rom: "Mercadante, Thomas F" <[EMAIL PROTECTED]>
> Date: 2003/09/29 Mon PM 12:29:40 EDT
> To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
> Subject: RE: RE: interesting sql question
>
> yeah! I think it *is* homework :)
>
> Tom
>
Title: RE: RE: interesting sql question
yeah! I think it *is* homework :)
Tom
-Original Message-From: Jamadagni, Rajendra
[mailto:[EMAIL PROTECTED]Sent: Monday, September 29, 2003
12:10 PMTo: Multiple recipients of list ORACLE-LSubject:
RE: RE: interesting sql question
you could do this, but i would have concerns over the indexing strategy.
select name
from person,
(select distinct sid, count(*) bid_count
from bids
group by sid
HAVING count(*) = (SELECT COUNT(BOAT_ID FROM BOATS)) bids
where person.sid = bids.sid;
Now yours bids table is an intersect
Title: RE: RE: interesting sql question
Hey ... the question wasn't complete ...
give us the full statement of the question ...
Raj
Rajendra dot Jamadagni at nospamespn dot com
All Views expressed in
a user may request the same boat more than once. not sure that work.
>
> From: "Jamadagni, Rajendra" <[EMAIL PROTECTED]>
> Date: 2003/09/29 Mon AM 10:34:53 EDT
> To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
> Subject: RE: RE: interesting
Title: RE: RE: interesting sql question
Here is an attempt ...
select p.*
from persons p
where sid in
(select sid, count(bid)
from bids
group by sid
having count(sid) = (select count(boad_id) from boats))
/
You wanted to find all persons who have booked all
>
> From: "Stephane Faroult" <[EMAIL PROTECTED]>
> Date: 2003/09/29 Mon AM 09:59:39 EDT
> To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
> Subject: RE: interesting sql question
>
>
>
> >- --- Original Message --- -
> >From: <[EMAIL PROTECTED]>
> >To: Multiple reci
