Question

我的表格的每一行都包含一个引用和一个日期。我试图找到一个日期和前一个日期之间的天数。

让我们来看下面的例子：

CREATE TABLE #Test
(   
    Ref NVARCHAR(100) NOT NULL,
    Dat DATETIME NOT NULL
)

INSERT INTO #Test ( Ref, Dat )
VALUES  ( N'ABC123', GETDATE() ), 
        ( N'DHI458', GETDATE() + 1 ),
        ( N'HGYU556', GETDATE() - 3 ),
        ( N'UUU021', GETDATE() + 17 )

我想得到类似的东西：

UUU021  2017-01-01 17:27:00.720  17 
DHI458  2016-12-16 17:27:00.720  1
ABC123  2016-12-15 17:27:00.720  3
HGYU556 2016-12-12 17:27:00.720  NULL

在一个选择中。可能吗？我正在使用SQL Server 2008.

Answer 1

SQL Server 2012+支持lag()。在SQL Server 2008中，您有几个选项 - 没有一个真正好。我会选择outer apply：

select t.*,
       datediff(day, tprev.date, t.date) as diff
from #test t outer apply
     (select top 1 t2.*
      from #test t2
      where t2.date < t.date
      order by t2.date desc
     ) tprev;

Answer 2

戈登的方法很扎实，但这是另一个

with CTE as
(
select Ref, Dat, row_number() over (order by Dat desc) as DatO
from #Test
)
select A1.Ref, A1.Dat, A1.Dat-A2.Dat as Diff
from CTE A1
left join CTE A2
    on A1.DatO = A2.DatO - 1;

Answer 3

我通常喜欢自我加入，直到有效的2个表被1行偏移：

select t1.Ref, t1.Dat. datediff(dd, t1.Dat, t2.Dat) as diff
from #Test as t1
inner join #Test as t2
on t2.Dat > t1.Dat
and not exists (
    select 1
    from #Test as t3
    where t3.Dat > t1.Dat
    and t3.Dat < t2.Dat
    )
order by t1.Dat desc

两个连续日期之间的差异，存储在单独的行中

3 个答案: