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我有一个数据集，我需要计算日期差异。我需要一个通用查询，因为数据集每天都会有所不同。

这是我的数据集

**Client Number    :  Status  :    StatusDate**
   123456           Admit       2012-01-18
   123456           Accepted    2012-01-15
   123456           Offered     2012-01-10

   567899           Admit       2012-01-11
   567899           Accepted    2012-01-07
   567899           Offered     2012-01-05

   987342           Admit       2012-01-24
   987342           Accepted    2012-01-24
   987342           Offered     2012-01-21

以上是示例数据，正如我所说，数据会每天发生变化。如您在上面看到的，始终会为特定的客户编号输入三个，并且始终按以下顺序输入三个状态。

我需要的是计算从提供到接受和接受到承认之间的不同天数。上面的结果集应该是这样的。三列

**ClientNumber : Date Diff Betw Offeres & Accepted : Date Diff Betw Accepted & Admited** 
   123456                     5 Days                           3 Days
   567899                     2 Days                           4 Days
   987342                     3 Days                           0 Days

我需要通用查询，正如我所说，每个客户编号将具有三种状态。

提前致谢。

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2 回答 2

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您必须进行自我加入。这是一个通用示例。

select t1.clientnumber, datediff(day, t1.statusdate, t2.statusdate) OfferToAccept
from yourtable t1 join yourtable t2 on t1.clientnumber = t2clientnumber
and t1.status = 'Offered' and t2.status = 'Accepted'

where whatever

对于您的问题中可能输入错误的另一个间隔，这也是类似的事情。它说“接受和接受的日期差异”

于 2013-02-08T02:21:12.020 回答
0

询问：

SQLFIDDLE示例

SELECT t1.[Client Number],
       MAX(datediff(DAY, t1.statusdate, t2.statusdate)) [Offeres & Accepted],
       MAX(datediff(DAY, t3.statusdate, t1.statusdate)) [Accepted & Admited]
FROM Table1 t1
LEFT JOIN Table1 t2 ON t1.[Client Number] = t2.[Client Number]
AND t1.status = 'Offered'
AND t2.status = 'Accepted'
LEFT JOIN Table1 t3 ON t1.[Client Number] = t3.[Client Number]
AND t1.status = 'Admit'
AND t3.status = 'Accepted'
GROUP BY t1.[Client Number]

结果：

| CLIENT NUMBER | OFFERES & ACCEPTED | ACCEPTED & ADMITED |
-----------------------------------------------------------
|        123456 |                  5 |                  3 |
|        567899 |                  2 |                  4 |
|        987342 |                  3 |                  0 |
于 2013-02-08T07:01:33.983 回答