存储过程时间问题
@d datetime

SELECT
ISNULL (SUM(CASE WHEN qfrq BETWEEN CONVERT(varchar(8),@d,120)+'01'
AND @d
THEN sjfl ELSE 0 END),0) AS 本月方量
这个是查询自然月的 可现在需要25-25的结账日
比如：2012-08-21 那应该查询 2012-07-26-------201208-21的数据
2012-07-25 那应该查询 2012-06-26------2012-07-25
如何做？

[解决办法]

SQL code

->?declare @dt datetimeset @dt='2012-08-21'select qfrq between convert(varchar(7),dateadd(month,-1,@dt)),120)+'-25' and @dt from tab
[解决办法]
SQL code
declare @d0 datetimeset @d0='1900-1-25'declare @time datetimeset @time='2011-1-1'--用偏移量计算select case when DAY(@time)>=25 then DATEADD(MONTH,DATEDIFF(month,@d0,@time),@d0) else DATEADD(MONTH,DATEDIFF(month,@d0,@time)-1,@d0) end
[解决办法]
SQL code
/*创建Nums表，1W行                                    */ CREATE TABLE dbo.Nums(n INT NOT NULL PRIMARY KEY);DECLARE @max AS INT, @rc AS INT;SET @max = 10000;SET @rc = 1;INSERT INTO Nums VALUES(1);WHILE @rc * 2 <= @maxBEGIN  INSERT INTO dbo.Nums SELECT n + @rc FROM dbo.Nums;  SET @rc = @rc * 2;ENDINSERT INTO dbo.Nums   SELECT n + @rc FROM dbo.Nums WHERE n + @rc <= @max;GO/*函数用于返回时间的月份序列，类似方法可以产生日序列，周序列 等等        参数：@s 开始时间                    @e 结束时间                     *磊仔 2012年5月29日创建                                       */   Create function dbo.fn_DayList(@s as datetime,@e as datetime) returns tableasreturnwith CET1 AS(        SELECT  cast(convert(varchar(6),dateadd(mm, n - 1,@s) ,112) as int) as [Month],--月序列            dateadd(dd,25,dateadd(mm,-1,dateadd(d,datediff(d,0,dateadd(mm, n - 1,@s)),0) - DATEPART(dd, dateadd(mm, n - 1,@s))+1))as StartT,                dateadd(dd,25,dateadd(d,datediff(d,0,dateadd(mm, n - 1,@s)),0) - DATEPART(dd, dateadd(mm, n - 1,@s))+1) as EndT        FROM dbo.Nums        WHERE n <= DATEDIFF(mm, @s, @e))SELECT * FROM CET1        GO--创建完以上表值函数后就可轻松调用，且性能极佳declare @time datetimeset @time='2011-7-25'select StartT, @Time as NowTime     from dbo.fn_DayList('19000101','20500101')         where StartT <= @Time and EndT > @Time