declare @dateBegin datetime, @dateEnd datetime, @dateTmp datetime
declare @day int, @i int, @j int
select @dateBegin = '20071001', @dateEnd = getDate()
select @day = datediff(dd,@dateBegin,@dateEnd)

select @i = 1, @j = 1
while (@i <= @day)
begin
  select @dateTmp = dateadd(dd,@i,@dateBegin)
  if (datePart(weekday,@dateTmp) > 1 and datePart(weekday,@dateTmp) < 7)
    select @j = @j + 1
  select @i = @i + 1
end
select @j '两个日期差距的天数'

------解决方案--------------------

按 hb_gx  的方法 写成一个函数

------解决方案--------------------
算法可以这样
1. 先求出差几天
2. 再除以 7 算出是 N = 几周,  -> 2N
3. 再判断 被7整除的余数有几天是周六，周日, m
4. select 2N + m
---------
这样快一点



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

declare @k table(datestr datetime)
insert into @k select '2007-10-01'
union all select '2007-10-02'
union all select '2007-10-03'
union all select '2007-10-04'
union all select '2007-10-05'
union all select '2007-10-06'
union all select '2007-10-07'
union all select '2007-10-08'
union all select '2007-10-09'
union all select '2007-10-10'
union all select '2007-10-11'
union all select '2007-10-12'
union all select '2007-10-13'
union all select '2007-10-14'
select * from @k

select count(1) from @k where convert(varchar(10),datestr,120) between '2007-10-01' and '2007-10-14' and datepart(dw,datestr)<>7