Reputation: 25
I have a table like the following:
------------------------------------
Id FId UId Version
1 1 1 1
2 1 2 1
3 1 3 1
4 1 2 2
5 1 3 2
6 1 3 2
7 1 4 2
8 2 1 1
9 2 2 1
then I want the result to be:
--------------------------
FId UId Version
1 2 2
1 3 2
1 4 2
2 1 1
2 2 1
How to write the query based on the max 'Version' of each FId-UId pair?
Upvotes: 1
Views: 7987
Reputation: 26782
select FId, UId, Version
from MyTable
join (select Fid, Max(Version) as MaxVersion group by Fid) x
on x.FId = MyTable.FId and x.MaxVersion = MyTable.Version
Upvotes: 1
Reputation: 9709
Is the result you show correct- 1,3,2 should appear twice.If you need only once use select distinct
The foll query is working
with t as(
select 1 as id, 1 as fid , 1 as uid, 1 as version union all
select 2 , 1 , 2 , 1 union all
select 3 , 1 , 3 , 1 union all
select 4 , 1 , 2 , 2 union all
select 5 , 1 , 3 , 2 union all
select 6 , 1 , 3 , 2 union all
select 7 , 1 , 4 , 2 union all
select 8 , 2 , 1 , 1 union all
select 9 , 2 , 2 , 1)
select distinct t.fid,t.uid,t.version from t
inner join(
select fid,max(version) as maxversion from t
group by fid)as grp
on t.fid=grp.fid
and t.version=grp.maxversion
Upvotes: 0
Reputation: 5357
The following gives the output requested.
select distinct t2.FId, t2.UId, t2.Version
from
(
select FId, max(Version) as "Version"
from MyTable
group by FId
) t1
inner join MyTable t2 on (t1.FId = t2.FId and t1.Version = t2.Version)
order by t2.FId, t2.UId
Upvotes: 4
Reputation: 21505
This will work on SQL 2005 and later:
DECLARE @t TABLE
(Id INT,
Fid INT,
[uid] INT,
[VERSION] INT
)
INSERT @t
SELECT 1,1,1,1
UNION ALL SELECT 2,1,2,1
UNION ALL SELECT 3,1,3,1
UNION ALL SELECT 4,1,2,2
UNION ALL SELECT 5,1,3,2
UNION ALL SELECT 6,1,3,2
UNION ALL SELECT 7,1,4,2
UNION ALL SELECT 8,2,1,1
UNION ALL SELECT 9,2,2,1
;WITH myCTE
AS
(
SELECT *,
RANK() OVER (PARTITION BY Fid
ORDER BY [VERSION] DESC
) AS rnk
FROM @t
)
SELECT DISTINCT Fid, [uid],[VERSION]
FROM myCTE
WHERE rnk = 1
ORDER BY Fid, [uid]
Upvotes: 2