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"Keith Boulton" <kboulton_at_ntlworld.com> wrote in message news:iH_38.17847$Ph2.2913491_at_news2-win.server.ntlworld.com...
> select a.a, a.b, a.c, a.d, a.e, sum(b.x)
> from a, b
> where a.a = b.a
> group by a.a, a.b, a.c, a.d, a.e
select a.a, a.b, a.c, a.d, a.e, sum(b.x)
from a, b
where a.a = b.a
or better
select sum(b.x) from b
where b.a = ( select a.a from a )
Received on Fri Jan 25 2002 - 03:34:18 CST