cod*_*ike 158 sql t-sql sql-server running-total
想象一下下表(称为TestTable):
id somedate somevalue
-- -------- ---------
45 01/Jan/09 3
23 08/Jan/09 5
12 02/Feb/09 0
77 14/Feb/09 7
39 20/Feb/09 34
33 02/Mar/09 6
我想要一个以日期顺序返回运行总计的查询,例如:
id somedate somevalue runningtotal
-- -------- --------- ------------
45 01/Jan/09 3 3
23 08/Jan/09 5 8
12 02/Feb/09 0 8
77 14/Feb/09 7 15
39 20/Feb/09 34 49
33 02/Mar/09 6 55
我知道在SQL Server 2000/2005/2008中有各种方法可以做到这一点.
我对使用aggregate-set-statement技巧的这种方法特别感兴趣:
INSERT INTO @AnotherTbl(id, somedate, somevalue, runningtotal)
SELECT id, somedate, somevalue, null
FROM TestTable
ORDER BY somedate
DECLARE @RunningTotal int
SET @RunningTotal = 0
UPDATE @AnotherTbl
SET @RunningTotal = runningtotal = @RunningTotal + somevalue
FROM @AnotherTbl
...这非常有效但我听说有这方面的问题,因为你不一定能保证UPDATE语句将以正确的顺序处理行.也许我们可以就这个问题得到一些明确的答案.
但也许人们可以建议其他方式？
编辑:现在使用带有设置的SqlFiddle和上面的"更新技巧"示例
Sam*_*ron 127
更新,如果您运行的是SQL Server 2012,请参阅:https://stackoverflow.com/a/10309947
问题是Over子句的SQL Server实现有些限制.
Oracle(和ANSI-SQL)允许您执行以下操作:
SELECT somedate, somevalue,
SUM(somevalue) OVER(ORDER BY somedate
ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)
AS RunningTotal
FROM Table
SQL Server为您提供此问题的干净解决方案.我的直觉告诉我,这是极少数情况下光标最快的情况之一,但我必须对大结果做一些基准测试.
更新技巧很方便但我觉得它相当脆弱.看来,如果要更新完整的表,那么它将按主键的顺序进行.因此,如果您将日期设置为主键升序,那么您将probably是安全的.但是你依赖于一个未记录的SQL Server实现细节(如果查询最终由两个proc执行,我想知道会发生什么,请参阅:MAXDOP):
完整的工作样本:
drop table #t
create table #t ( ord int primary key, total int, running_total int)
insert #t(ord,total) values (2,20)
-- notice the malicious re-ordering
insert #t(ord,total) values (1,10)
insert #t(ord,total) values (3,10)
insert #t(ord,total) values (4,1)
declare @total int
set @total = 0
update #t set running_total = @total, @total = @total + total
select * from #t
order by ord
ord total running_total
----------- ----------- -------------
1 10 10
2 20 30
3 10 40
4 1 41
你问了一个基准,这是低点.
执行此操作的最快SAFE方式是Cursor,它比交叉连接的相关子查询快一个数量级.
绝对最快的方法是UPDATE技巧.我唯一担心的是,我不确定在任何情况下更新都会以线性方式进行.查询中没有明确说明的内容.
底线,对于生产代码,我会使用光标.
测试数据:
create table #t ( ord int primary key, total int, running_total int)
set nocount on
declare @i int
set @i = 0
begin tran
while @i < 10000
begin
insert #t (ord, total) values (@i, rand() * 100)
set @i = @i +1
end
commit
测试1:
SELECT ord,total,
(SELECT SUM(total)
FROM #t b
WHERE b.ord <= a.ord) AS b
FROM #t a
-- CPU 11731, Reads 154934, Duration 11135
测试2:
SELECT a.ord, a.total, SUM(b.total) AS RunningTotal
FROM #t a CROSS JOIN #t b
WHERE (b.ord <= a.ord)
GROUP BY a.ord,a.total
ORDER BY a.ord
-- CPU 16053, Reads 154935, Duration 4647
测试3:
DECLARE @TotalTable table(ord int primary key, total int, running_total int)
DECLARE forward_cursor CURSOR FAST_FORWARD
FOR
SELECT ord, total
FROM #t
ORDER BY ord
OPEN forward_cursor
DECLARE @running_total int,
@ord int,
@total int
SET @running_total = 0
FETCH NEXT FROM forward_cursor INTO @ord, @total
WHILE (@@FETCH_STATUS = 0)
BEGIN
SET @running_total = @running_total + @total
INSERT @TotalTable VALUES(@ord, @total, @running_total)
FETCH NEXT FROM forward_cursor INTO @ord, @total
END
CLOSE forward_cursor
DEALLOCATE forward_cursor
SELECT * FROM @TotalTable
-- CPU 359, Reads 30392, Duration 496
测试4:
declare @total int
set @total = 0
update #t set running_total = @total, @total = @total + total
select * from #t
-- CPU 0, Reads 58, Duration 139
Mik*_*son 117
在SQL Server 2012中,您可以将SUM()与OVER()子句一起使用.
select id,
somedate,
somevalue,
sum(somevalue) over(order by somedate rows unbounded preceding) as runningtotal
from TestTable
Rom*_*kar 40
虽然Sam Saffron在这方面做了很多工作,但他仍然没有提供这个问题的递归公用表表达式代码.对于使用SQL Server 2008 R2而不是Denali的我们来说,它仍然是获得总计运行速度的最快方式,它比我的工作计算机上的光标快10倍,而且它也是内联查询.
所以,这里是(我假设表中有一ord列,它的序列号没有间隙,为了快速处理,这个数字也应该有唯一约束):
;with
CTE_RunningTotal
as
(
select T.ord, T.total, T.total as running_total
from #t as T
where T.ord = 0
union all
select T.ord, T.total, T.total + C.running_total as running_total
from CTE_RunningTotal as C
inner join #t as T on T.ord = C.ord + 1
)
select C.ord, C.total, C.running_total
from CTE_RunningTotal as C
option (maxrecursion 0)
-- CPU 140, Reads 110014, Duration 132
更新
我也很好奇这个更新与变量或奇怪的更新.所以通常它工作正常,但我们如何确保它每次都有效？好吧,这里有一个小技巧(在这里找到 - http://www.sqlservercentral.com/Forums/Topic802558-203-21.aspx#bm981258) - 你只需检查当前和之前的情况ord并使用1/0任务,以防它们与什么不同你期待:
declare @total int, @ord int
select @total = 0, @ord = -1
update #t set
@total = @total + total,
@ord = case when ord <> @ord + 1 then 1/0 else ord end,
------------------------
running_total = @total
select * from #t
-- CPU 0, Reads 58, Duration 139
从我所看到的,如果你的表上有适当的聚集索引/主键(在我们的例子中它将是索引ord_id),更新将一直以线性方式进行(从未遇到除以零).也就是说,由您来决定是否要在生产代码中使用它:)
小智 28
SQL 2005及更高版本中的APPLY运算符适用于此:
select
t.id ,
t.somedate ,
t.somevalue ,
rt.runningTotal
from TestTable t
cross apply (select sum(somevalue) as runningTotal
from TestTable
where somedate <= t.somedate
) as rt
order by t.somedate
Sam*_*Axe 11
SELECT TOP 25 amount,
(SELECT SUM(amount)
FROM time_detail b
WHERE b.time_detail_id <= a.time_detail_id) AS Total FROM time_detail a
您还可以使用ROW_NUMBER()函数和临时表来创建在内部SELECT语句的比较中使用的任意列.
使用相关的子查询.很简单,你走了:
SELECT
somedate,
(SELECT SUM(somevalue) FROM TestTable t2 WHERE t2.somedate<=t1.somedate) AS running_total
FROM TestTable t1
GROUP BY somedate
ORDER BY somedate
代码可能不完全正确,但我确信这个想法是正确的.
如果日期出现不止一次,则GROUP BY只会在结果集中看到一次.
如果您不介意重复日期,或者想要查看原始值和ID,那么以下是您想要的:
SELECT
id,
somedate,
somevalue,
(SELECT SUM(somevalue) FROM TestTable t2 WHERE t2.somedate<=t1.somedate) AS running_total
FROM TestTable t1
ORDER BY somedate
您还可以非规范化 - 在同一个表中存储运行总计:
选择工作的速度比任何其他解决方案快得多,但修改可能会更慢
小智 5
如果您使用的是 Sql server 2008 R2 以上。那么，这将是最短的方法；
Select id
,somedate
,somevalue,
LAG(runningtotal) OVER (ORDER BY somedate) + somevalue AS runningtotal
From TestTable
LAG用于获取前一行值。你可以做谷歌了解更多信息。
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