帮我看看datadiff这个函数！

[yujiang930]

Private Sub Form_Load()
Dim aa, ii, ss
For ii = 1 To 31
If Month(DateSerial(Year(Date), Month(Date), ii)) > Month(Date) Then Exit For
ss = Year(Date) & "-" & Month(Date) & "-" & ii
If Weekday(CDate(ss), vbMonday) = "1" Then
aa = aa & ss & ","
End If
ii = ii + 1
Next
Debug.Print aa
End Sub
==========================================
为什么我执行出来的结果是2007-2-5,2007-2-19, 2月12号和26号为什么没有？？

[feiyun0112]

去掉 ii = ii + 1

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最新版本：20070130

http://www.cnblogs.com/feiyun0112/archive/2006/09/20/509783.html

[hupeng213]

DATEDIFF ( datepart , startdate , enddate )

DateDiff(year,'2007-02-01 15:30:00','2007-02-02 17:30:00')=0

DateDiff(month,'2007-02-01 15:30:00','2007-02-02 17:30:00')=0

DateDiff(day,'2007-02-01 15:30:00','2007-02-02 17:30:00')=1

select DateDiff(hour,'2007-02-01 15:30:00','2007-02-02 17:30:00')

没必要那么复杂吧

[yujiang930]

我要得到这个月所以星期一的日期

[yujiang930]

谢谢
但是单独执行weekday(cdate(2007-2-12),vbMonday)这个怎么返回的4？
我就是应为这个才忘记看ii=ii+1了

[feiyun0112]

cdate("2007/02/12")

用字符串

[wxrwan]

weekday( #2007-2-12#,vbMonday)=1

[wxrwan]

cdate(2007-2-12)=1905-6-15

[guyehanxinlei]

呵呵！