Question

我试图计算运行总数。但是当累积和大于另一个列值时，它应该重置

create table #reset_runn_total
(
id int identity(1,1),
val int, 
reset_val int
)

insert into #reset_runn_total
values 
(1,10),
(8,12),(6,14),(5,10),(6,13),(3,11),(9,8),(10,12)

示例数据

+----+-----+-----------+
| id | val | reset_val |
+----+-----+-----------+
|  1 |   1 |        10 |
|  2 |   8 |        12 |
|  3 |   6 |        14 |
|  4 |   5 |        10 |
|  5 |   6 |        13 |
|  6 |   3 |        11 |
|  7 |   9 |         8 |
|  8 |  10 |        12 |
+----+-----+-----------+

预期结果

+----+-----+-----------------+-------------+
| id | val |    reset_val    | Running_tot |
+----+-----+-----------------+-------------+
|  1 |   1 | 10              |       1     |  
|  2 |   8 | 12              |       9     |  --1+8
|  3 |   6 | 14              |       15    |  --1+8+6 -- greater than reset val
|  4 |   5 | 10              |       5     |  --reset 
|  5 |   6 | 13              |       11    |  --5+6
|  6 |   3 | 11              |       14    |  --5+6+3 -- greater than reset val
|  7 |   9 | 8               |       9     |  --reset -- greater than reset val 
|  8 |  10 | 12              |      10     |  --reset
+----+-----+-----------------+-------------+

查询：

;WITH cte
     AS (SELECT id,
                val,
                reset_val,
                val AS running_total
         FROM   #reset_runn_total
         WHERE  id = 1
         UNION ALL
         SELECT r.*,
                CASE
                  WHEN lag(c.running_total + r.val) over(order by r.id) > lag(r.reset_val) over(order by r.id) THEN r.reset_val 
                  ELSE c.running_total + r.val
                END
         FROM   cte c
                JOIN #reset_runn_total r
                  ON r.id = c.id + 1)
SELECT *
FROM   cte

显然滞后不会得到以前的任何想法吗？

Answer 1

尝试标记前一行与cte AS（SELECT id， VAL， reset_val， val AS running_total，例如，当val> reset_val THEN 1 ELSE 0 END作为标志来自#reset_runn_total WHERE id = 1 UNION ALL SELECT r。*， CASE c.flag 1然后r.val ELSE c.running_total + r.val 结束，案例c.flag 1然后r.val ELSE c.running_total + r.val 结束＆gt; r.reset_val 那么1结束0结束来自cte c 加入#reset_runn_total r ON r.id = c.id + 1）选择 * 来自cte

Answer 2

您可以尝试使用像这样的奇怪更新

--- setup
IF OBJECT_ID('tempdb..#reset_runn_total') IS NOT NULL DROP TABLE #reset_runn_total
create table #reset_runn_total(id int identity(1,1) PRIMARY KEY, val int, reset_val int, running_sum int)

insert into #reset_runn_total(val, reset_val) values (1,10),(8,12),(6,14),(5,10),(6,13),(3,11),(9,8),(10,12)

--- use quirky update
DECLARE     @running_sum    INT
            , @temp         INT

UPDATE      #reset_runn_total
SET         @temp = running_sum = COALESCE(@running_sum, 0) + val
            , @running_sum = CASE WHEN @temp < reset_val THEN @temp ELSE 0 END
OPTION      (FORCE ORDER)

--- dump result
SELECT * FROM #reset_runn_total

请注意，临时表上的CLUSTERED INDEX是必需的（PK的默认类型），OPTION (FORCE ORDER)才有意义。

根据其他列重置“运行总计”

2 个答案: