced(?) query question

In article <ZWuab.6834$nU6.1269431_at_twister.nyc.rr.com>, jag_at_acm.org says...
>
> SELECT CNT.country_id
> FROM Countries AS CNT
> INNER JOIN
> Cities AS CIT
> ON CNT.country_id = CIT.country_id
> WHERE CNT.password IS NULL
> GROUP BY CNT.country_id
> HAVING COUNT(*) <= 10

Thanks, it worked out great. Maybe about time to brush up on the GROUP BY / HAVING clauses. =)

\Anders Received on Fri Sep 19 2003 - 20:20:03 CEST