如何获取最大版本记录？

Question

我有一个表，如下所示：

------------------------------------ 
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 

然后我想要的结果是：

-------------------------- 
FId UId Version 
1 2  2 
1 3  2 
1 4  2 
2 1  1 
2 2  1 

如何基于最大各个FID-UID对“版本”编写查询？

Answer 1

下面给出要求的输出。

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 

Answer 2

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 

Answer 3

这将在SQL 2005和更高的工作：

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] 

Answer 4

是告诉你正确─1,3,2应该出现twice.If你只需要使用一次选择不同

的FOLL查询工作

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