Question

表格中有一些数据：

    date   finance
    1      5 (income)
    3     -3 (spending)
    4     -1 (spending)
    5     10 (income)
    9     -2 (spending)
   10     -3 (spending)
   15     -3 (spending)

每个正值构成一个组，每个组以正值开头，并包括下一个posivitve值之前的所有负值。

所以我必须识别这样的群体：

    date  finance  rank
    1      5       group 1  
    3     -3       group 1
    4     -1       group 1

    5     10       group 2
    9     -2       group 2
   10     -3       group 2
   15     -3       group 2

日期总是在增长，而且永远不会相同。

窗口函数t-sql可以吗？得到这个结果的最佳方法是什么？

Answer 1

这有点无用，但似乎有用

SELECT [date], finance 
    , 'group ' 
        + rtrim((SELECT count(1) FROM #Tmp WHERE finance > 0 AND [date] < t.[date])  
            + CASE WHEN finance > 0 THEN 1 ELSE 0 END) 'Group'
FROM #Tmp t

Answer 2

您的分组似乎非常依赖于行的顺序。

假设您的源表中有一个Row＃[R]，则以下代码应该有效。

--FIND ROWS WITH POSITIVE VALUE
SELECT R2=DENSE_RANK() OVER(ORDER BY R), R
INTO #TEMP
FROM <SOURCE_TABLE>
WHERE FINANCE>0

--CREATE GROUPS BY DETECTING STARTING ROW AND ENDING ROW (ENDING ROW BEING 1 LESS THAN NEXT STARTING ROW)
SELECT GROUP_N=ROW_NUMBER() OVER(ORDER BY LOW.R)    --ASSIGN GROUP #S IN THE SAME ORDER AS THEY APPEAR IN THE SOURCE TABLE
    ,R_START=LOW.R, R_END=(HIG.R-1)
INTO #TEMP2
FROM #TEMP LOW
JOIN #TEMP HIG
    ON LOW.R2=(HIG.R2-1)
UNION
--LAST GROUP STARTS AT THE LAST DETECTED STARTING ROW AND ENDS AT THE LAST ROW OF THE SOURCE TABLE
SELECT R_START_LAST_GROUP=(SELECT MAX(R) FROM #TEMP), R_END_LAST_GROUP=(SELECT MAX(R) FROM <SOURCE_TABLE>)

--GET SOURCE DATA AND APPEND GROUP #
SELECT DAT.*, GRP.GROUP_N
FROM #TEMP2 GRP
JOIN <SOURCE_TABLE> DAT
    ON DAT.R BETWEEN GRP.R_START AND GRP.R_END

T-sql：通过改变正值排名组

2 个答案: