如何计算结果集中前一行的工作天数？

Question

有人可以帮助我解决这个令人困惑的情况吗？我有一个在SQL Server 2008中生成的表。

Number Date 
------ --------- 
22619 30/06/11 
22617 01/07/11 
22556 03/07/11 
22556 03/07/11 
22560 04/07/11 
22560 05/07/11 
22560 08/07/11 
22561 07/07/11 
22561 17/07/11 
22564 09/07/11 
22564 20/07/11 

我想从以前的角色中为每个数字生成以下相同的日子。

而且我想返回以下内容：

Number Date  Days 
------ ---------- ---- 
22619 30/06/11  0 
22617 01/07/11  0 
22556 03/07/11  0 
22556 03/07/11  0 
22560 04/07/11  0 
22560 05/07/11  1 
22560 08/07/11  3 
22561 07/07/11  0 
22561 17/07/11  10 
22564 09/07/11  0 
22564 20/07/11  11 

任何援助，高度赞赏

Answer 1

我想你想要这样的：

with data as (
    select number = 22619, date = convert(datetime, '20110630') union all 
    select 22617, convert(datetime, '20110701') union all 
    select 22556, convert(datetime, '20110703') union all 
    select 22556, convert(datetime, '20110703') union all 
    select 22560, convert(datetime, '20110704') union all 
    select 22560, convert(datetime, '20110705') union all 
    select 22560, convert(datetime, '20110708') union all 
    select 22561, convert(datetime, '20110707') union all 
    select 22561, convert(datetime, '20110717') union all 
    select 22564, convert(datetime, '20110709') union all 
    select 22564, convert(datetime, '20110720') 
) 
select next.number, next.date, days = isnull(DATEDIFF(dd, prev.date, next.date),0) 
from (
    select *, sequence_no = ROW_NUMBER() OVER(PARTITION BY number ORDER BY date ASC) 
    from data 
) next 
left join (
    select *, sequence_no = ROW_NUMBER() OVER(PARTITION BY number ORDER BY date ASC) 
    from data 
) prev 
on prev.number = next.number 
and prev.sequence_no + 1 = next.sequence_no