declare @date1 datetime,@date2 datetime
set @date1='2011-05-12'
set @date2='2011-05-15'
select 'yes' where datediff(d,@date1,@date2)<3+(case when datepart(dw,@date1)>datepart(dw,@date2) then 1 else 0 end)
/*
----
yes

(1 行受影响)

*/

------解决方案--------------------
SQL code

/*
*********************************************************************************
Program ID: FUN_GetDaysOfWeeks
Purpose   : 求某段时间内某一星期日期的天数
Author    : Jesse
Date      : 2011.02.19
*********************************************************************************
*/

Create Function dbo.FUN_GetDaysOfWeeks(    @DateS smalldatetime,
                                        @DateE smalldatetime,
                                        @WeekValue int)    --星期的某一天：1-周日、2-周一...7-周六
Returns int 
As
Begin
    declare
        @nResult int,            --结果天数
        @nWeekValue int,        --起始日期是星期几(1-周日、2-周一...7-周六)
        @nDays int,                --时间段内的天数
        @nBeforeDays int,        --起始日期到第一次相符日期的天数
        @sWeekName varchar(20),    --所求日期的星期名称
        @tmpDate smalldatetime
    
    set @WeekValue=@WeekValue % 7
    if @WeekValue=0 
        set @WeekValue=7

    --如果起始日期大于终止日期，则对换
    if @DateS>@DateE
        select @tmpDate=@DateE,@DateE=@DateS,@DateS=@tmpDate

    --取得起始日期的星期数，与设置无关(1-周日、2-周一...7-周六)
    select @nWeekValue=(Datepart(dw,@DateS)+(@@Datefirst %7))%7
    if @nWeekValue=0
        set @nWeekValue=7
    
    set @sWeekName=DateName(dw,DateAdd(d,@WeekValue-@nWeekValue,@DateS))

    if @WeekValue>=@nWeekValue
        set @nBeforeDays=@WeekValue-@nWeekValue
    else
        set @nBeforeDays=7-abs(@WeekValue-@nWeekValue)

    --取得时间段内的天数
    select @nDays=Datediff(d,@DateS,@DateE)+1

    if @nDays<@nBeforeDays
        select @nResult=0
    else
        select @nResult=ceiling((@nDays-@nBeforeDays)/7.0)


    Return @nResult
end    

/*
    --select @@datefirst
    select dbo.FUN_GetDaysOfWeeks('2011-03-01','2011-05-31',7)
*/

--先建一函数，再利用函数计算得出
select * from t where datediff(d,date1,date2)- dbo.FUN_GetDaysOfWeeks('2011-03-01','2011-05-31',7)<=3

------解决方案--------------------
SQL code

select (case when date2-date1>4 then 0 else when datename(dw,date1) <3 then 0 else then 1) as Complete from tb

------解决方案--------------------
这个是需要写函数来判断时间在所在区间内且是不是星期六

------解决方案--------------------

楼主需求不太明确,猜测如下:
SQL code
create table table1(date1 datetime,date2 datetime)
insert into table1 select '2011-5-9','2011-5-10'
insert into table1 select '2011-5-9','2011-5-15'
insert into table1 select '2011-5-13','2011-5-10'
insert into table1 select '2011-5-12','2011-5-15'
go
select date1,date2,'满足要求' from table1
 where datediff(d,date1,date2)<3+(case when datepart(dw,date1)>datepart(dw,date2) then 1 else 0 end)
go
drop table table1
/*
date1                   date2                   
----------------------- ----------------------- --------
2011-05-09 00:00:00.000 2011-05-10 00:00:00.000 满足要求
2011-05-13 00:00:00.000 2011-05-10 00:00:00.000 满足要求
2011-05-12 00:00:00.000 2011-05-15 00:00:00.000 满足要求

(3 行受影响)

*/