Question

我需要通过查询编写一个包含多个日期和seq列的表的T-SQL组：

DROP TABLE #temp
CREATE TABLE #temp(
       id char(1),
       dt DateTime,
       seq int)

Insert into #temp values('A','2015-03-31 10:00:00',1)
Insert into #temp values('A','2015-08-31 10:00:00',2)
Insert into #temp values('A','2015-03-31 10:00:00',5)
Insert into #temp values('B','2015-09-01 10:00:00',1)
Insert into #temp values('B','2015-09-01 10:00:00',2)

我希望结果只包含项目A，B及其最新日期和对应的序列号，例如：

id  MaxDate CorrespondentSeq
A   2015-08-31 10:00:00.000 2
B   2015-09-01 10:00:00.000 2

我正在尝试（显然是错误的！）：

select id, max(dt) as MaxDate, max(seq) as CorrespondentSeq
from #temp
group by id

返回：

id  MaxDate CorrespondentSeq
A   2015-08-31 10:00:00.000 5  <-- 5 is wrong
B   2015-09-01 10:00:00.000 2

我怎样才能做到这一点？

修改

dt datetime列有重复的值（完全相同的日期！）

我正在使用SQL Server 2005

Answer 1

您可以使用排名子选项来仅获取ID最高排名的条目：

select id, dt, seq 
from (
        select id, dt, seq, rank() over (partition by id order by dt desc, seq desc) as r
        from #temp            
     ) ranked
where r=1;

Answer 2

SELECT ID, DT, SEQ 
FROM (
        SELECT ID, DT, SEQ, Row_Number()
            OVER (PARTITION BY id ORDER BY dt DESC, seq DESC) AS row_number
        FROM temp            
     ) cte
WHERE row_number = 1;

演示：http://www.sqlfiddle.com/#!3/3e3d5/5

Answer 3

Select id, dt, seq
From #temp t
where dt = (Select Max(dt) from #temp
            Where id = t.Id)

如果存在重复的行，则还需要指定查询处理器应使用哪些来确定要返回的重复项。假设您想要seq的最低值，

然后你可以写：

Select id, dt, seq
From #temp t
where dt = (Select Max(dt) from #temp
            Where id = t.Id)
   and seq = (Select Min(Seq) from #temp
              where id = t.Id
                 and dt = t.dt)

Answer 4

通过试验和错误，我可能找到了解决方案，但我不完全确定这是正确的：

select A.id, B.dt, max(B.seq) 
from (select id, max(dt) as maxDt
      from #temp
      group by id) as A
inner join #temp as B on A.id = B.id AND A.maxDt = B.dt
group by A.id, B.dt

SQL Server：分组与相应的行值

4 个答案: