Question

SELECT [DATE], [AMOUNT], SUM(AMOUNT) OVER (ORDER BY DATE) AS 'Running Total' 
FROM PeopleActi
WHERE INSTANCE = 'Bank' 
AND DATE IS NOT NULL 
GROUP BY [DATE], [AMOUNT];

在上面的代码中，我选择一个用户的日期，金额和"SUM(AMOUNT) OVER (ORDER BY DATE) AS 'Running Total'"是他们在一段时间内的费用的总计。运行此代码时，我得到以下结果：

DATE       AMOUNT  Running Total
2018-10-05  100     100
2018-10-06  1000    1100
2018-10-07  5000    6100
2018-10-08  2000    8100
2018-10-09  1000    9100
2018-10-10  5000    14100
2018-10-11  3000    25100
2018-10-11  8000    25100

这很好用，但是我的问题是最后两行。我希望按日期将它们分组，并具有同一天的总数，所以应该是：

 Date       Amount   Running Total
2018-10-11   11000    25100

有人知道如何实现吗？我的[DATE]的类型为DATE。

更新！！！

我看过您的一些解决方案，它们很好，但重要的是，我也显示了AMOUNT和Running Total金额，所以最终结果应该是...

DATE       AMOUNT  Running Total
2018-10-05  100     100
2018-10-06  1000    1100
2018-10-07  5000    6100
2018-10-08  2000    8100
2018-10-09  1000    9100
2018-10-10  5000    14100
2018-10-11  11000   25100

谢谢大家的帮助！

Answer 1

将金额分组，然后进行累计总计

WITH CTE
AS
(
    SELECT  A.Dt,
            SUM(A.Amount) AS Amount
    FROM    (
                VALUES  ('2018-10-05',100),
                        ('2018-10-06',1000),
                        ('2018-10-07',5000),
                        ('2018-10-08',2000),
                        ('2018-10-09',1000),
                        ('2018-10-10',5000),
                        ('2018-10-11',3000),
                        ('2018-10-11',8000)
            ) AS A(Dt,Amount)
    GROUP   BY A.Dt
)
SELECT  C.Dt,
        C.Amount,
        SUM(C.Amount) OVER (ORDER BY C.Dt) AS CumTotal
FROM    CTE AS C;

Answer 2

尝试如下

SELECT [DATE],sum( [AMOUNT]), SUM(AMOUNT) OVER (ORDER BY DATE) AS 'Running Total' 
FROM PeopleActi
WHERE INSTANCE = 'Bank' 
AND DATE IS NOT NULL 
GROUP BY [DATE]

Answer 3

如果您需要摸索总和，那么为什么要使用窗口函数，仅聚合就足够了：

SELECT [DATE], SUM([AMOUNT])
FROM PeopleActi
WHERE INSTANCE = 'Bank' AND DATE IS NOT NULL 
GROUP BY [DATE];

Answer 4

group by中的每个字段都将导致其潜在地创建新行。如果您对金额字段求和并将其从分组中删除，那应该可以解决问题。编辑：我看到了这个问题，我在下面提供了一个完全独立的查询示例，您可以对其进行调整。

DECLARE @PeopleActi TABLE ([DATE] DATE,[AMOUNT] MONEY)
INSERT INTO @PeopleActi SELECT '2018-10-05',100 
INSERT INTO @PeopleActi SELECT '2018-10-06',1000
INSERT INTO @PeopleActi SELECT '2018-10-07',5000
INSERT INTO @PeopleActi SELECT '2018-10-08',2000
INSERT INTO @PeopleActi SELECT '2018-10-09',1000
INSERT INTO @PeopleActi SELECT '2018-10-10',5000
INSERT INTO @PeopleActi SELECT '2018-10-11',3000
INSERT INTO @PeopleActi SELECT '2018-10-11',8000

SELECT *, SUM(AMOUNT) OVER (ORDER BY DATE) AS 'Running Total' 
FROM (
    SELECT [DATE], SUM([AMOUNT]) AS AMOUNT
    FROM @PeopleActi
    WHERE DATE IS NOT NULL 
    GROUP BY [DATE]
) a
GROUP BY [DATE],Amount

Answer 5

尝试一下

;WITH CTe([DATE],AMOUNT)
AS
(
SELECT '2018-10-05',  100   UNION ALL
SELECT '2018-10-06',  1000  UNION ALL
SELECT '2018-10-07',  5000  UNION ALL
SELECT '2018-10-08',  2000  UNION ALL
SELECT '2018-10-09',  1000  UNION ALL
SELECT '2018-10-10',  5000  UNION ALL
SELECT '2018-10-11',  3000  UNION ALL
SELECT '2018-10-11',  8000  
)
SELECT DISTINCT  [DATE],SUM(AMOUNT)OVER(PARTITION BY [DATE] ORDER BY [DATE]) AMOUNT , SUM(AMOUNT)OVER( ORDER BY [DATE]) AS RuningTot FROM CTe

脚本

SELECT  DISTINCT [DATE], 
        SUM(AMOUNT)OVER(PARTITION BY [DATE] ORDER BY [DATE]) AS AMOUNT, 
        SUM(AMOUNT) OVER (ORDER BY DATE) AS 'Running Total' 
FROM PeopleActi
WHERE INSTANCE = 'Bank' 
AND DATE IS NOT NULL

Answer 6

我会使用CTE首先按日期分组，然后再进行累计。所以像

with myAmounts AS
(
SELECT [DATE], SUM([AMOUNT]) AS Amount 
FROM PeopleActi
WHERE INSTANCE = 'Bank' 
AND DATE IS NOT NULL 
GROUP BY [DATE]
)
SELECT [DATE], [AMOUNT], SUM(AMOUNT) OVER (ORDER BY DATE) AS 'Running Total' 
FROM myAmounts

GROUP BY [DATE], [AMOUNT]
;

HTH，

B

ps;刚刚看到它的答案与另一个..实际解码中的答案相同

Answer 7

尝试子选择：

SELECT p.[DATE], p.[AMOUNT], SUM(AMOUNT) OVER (ORDER BY DATE) AS 'Running Total' 
FROM 
(
    select [date], sum([amount]) as Amount from PeopleActi
    WHERE INSTANCE = 'Bank' 
    AND DATE IS NOT NULL
    group by [date]
) p
GROUP BY p.[DATE], p.[AMOUNT]

SQL Server按日期分组

7 个答案: