Re: Query Help Needed

From: Matthias Gresz <nospam_at_newsranger.com>
Date: Fri, 07 Sep 2001 08:16:12 GMT
Message-ID: <gn%l7.7656$4z.32485_at_www.newsranger.com>


Hi Greg,

the second issue may be solved by something like:

Select
e.name,
e.empno
from
emp e,
(

select
empno
from
(select

count(deptno) numdept,
empno
from
yourtable
group by
emp,deptno
) a
where
a.numdept > 1
) b
where
e.empno=b.empno
;

hth

Matthias Received on Fri Sep 07 2001 - 10:16:12 CEST

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