Question

我不确定哪里出错了，但似乎LAST_VALUE函数没有返回所需的结果。我想要做的是如下

Create table #temp(
   a varchar(10),
   b varchar(10),
   c datetime
   )

   insert into #temp
   Values('aa','bbb','2014-10-15 16:39:41.000'),
  ('aa','bbb','2014-10-16 06:00:04.000')

  select a,b,c,
  FIRST_VALUE(c) over (partition by a, b order by c asc) as first_date,
  LAST_VALUE(c) over (partition by a, b order by c asc) as last_date,
  row_number() over (partition by a, b order by c asc) as rn
   from #temp

我得到的结果如下，它具有不同的最后值。

a | b | c | first_date | last_date | RN

aa | bbb | 2014-10-15 16：39：41.000 | 2014-10-15 16：39：41.000 | 2014-10-15 16：39：41.000 | 1

aa | bbb | 2014-10-16 06：00：04.000 | 2014-10-15 16：39：41.000 | 2014-10-16 06：00：04.000 | 2

Answer 1

您需要告诉SQL Server哪些行要包含在窗口中，默认情况下这些功能将是“无限制的预先行和当前行之间的范围”或简写为“ROWS UNBOUNDED PRECEDING”，意味着包括所有行窗口的开始直到当前行。所以知道这一点，以下将会产生你期望的东西。

select a,b,c,
  FIRST_VALUE(c) over (partition by a, b order by c asc 
    ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as first_date,
  LAST_VALUE(c) over (partition by a, b order by c asc 
    ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as last_date,
  row_number() over (partition by a, b order by c asc) as rn
from #temp

PS：这给出了相同的结果，但是更具可读性，可能更快。

select a,b,c,
  min(c) over (partition by a, b ) as first_date,
  max(c) over (partition by a, b) as last_date,
  row_number() over (partition by a, b order by c asc) as rn
 from #temp

sql server 2012中的LAST_VALUE函数

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