如何获取列中不同值的运行总计

Question

我希望按日期范围分组，如下例

    Date        ItemNo  Qty
==================================
1/1/2014         101    20
2/1/2014         102    10
3/1/2014         103    5
4/1/2014         104    10
1/1/2014         101    5
2/1/2014         101    10
3/1/2014         102    15
4/1/2014         104    20

我希望每天通过将数量加到按ItemNo分组的数量来得到余额

Date        ItemNo  Qty
==================================
1/1/2014      101   25
2/1/2014      101   35
2/1/2014      102   10
3/1/2014      102   25
3/1/2014      103   5
4/1/2014      104   30

我知道我可以通过使用游标解决问题，但我需要另一种解决方案

感谢

Answer 1

所以只需使用SUM

SELECT Date, ItemNo, SUM(Qty)
FROM table
GROUP BY Date, ItemNo

请阅读agregate function和sum

修改

我接受了你的评论并做了这个：

SELECT a.Date, a.ItemNo, tmp.qty + a.ItemNo
FROM table a
JOIN (SELECT TOP 1 * FROM table t WHERE t.date < a.Date ORDER BY t.date DESC) tmp ON a.ItemNo = tmp.ItemNo 

我现在正在检查它，所以可能需要一些调整，但我想立即发布它，以便你有一般的想法

Answer 2

这是您的示例表

SELECT * INTO #TEMP
FROM
(
SELECT  '1/1/2014' [DATE],         101 [ItemNo],    20 QTY
UNION ALL
SELECT '2/1/2014',         102,    10
UNION ALL
SELECT '3/1/2014',         103,    5
UNION ALL
SELECT '4/1/2014',         104,    10
UNION ALL
SELECT '1/1/2014',         101,    5
UNION ALL
SELECT '2/1/2014',        101,    10
UNION ALL
SELECT '3/1/2014',         102,    15
UNION ALL
SELECT '4/1/2014',         104,    20
)TAB

使用Row_Number为每个项目的日期获取sum CTE内的;WITH CTE1 AS ( SELECT ROW_NUMBER() OVER(PARTITION BY [ItemNo] ORDER BY CAST([DATE] AS DATE))RNO, [DATE],[ItemNo],SUM(Qty)Qty FROM #TEMP GROUP BY [DATE],[ItemNo] ) SELECT A.RNO,[DATE],[ItemNo], CASE WHEN RNO=1 THEN Qty ELSE (SELECT SUM(b.Qty) FROM CTE1 b WHERE A.ItemNo=B.ItemNo AND B.RNO<=A.RNO) END QTY FROM CTE1 A ORDER BY A.itemno,CAST(A.[DATE] AS DATE);

{{1}}

<强> RESULT

Answer 3

这是一个使用递归公用表表达式的解决方案。

不确定它是否会比Sarath Avanavu的答案更快或更快，但你可以试试！

示例数据：

DECLARE @t TABLE([Date] DATETIME, ItemNo INT, QTY INT)
INSERT @t
        ( Date, ItemNo, QTY )
          SELECT  '1/1/2014',         101,    20
UNION ALL SELECT '2/1/2014',         102,    10
UNION ALL SELECT '3/1/2014',         103,    5
UNION ALL SELECT '4/1/2014',         104,    10
UNION ALL SELECT '1/1/2014',         101,    5
UNION ALL SELECT '2/1/2014',        101,    10
UNION ALL SELECT '3/1/2014',         102,    15
UNION ALL SELECT '4/1/2014',         104,    20

查询：

;WITH dSum AS (
    SELECT [Date], ItemNo, SUM(QTY) AS QTY 
    FROM @t AS t
    GROUP BY [Date], [ItemNo]
), dSumRN AS (
    SELECT [Date], ItemNo, QTY, ROW_NUMBER() OVER(PARTITION BY ItemNo ORDER BY [Date]) AS rn 
    FROM dSum
), cte AS (
        SELECT [Date], ItemNo, QTY, rn
        FROM  dSumRN
        WHERE rn = 1
    UNION ALL SELECT
        dSumRN.[Date], dSumRN.ItemNo, cte.QTY + dSumRN.QTY AS QTY, cte.rn + 1 AS rn
    FROM cte
        JOIN dSumRN ON cte.ItemNo = dSumRN.ItemNo AND cte.rn + 1 = dSumRN.rn

)
SELECT [Date], [ItemNo], QTY FROM cte
ORDER BY [Date], [ItemNo]
OPTION (MAXRECURSION 1000) -- maximum this can be set to is 32767

Answer 4

以下最简单的查询代码：

select Date,itemno,
(select sum(Qty) from #temp where date<=T.date and itemno=T.itemno)
from #temp T 
group by Date,itemno order by date