BY AID
HAVING count(distinct BID) =2
Donna
Peter Brawley <[EMAIL PROTECTED]>
11/28/2006 10:53 AM
Please respond to
[EMAIL PROTECTED]
To
James Northcott / Chief Systems <[EMAIL PROTECTED]>,
"mysql@lists.mysql.com"
cc
Subject
Re: Many-Many relation, matching all
James Northcott / Chief Systems wrote:
>SELECT AID
>FROM AhasB WHERE BID in (1,2)
>GROUP BY AID
>HAVING count(BID) =2
Not quite, since that will catch aid's with two bid=1 rows or bid=2 rows:
SELECT * FROM t;
+--+--+
| i| j|
+--+--+
|1 |4 |
|1 |5 |
|3
Peter Brawley wrote:
>I want to find all A's such that
>they have exactly B's 1 and 2
>SELECT A.ID, group_concat(BID ORDER BY BID) as Bs
>FROM A INNER JOIN AhasB ON A.ID=AID
>GROUP BY A.ID
>HAVING Bs='1,2'
Why the join? Doesn't your ahasb bridge table already incorporate the
join logic? If your
I think this will work:
SELECT AID
FROM AhasB
WHERE BID in (1,2)
GROUP BY AID
HAVING count(BID) =2
Donna
James Northcott / Chief Systems <[EMAIL PROTECTED]>
11/27/2006 04:35 PM
To
mysql@lists.mysql.com
cc
Subject
Many-Many relation, matching all
Hello,
I'm having a conceptual iss
>I want to find all A's such that
>they have exactly B's 1 and 2
>SELECT A.ID, group_concat(BID ORDER BY BID) as Bs
>FROM A INNER JOIN AhasB ON A.ID=AID
>GROUP BY A.ID
>HAVING Bs='1,2'
Why the join? Doesn't your ahasb bridge table already incorporate the
join logic? If your requirement is to ret
