两个连续日期之间的差异，存储在不同的行中

Question

我的表格的每一行都包含一个引用和一个日期。我试图找出一个日期和前一个日期之间的天数。

让我们在下面的示例工作：

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) 

我想要得到的东西，如：

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 

在一个单一的选择。可能吗？我正在使用SQL Server 2008.

Answer 1

SQL Server 2012+支持lag()。在SQL Server 2008中，你有几个选项 - 没有一个真的很好。我会去与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

戈登的做法是固体，但这里的另一个

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

我一般喜欢自联接，直到你有有效的2台由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