Question

我有一张桌子，如果它们在特定时间段内，我想对其进行分类

我现在所拥有的：

SELECT RelatieId,
   AccountId,
   [UaId],
   [VisitDate],
   [Titel],
   [Editie],
   CASE
      WHEN DATEDIFF(SECOND, v.LastVisit, v.VisitDate) > (10 * 60)
       THEN 0
       ELSE 1
   END AS reset
   FROM rfv.click v

结果是这样的

RelatieId   AccountId   UaId    VisitDate   Titel   reset
70409280    120211  8408    2019-04-01 09:15:52.000 Nx  1
70409280    120211  14531   2019-04-01 11:45:41.000 Nx  0
70409280    120211  14531   2019-04-01 11:45:55.000 Nx  1
70409280    120211  14531   2019-04-01 11:46:10.000 Nx  1
70409280    120211  14531   2019-04-01 11:47:16.000 Nx  1
70409280    120211  14531   2019-04-01 11:52:52.000 Nx  1
70409280    120211  14531   2019-04-01 11:53:14.000 Nx  1
70409280    120211  14531   2019-04-01 12:02:44.000 Nx  1
70409280    120211  14531   2019-04-01 12:03:08.000 Nx  1
70409280    120211  14531   2019-04-02 08:06:42.000 Nx  0
70409280    120211  14531   2019-04-02 08:07:07.000 Nx  1
70409280    120211  14531   2019-04-02 08:36:15.000 Nx  0
70409280    120211  14531   2019-04-02 08:36:26.000 Nx  1

但是我真正需要的是

RelatieId   AccountId   UaId    VisitDate   Titel   needed 
70409280    120211  8408    2019-04-01 09:15:52.000 Nx  1
70409280    120211  14531   2019-04-01 11:45:41.000 Nx  2
70409280    120211  14531   2019-04-01 11:45:55.000 Nx  3
70409280    120211  14531   2019-04-01 11:46:10.000 Nx  3
70409280    120211  14531   2019-04-01 11:47:16.000 Nx  3
70409280    120211  14531   2019-04-01 11:52:52.000 Nx  3
70409280    120211  14531   2019-04-01 11:53:14.000 Nx  3
70409280    120211  14531   2019-04-01 12:02:44.000 Nx  3
70409280    120211  14531   2019-04-01 12:03:08.000 Nx  3
70409280    120211  14531   2019-04-02 08:06:42.000 Nx  4
70409280    120211  14531   2019-04-02 08:07:07.000 Nx  5
70409280    120211  14531   2019-04-02 08:36:15.000 Nx  6
70409280    120211  14531   2019-04-02 08:36:26.000 Nx  7

Answer 1

我不确定何时要重置此值（从1开始），例如针对不同的RelatieId，针对不同的AccountId还是针对两者。您需要同时调整PARTITION BY和LAG窗口函数中的SUM。这是基本版本：

;WITH resetChangeCTE AS
(
    SELECT *
         , IIF(reset != LAG(reset, 1, 0) OVER (PARTITION BY RelatieId, AccountId ORDER BY VisitDate), 1, 0) AS hasResetChanged
     FROM (VALUES (70409280,    120211,  8408 ,   CONVERT(DATETIME, '2019-04-01 09:15:52.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 11:45:41.000'), 0),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 11:45:55.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 11:46:10.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 11:47:16.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 11:52:52.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 11:53:14.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 12:02:44.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-01 12:03:08.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-02 08:06:42.000'), 0),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-02 08:07:07.000'), 1),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-02 08:36:15.000'), 0),
                  (70409280,    120211,  14531,   CONVERT(DATETIME, '2019-04-02 08:36:26.000'), 1)) x(RelatieId  , AccountId   , UaId    ,VisitDate   ,reset)
)
SELECT *, SUM(hasResetChanged) OVER (PARTITION BY RelatieId, AccountId ORDER BY VisitDate) AS needed 
  FROM resetChangeCTE

因此，我首先检查reset值在当前记录和上一个记录之间是否发生了变化（使用LAG，按访问日期排序），然后将变化量求和从第一个记录到使用SUM的当前记录。这些是window functions，非常有用。

sqlfiddle with the result。

Answer 2

我认为您确实想要：

SELECT c.*,
       SUM(CASE WHEN DATEDIFF(SECOND, v.LastVisit, v.VisitDate) > (10 * 60)
                THEN 0
                ELSE 1
           END) AS needed
FROM (SELECT c.*,
             LAG(VisitDate) OVER (PARTITION BY RelatieId, AccountId, UaId ORDER BY VisitDate) as LastVisit
      FROM rfv.click c
     ) c;

这不会完全返回您指定的结果。但是，这似乎更有用，可能正是您真正想要的。

按具有相同值的列分组

2 个答案: