Question

您好我有以下数据：

 Item daytime               sub_hours  ch_size
 -------------------------------------------------------
 t23  18-Aug-17 03:30:00    0.5        56.8
 t23  18-Aug-17 04:00:00    0.5  
 t24  18-Aug-17 02:30:00    0.5        40.8
 t24  18-Aug-17 04:40:00    0.5           
 t24  18-Aug-17 02:30:00    0          45.4
 t23  19-Aug-17 03:30:00    0.5        70.2
 t23  19-Aug-17 04:00:00    0.5  
 t24  19-Aug-17 02:30:00    0.5        40.8
 t24  19-Aug-17 04:40:00    0.5           
 t24  19-Aug-17 02:30:00    0.2

我想得到这个结果：

 Item daytime               sub_hours  ch_size
 -------------------------------------------------------
 t23  18-Aug-17 03:30:00    0.5        56.8
 t23  18-Aug-17 04:00:00    0.5        56.8
 t24  18-Aug-17 02:30:00    0.5        40.8
 t24  18-Aug-17 04:40:00    0.5        40.8   
 t24  18-Aug-17 02:30:00    0          45.4
 t23  19-Aug-17 03:30:00    0.5        70.2
 t23  19-Aug-17 04:00:00    0.5        70.2
 t24  19-Aug-17 02:30:00    0.5        40.8
 t24  19-Aug-17 04:40:00    0.5        40.8   
 t24  19-Aug-17 02:30:00    0.2

我需要更新ch_size列，将其设置为前一个值，其中
ch_size为null，sub_hours = 0.5：

这是我使用的代码：

update day_status a set ch_size = 
( select prev_choke from
  (
     select LAG(ch_size) OVER (order by item, daytime) as prev_choke
     from day_status  
     where trunc(daytime) between to_date('1 Aug 2017') and to_date('31 Aug 2017')
   ) b where a.item = b.item and a.daytime=b.daytime
 )

我还没有测试过这段代码，你认为它是正确的并且达到我想要的结果吗？

感谢，小号

Answer 1

您可以使用LAG功能。它将为您提供上一行的值。

SELECT empno,
       ename,
       job,
       sal,
       LAG(sal, 1, 0) OVER (ORDER BY sal) AS sal_prev,
       sal - LAG(sal, 1, 0) OVER (ORDER BY sal) AS sal_diff
FROM   emp;

     EMPNO ENAME      JOB              SAL   SAL_PREV   SAL_DIFF
---------- ---------- --------- ---------- ---------- ----------
      7369 SMITH      CLERK            800          0        800
      7900 JAMES      CLERK            950        800        150
      7876 ADAMS      CLERK           1100        950        150
      7521 WARD       SALESMAN        1250       1100        150
      7654 MARTIN     SALESMAN        1250       1250          0
      7934 MILLER     CLERK           1300       1250         50
      7844 TURNER     SALESMAN        1500       1300        200
      7499 ALLEN      SALESMAN        1600       1500        100
      7782 CLARK      MANAGER         2450       1600        850
      7698 BLAKE      MANAGER         2850       2450        400
      7566 JONES      MANAGER         2975       2850        125
      7788 SCOTT      ANALYST         3000       2975         25
      7902 FORD       ANALYST         3000       3000          0
      7839 KING       PRESIDENT       5000       3000       2000

Answer 2

假设item和daytime是主键

UPDATE day_status a 
SET a.ch_size = 
( 
  SELECT LAG(ch_size) OVER (ORDER BY item, daytime) as prev_choke
  FROM day_status b
  WHERE a.item = b.item AND a.daytime=b.daytime
) 
WHERE a.ch_size IS NULL AND a.sub_hours = 0.5

从上一行值更新值

2 个答案: