根据时间查询，时间最大的所有数据。该如何处理_Sql Server

根据时间查询，时间最大的所有数据。
比如A表：

name date
A 2012.01
A 2012.02
A 2012.03

B 2012.02
B 2012.03
B 2012.04

C 2012.03
C 2012.04
C 2012.05

最终得到数据为：
A 2012.03
B 2012.04
C 2012.05

得到每个人时间最大的数据。
Sql语句该如何写呢？谢谢。

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SQL code

select a.name,date a.from(select row_number()over(partition by name order by cast(right(date,2) as int) desc) as num,*from tbl)awhere num=1
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SQL codeCREATE TABLE #t(NAME CHAR(2),dates DATE)insert into #tselect 'A', '201201' union allselect 'A', '201202' union allselect 'A', '201203' union allselect 'B', '201202' union allselect 'B', '201203' union allselect 'B', '201204' union allselect 'C', '201203' union allselect 'C', '201204' union allselect 'C', '201205'SELECT NAME,MAX(dates) FROM #tGROUP BY NAME
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SQL codeselect * from A twhere not exists(select 1 from name=t.name and [date]>t.[date])
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SQL code
select * from A t
where not exists(select 1 from name=t.name and [date]>t.[date])