Question

我有多个应用程序，每个应用程序都有唯一的应用程序ID，每个应用程序都有一个存储在表格中的日志。我想知道如何计算日期差异例如：

    AppID Start           Loggedon   change 
    A1    08/07/2010      08/09/2010 Xchange
    A1    08/07/2010      08/20/2010 Ychange
    A1    08/07/2010      08/30/2010 Zchange
    A2    07/07/2010      07/13/2010 Ychange
    A3    09/07/2010      09/09/2010 Xchange

所以我想要值

Difference
2   (Difference between the application start date and 1st loggedon date)
11  (difference between 08/20 and 08/09, the prior row because AppID stayed the same)
10  (difference between 08/30 and 08/20, the prior row because AppID stayed the same)
6   (Difference between the application start date and 1st loggedon date)
2   (Difference between the application start date and 1st loggedon date)

希望我很清楚。我怎样才能实现这一点，我尝试了排名和Row_Number。但我可能在某处错了。我使用的是SQL Server，无法使用LAG（）

Answer 1

好的，这会奏效。享受。

现在测试了！谢谢sgeddes - http://sqlfiddle.com/#!3/e4d28/19

WITH tableNumbered AS
(
   SELECT *,
      ROW_NUMBER() OVER (PARTITION BY AppID ORDER BY AppID, Start , Loggedon ) AS row
   FROM Apps
)
SELECT t.*,
  CASE WHEN t2.row IS NULL THEN DATEDIFF(day,t.Start,t.LoggedOn)
       ELSE DATEDIFF(day,t2.LoggedOn,t.Loggedon)
  END as diff
FROM tableNumbered t
LEFT JOIN tableNumbered t2 ON t.AppID = t2.AppID AND t2.row+1 = t.row

我仍然认为你应该在用户界面中这样做。

Answer 2

@Hogan有正确的方法，但由于某种原因，我无法让它完全运作。这是一个微小的变化，似乎产生了正确的结果 - 但是，请接受@Hogan的回答，因为他打败了我：）

WITH cte AS (
  SELECT AppId, Start, LoggedOn, 
  ROW_NUMBER() OVER (ORDER BY AppId) rn
FROM Apps
  ) 
SELECT c.AppId,
  CASE 
    WHEN c2.RN IS NULL
    THEN DATEDIFF(day,c.start,c.Loggedon)
    ELSE
         DATEDIFF(day,c2.Loggedon,c.Loggedon)
    END as TimeWanted
FROM cte c
   LEFT JOIN cte c2 on c.AppId = c2.AppId
      AND c.rn = c2.rn + 1

这是Fiddle。

祝你好运。

如何计算具有特殊第一行规则的行之间的日期差异

2 个答案: