Question

我有一个表，其中不同时期插入记录（每个记录包含一个名为'Amount'的列）。

我想在每5秒后显示累计总量。我尝试了以下查询但没有成功：

SELECT Sum(totalamount) AS RealTimeTotalAmount,
       Datepart(second, createstamp) / 5 AS dp
FROM   [order]
WHERE 
       createstamp BETWEEN Dateadd(s, -5, Getdate()) AND Getdate()
GROUP  BY Datepart(second, createstamp) / 5

我面临的问题是，它向我显示“每秒的累积金额”，我希望看到它像'（每秒的累积金额+累计金额直到那一秒）'

以下是源数据的外观：

-----------------------------------------------------------
|OrderID  | CreateStamp             | TotalAmount         |
-----------------------------------------------------------
|1        |2015-03-22 15:26:05.620  | 10                  |
-----------------------------------------------------------
|2        |2015-03-22 15:26:05.653  | 20                  |
-----------------------------------------------------------
|3        |2015-03-22 15:26:05.660  | 10                  |
-----------------------------------------------------------
|4        |2015-03-22 15:26:06.663  | 10                  |
-----------------------------------------------------------
|5        |2015-03-22 15:26:06.670  | 30                  |
-----------------------------------------------------------

基本上，我希望生成的查询返回如下：

----------------------------------------
|Period          | Accumulative Amount | 
----------------------------------------
|0 to 5 seconds  | 30                  |
----------------------------------------
|0 to 10 seconds | 80                  |
----------------------------------------

这基本上是从0时间到5的倍数的累积。对于最后5秒基本上我计算一整天的数量直到我执行此查询的时间，例如在此时间之前的整天的数量是50，所以结果表应该看起来像

----------------------------------------
|0 to 5 seconds  | 30 + 50 = 80        |
----------------------------------------
|0 to 10 seconds | 80 + 80 = 160       |
----------------------------------------

Answer 1

你可以尝试这样的事情。

输入数据

DECLARE @Orders TABLE
(
    OrderId INT,
    CreateStamp DATETIME,
    TotalAmount NUMERIC(9,2)
)
INSERT INTO @Orders
SELECT 1,'2015-03-22 15:26:05.620',400
UNION ALL SELECT 2,'2015-03-22 15:26:04.653',500
UNION ALL SELECT 3,'2015-03-22 15:26:05.660',600
UNION ALL SELECT 4,'2015-03-22 15:26:06.663',700
UNION ALL SELECT 5,'2015-03-22 15:26:06.670',900
UNION ALL SELECT 6,'2015-03-22 15:26:05.660',600
UNION ALL SELECT 7,'2015-03-22 15:26:09.663',700
UNION ALL SELECT 8,'2015-03-22 15:26:12.670',900

<强>查询

;WITH CTE as 
(
    SELECT DATEDIFF(minute,0,CreateStamp)totalminutes,Datepart(second, CreateStamp ) / 5 sec,SUM(TotalAmount) TotalAmount
    FROM @Orders
    GROUP BY DATEDIFF(minute,0,CreateStamp),Datepart(second, CreateStamp) / 5
)
SELECT  DATEADD(minute,totalminutes,0) dt,sec,(SELECT SUM(TotalAmount) FROM cte WHERE totalminutes <=c2.totalminutes and sec <=c2.sec)
FROM CTE c2
ORDER BY sec;

我为不同的日期和分钟添加了GROUP BY DATEDIFF(minute,0,CreateStamp)来分隔秒数。

Answer 2

如果我理解正确的话：

DECLARE @t TABLE
    (
      ID INT ,
      D DATETIME ,
      A MONEY
    )
DECLARE @mind DATETIME ,
    @maxd DATETIME


INSERT  INTO @t
VALUES  ( 1, '2015-04-07 13:49:15.000', 5 ),
        ( 2, '2015-04-07 13:49:17.000', 15 ),
        ( 3, '2015-04-07 13:49:35.000', 2 ),
        ( 4, '2015-04-07 13:49:45.000', 4 ),
        ( 5, '2015-04-07 13:49:49.000', 20 ),
        ( 6, '2015-04-07 13:50:05.000', 20 ),
        ( 7, '2015-04-07 13:50:09.000', 3 ),
        ( 8, '2015-04-07 13:50:09.000', 3 ),
        ( 9, '2015-04-07 13:50:10.000', 1 ),
        ( 10, '2015-04-07 13:50:15.000', 1 )


SELECT  @mind = MIN(d) ,
        @maxd = MAX(d)
FROM    @t;

WITH    cte
          AS ( SELECT   @mind AS d
               UNION ALL
               SELECT   DATEADD(ss, 5, d)
               FROM     cte
               WHERE    cte.d <= @maxd
             )

SELECT cte.d, SUM(A) AS A FROM cte
JOIN @t t ON t.D < cte.d
GROUP BY cte.d

输出：

d                       A
2015-04-07 13:49:20.000 20.00
2015-04-07 13:49:25.000 20.00
2015-04-07 13:49:30.000 20.00
2015-04-07 13:49:35.000 20.00
2015-04-07 13:49:40.000 22.00
2015-04-07 13:49:45.000 22.00
2015-04-07 13:49:50.000 46.00
2015-04-07 13:49:55.000 46.00
2015-04-07 13:50:00.000 46.00
2015-04-07 13:50:05.000 46.00
2015-04-07 13:50:10.000 72.00
2015-04-07 13:50:15.000 73.00
2015-04-07 13:50:20.000 74.00

总和列作为基于时间段的累积

2 个答案: