August 18, 2004 at 2:47 pm
I am trying to retrieve distinct ClientID's that match more than 1 item for example my query is currently
Select distinct ClientID from ClientRegs where RegionID = 1 and RegionID = 28
The Table structure example is
ClientID RegionID
222 1
222 28
444 1
444 44
So I want the above query to return only 222 because it is the only client id that has a record for RegionID 1 and 28.
The above query doesn't work because the RegionID column will never be both 1 and 28 on the same record.
I've also tried
Select distinct ClientID from ClientRegs where RegionID = 1 or RegionID = 28
This query does not work because it will return 444 and 222.
Does anyone know if what I'm trying to do is possible and how it would be done?
Thanks,
Brian
August 18, 2004 at 4:44 pm
Here's a couple of methods:
create table clients (clientid int, regionid int)
insert into clients
select 222,1
union
select 222,28
union
select 444,1
union
select 444,44
select clientid from
clients
where regionid = 28
and clientid in
(
select clientid from clients where regionid = 1
)
select clients.clientid
from clients
inner join
(
select clientid from clients where regionid = 1
) x
on clients.clientid = x.clientid
and regionid = 28
If however, you might have any number of regionid's then you're going to need to use dynamic sql. Let me know and I'll post you a version with dynamic sql - not very fast though.
Dave Hilditch.
August 18, 2004 at 6:35 pm
Here's another version that does not require any joins:
Select ClientID from ClientRegs
where RegionID = 1 or RegionID = 28
group by ClientID
having count(ClientID) > 1
Regards
Phil
The absence of evidence is not evidence of absence.
Martin Rees
You can lead a horse to water, but a pencil must be lead.
Stan Laurel
August 18, 2004 at 6:42 pm
Thanks to everyone that posted. I used
select clientid from
clients
where regionid = 28
and clientid in
(
select clientid from clients where regionid = 1
)
and it works
Enthusiast it seems your suggestion is a simple solution I may try it although I have already wrote the code using the other statement.
Thanks For all the information.
August 19, 2004 at 3:17 am
I like this method! My usual solution to such requirements is to use selfjoins or EXISTS subqueries, but I think in future, at least when doing quick-n-dirty stuff in QA, I will use this method. If we switch to using IN for the (here) RegionID list, this is the quickest method (in typing terms ) to extend:
select ID from Thing where OtherID IN ( 3,4,5 ) -- just type new ones here group by ID having count(ID) > 2 -- and make sure this is the number of OtherIDs to match, less one
August 19, 2004 at 7:08 am
TRY THIS:
SELECT ClientID FROM ClientRegs WHERE RegionID=1 AND ClientID in
(SELECT ClientID FROM ClientRegs WHERE RegionID='2')
IT WORK FINE FORM ME.
August 19, 2004 at 7:15 pm
Yeah, I usually use a self-join for something like this. I tested it out, though, and using the group by costs half of using the self-join.
Nice technique! I would not have expected this result.
create table #clients (clientid int, regionid int)
insert into #clients
select 222,1
union
select 222,28
union
select 444,1
union
select 444,44
--cost = .0931
select c1.ClientID
From #Clients c1
JOIN #Clients c2 on c1.ClientID = c2.ClientID
where c1.RegionID = 1 and c2.RegionID = 28
--cost = .0490
Select ClientID
from #Clients
where RegionID in( 1, 28)
group by ClientID
having count(ClientID) > 1
Signature is NULL
August 19, 2004 at 9:11 pm
If you add a clustered index to the table though....
create clustered index idx_foo on #clients (clientid)
--cost = .0443
select c1.ClientID
From #Clients c1
JOIN #Clients c2 on c1.ClientID = c2.ClientID
where c1.RegionID = 1 and c2.RegionID = 28
--cost = .0376
Select ClientID
from #Clients
where RegionID in( 1, 28)
group by ClientID
having count(ClientID) > 1
Which should mean that the group by is still quicker if you trust Execution Plan cost estimation as a means of measuring the speed something will run at. It's really just a cost for internal use - for true time estimates you should look at the statistics provided by the server.
I ran the statistics and it looks like the group by is still quicker though.
Dave Hilditch.
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