 --方法一 declare @month int, @year int select @month = 6, @year = 2004 select datepart(dd,dateadd(m,1,cast(str(@month) + /01/ + str(@year) as datetime)) - 1) --方法二 declare @date datetime set @date = getdate() select cast( -- first day of next month dateadd(month,datediff(month,0,@date)+1,0) - -- first day of current month dateadd(month,datediff(month,0,@date),0) as int ) --方法三 这个您也能够写一个udf来实现： create function fn_numofdaysinmon (@date datetime) returns int as begin declare @begmonth datetime,@endmonth datetime,@days int select @begmonth=dateadd(month,datediff(month,0,@date),0) select @endmonth=dateadd(dd,-datepart(dd,@date),dateadd(mm,1,@date)) select @days=datediff(dd,@begmonth,@endmonth)+1 return @days end |
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