假设我有一个奖励的SQL表,其中包含日期和金额字段。我需要生成一个包含连续日期序列的表,每天授予的金额以及运行(累计)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
这个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)
我尝试为Awards.Date添加索引,但它的差异非常小。
在采用其他策略(如缓存)之前,是否有更有效的方法来编写运行总计算?
答案 0 :(得分:3)
我通常使用临时表:
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
如果您无法将日期列设为主键,则需要在ORDER BY [Date]
语句中加入INSERT
。
此外,此问题曾被问过几次。请参阅here或搜索“sql running total”。据我所知,我发布的解决方案仍然是性能最佳的解决方案,并且易于编写。
答案 1 :(得分:0)
我没有在我面前设置数据库,所以我希望以下作品能够首先拍摄。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:我将表格表达式更改为顶部,它在前一天遗漏了...如果这是故意的,只需要在其上打一个where子句来排除它。如果这不起作用或不合适,请在评论中告诉我,我会做出任何调整。
答案 2 :(得分:0)
这是一个基于@Aaronaught答案的工作解决方案。我必须在T-SQL中克服的唯一问题是@RunningTotal
等不能为空(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