oracle 中的 Lead & LAG 聚合

Posted 2023-03-31

【中文标题】oracle 中的 Lead & LAG 聚合

【英文标题】：Aggregations on Lead & LAG in oracle

【发布时间】：2016-03-04 12:12:49

【问题描述】：

这是关于甲骨文的 输入

CUSTID          FROMDT         ACTIVITY     NEXTDATE
100000914   31/01/2015 14:23:51 Bet         3.999996
100000914   31/01/2015 14:29:07 Bet         3.999996
100000914   31/01/2015 14:32:59 Bet         2
100000914   31/01/2015 14:35:35 Bet         1.999998
100000914   31/01/2015 16:52:32 Settlement  3.999996
100000914   31/01/2015 16:54:39 Settlement  1.999998
100000914   31/01/2015 16:55:04 Settlement  2
100000914   31/01/2015 16:57:00 Settlement  3.999996
100000914   31/01/2015 16:57:10 Bet         3
100000914   31/01/2015 19:21:15 Settlement  3

结果

CUSTID      ACTIVITY    AMOUNT      
100000914   Bet         11.99999    
100000914   Settlement  11.99999    
100000914   Bet          3          
100000914   Settlement   3

结果应该有每次活动变化的总和

谢谢

【问题讨论】：

这不是问题，这是代码请求。 SO 不是代码编写服务。展示你的努力。

不是特定的编程！！

【参考方案1】：

SELECT CUSTID,
       ACTIVITY,
       total - LAG( total, 1, 0 ) OVER ( PARTITION BY CUSTID ORDER BY FROMDT ) AS total
FROM   (
  SELECT CUSTID,
         FROMDT,
         ACTIVITY,
         SUM( NEXTDATE ) OVER ( PARTITION BY CUSTID ORDER BY FROMDT ) AS total,
         CASE ACTIVITY
              WHEN LEAD( ACTIVITY ) OVER ( PARTITION BY CUSTID ORDER BY FROMDT )
              THEN 0
              ELSE 1
              END AS has_changed
  FROM   your_table
)
WHERE  has_changed = 1;

输出：

CUSTID    ACTIVITY   TOTAL
--------- ---------- --------
100000914 Bet        11.99999
100000914 Settlement 11.99999
100000914 Bet               3
100000914 Settlement        3

【参考方案2】：

select custid, activity, sum(amount)
from (select jg_dig_test.*,
             (row_number() over (partition by custid order by fromdate) - row_number() over (partition by custid, activity order by fromdate)
             ) as grp
      from jg_dig_test
     ) jg_dig_test
group by custid, grp, activity
ORDER BY CUSTID, MAX( FROMDaTe )
;

【参考方案3】：

您需要识别连续值的组。剩下的就是聚合。

一种方法是行数差法：

select custid, activity, sum(amount)
from (select t.*,
             (row_number() over (partition by custid order by fromdt) -
              row_number() over (partition by custid, activity order by fromdt)
             ) as grp
      from t
     ) t
group by custid, grp, activity
order by custid, max(fromdt);

【讨论】：

HI GORDON结果需求活动量订单100000914投注11.99999 1 100000914结算11.9999 2 100000914 BET 3 3 100000914结算3 4押卡活动金额100000914下注11.99999 100000914 BET 3 100000914结算3 100000914结算11.99999答案应该是这一点基于活动变化的方式

ORDER BY CUSTID, MAX( FROMDT )

select custid, activity, sum(amount) from (select jg_dig_test.*, (row_number() over (partition by custid order by fromdate) - row_number() over (partition by custid, activity order by fromdate) ) as grp from jg_dig_test ) jg_dig_test group by custid, grp, activity ORDER BY CUSTID, MAX( FROMDaTe ) ;

@nkalis 。 . .如果区别在于输出中的顺序，那么您只需要适当的order by 子句。