SQL Group By One Field and Order By Another Field - sql-server

Question

The Table:
declare #Table table (
id int,
ticketid int,
sponsor int,
dev int,
qa int,
savedate datetime
)
insert into #Table values (1,100,22,0, 0, '2008-10-29 11:17:59.527')
insert into #Table values (2,100,5,0, 0, '2008-10-29 11:00:37.030')
insert into #Table values (3,101,22,0, 0, '2009-10-29 11:10:27.687')
insert into #Table values (5,101,44,0, 0, '2008-10-31 12:07:52.917')
insert into #Table values (6,101,32,0, 0, '2009-06-30 08:16:12.343')
insert into #Table values (7,101,44,0, 0, '2009-10-31 10:12:11.369')
I'm trying to select the top 1 max of savedate where recordid is a certain record, grouped by sponsor.
My progress:
select max(savedate)
from #Table
where ticketid = 101
group by sponsor
Returns
2009-10-29 11:10:27.687
2009-06-30 08:16:12.343
2009-10-31 10:12:11.370
Close, I'm grouped correctly but I want the top 1 most recent date. So I do this:
select top 1 max(savedate)
from #Table
where ticketid = 101
group by sponsor
Returns
2009-10-29 11:10:27.687
Woohoo, got it, time for a break.. wait.. that's not thie most recent date! Let's try to order these by savedate
select top 1 max(savedate)
from #Table
where ticketid = 101
group by sponsor
order by savedate desc
Oh no! The dreaded:
"Column "#Table.savedate" is invalid
in the ORDER BY clause because it is
not contained in either an aggregate
function or the GROUP BY clause."
But savedate IS aggregated in the select list! How do I do what I want to do?

Answer 1

After typing this thing out and making sure I don't miss a detail, I came up with the answer right at the end. I figured I'd add it anyway so I or anyone else can find it later if they are as short sighted as I was in this instance.
select top 1 max(savedate) as date
from #Table
where ticketid = 101
group by sponsor
order by date desc
max(savedate) is not the same as savedate! Aliasing then refering to the aggregate worked perfectly:
2009-10-31 10:12:11.370
Hope this helps someone.

Answer 2

select top 1 max(savedate)
from #Table
where ticketid = 101
group by sponsor
order by max(savedate) desc

Answer 3

You probably wanted to return the sponsor along with the MAX(savedate).
If you didn't then note that this query:
SELECT MAX(savedate)
FROM #Table
WHERE ticketid = 101
is completely identical to your solution
select top 1 max(savedate) as date
from #Table
where ticketid = 101
group by sponsor
order by date desc
, while being much more efficient.
Alternatively, you can just do the following:
SELECT TOP 1 sponsor, savedate
FROM #Table
WHERE ticketid = 101
ORDER BY
savedate DESC
Update:
This query will return the date when the task was last assigned to a current sponsor:
declare #Table table (
id int PRIMARY KEY,
ticketid int,
sponsor int,
dev int,
qa int,
savedate datetime
)
insert into #Table values (1,100,22,0, 0, '2008-10-29 11:17:59.527')
insert into #Table values (2,100,5,0, 0, '2008-10-29 11:00:37.030')
insert into #Table values (3,101,22,0, 0, '2009-10-29 11:10:27.687')
insert into #Table values (5,101,44,0, 0, '2008-10-31 12:07:52.917')
insert into #Table values (6,101,32,0, 0, '2009-06-30 08:16:12.343')
insert into #Table values (7,101,44,0, 0, '2009-10-30 10:12:11.369')
insert into #Table values (8,101,44,1, 0, '2009-10-31 10:12:11.369')
;WITH rows AS
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY ticketid ORDER BY savedate DESC) AS rn
FROM #Table
)
SELECT rl.sponsor, ro.savedate
FROM rows rl
CROSS APPLY
(
SELECT TOP 1 rc.savedate
FROM rows rc
JOIN rows rn
ON rn.ticketid = rc.ticketid
AND rn.rn = rc.rn + 1
AND rn.sponsor <> rc.sponsor
WHERE rc.ticketid = rl.ticketid
ORDER BY
rc.rn
) ro
WHERE rl.rn = 1
I added one more record for sponsor 44 to the test data.
The query will return the rows 1 and 7, since the sponsor did not change in row 8.