2

假设我有一个 SQL 奖项表，其中包含日期和金额字段。我需要生成一个表格，其中包含一系列连续日期、每天奖励的金额以及运行（累积）总数。

Date         Amount_Total   Amount_RunningTotal
----------   ------------   -------------------
1/1/2010              100                   100
1/2/2010              300                   400
1/3/2010                0                   400
1/4/2010                0                   400
1/5/2010              400                   800
1/6/2010              100                   900
1/7/2010              500                  1400
1/8/2010              300                  1700

此 SQL 有效，但不如我想的那么快：

Declare @StartDate datetime, @EndDate datetime 
Select @StartDate=Min(Date), @EndDate=Max(Date) from Awards 

; With 

/* Returns consecutive from numbers 1 through the 
number of days for which we have data */
Nbrs(n) as (
   Select 1 Union All 
   Select 1+n 
   From Nbrs 
   Where n<=DateDiff(d,@StartDate,@EndDate)),

/* Returns all dates @StartDate to @EndDate */
AllDays as (
   Select Date=DateAdd(d, n, @StartDate) 
   From Nbrs ) 

/* Returns totals for each day */
Select 
 d.Date,
 Amount_Total = (
        Select Sum(a.Amount) 
        From Awards a 
        Where a.Date=d.Date),
 Amount_RunningTotal = (
        Select Sum(a.Amount) 
        From Awards a 
        Where a.Date<=d.Date)
From AllDays d
Order by d.Date 
Option(MAXRECURSION 1000)

我尝试向 Awards.Date 添加索引，但它产生的差异非常小。

在我诉诸缓存等其他策略之前，是否有更有效的方法来编写运行总计计算？

4

3 回答 3

3

我通常为此使用一个临时表：

DECLARE @Temp TABLE
(
    [Date] date PRIMARY KEY,
    Amount int NOT NULL,
    RunningTotal int NULL
)

INSERT @Temp ([Date], Amount)
    SELECT [Date], Amount
    FROM ...

DECLARE @RunningTotal int

UPDATE @Temp
SET @RunningTotal = RunningTotal = @RunningTotal + Amount

SELECT * FROM @Temp

如果您不能将日期列作为主键，那么您需要ORDER BY [Date]在INSERT语句中包含一个。

另外，这个问题之前已经被问过几次了。请参见此处或搜索“sql running total”。据我所知，我发布的解决方案仍然是性能最好的解决方案，并且也易于编写。

于 2010-01-10T17:54:29.113 回答
0

这是基于@Aaronaught 答案的工作解决方案。我在 T-SQL 中必须克服的唯一问题是@RunningTotaletc. 不能为空（需要转换为零）。

Declare @StartDate datetime, @EndDate datetime 
Select @StartDate=Min(StartDate),@EndDate=Max(StartDate) from Awards

/* @AllDays: Contains one row per date from @StartDate to @EndDate */
Declare @AllDays Table (
    Date datetime Primary Key)
; With 
Nbrs(n) as (
    Select 0 Union All 
    Select 1+n from Nbrs 
    Where n<=DateDiff(d,@StartDate,@EndDate) 
    )
Insert into @AllDays
Select Date=DateAdd(d, n, @StartDate) 
From Nbrs
Option(MAXRECURSION 10000) /* Will explode if working with more than 10000 days (~27 years) */

/* @AmountsByDate: Contains one row per date for which we have an Award, along with the totals for that date */ 
Declare @AmountsByDate Table (
    Date datetime Primary Key,
    Amount money)
Insert into @AmountsByDate
Select 
    StartDate, 
    Amount=Sum(Amount) 
from Awards a
Group by StartDate

/* @Result: Joins @AllDays and @AmountsByDate etc. to provide totals and running totals for every day of the award */
Declare @Result Table (
    Date datetime Primary Key,
    Amount money,
    RunningTotal money)
Insert into @Result 
Select 
    d.Date,
    IsNull(bt.Amount,0),
    RunningTotal=0
from @AllDays d
Left Join @AmountsByDate bt on d.Date=bt.Date
Order by d.Date

Declare @RunningTotal money Set @RunningTotal=0
Update @Result Set @RunningTotal = RunningTotal = @RunningTotal + Amount

Select * from @Result
于 2010-01-10T20:54:13.220 回答
0

我面前没有数据库设置，所以我希望下面的作品首先拍摄。像这样的模式应该会导致更快的查询......你只是加入了两次，类似的聚合量：

Declare @StartDate datetime, @EndDate datetime 
Select @StartDate=Min(Date), @EndDate=Max(Date) from Awards 
; 
WITH AllDays(Date) AS (SELECT @StartDate UNION ALL SELECT DATEADD(d, 1, Date) 
                       FROM AllDays 
                       WHERE Date < @EndDate)

SELECT d.Date, sum(day.Amount) Amount_Total, sum(running.Amount) Amount_RunningTotal
FROM AllDays d  
     LEFT JOIN (SELECT date, SUM(Amount) As Amount
                FROM Awards 
                GROUP BY Date) day
          ON d.Date = day.Date
     LEFT JOIN (SELECT date, SUM(Amount) As Amount
                FROM Awards 
                GROUP BY Date) running 
                ON (d.Date >= running.Date)
Group by d.Date
Order by d.Date

注意：我在顶部更改了您的表格表达式，它在前一天被遗漏了......如果这是故意的，只需在其上添加一个 where 子句以排除它。如果这不起作用或不适合，请在评论中告诉我，我会做出任何调整。

于 2010-01-10T19:24:46.740 回答