帮我看看datadiff这个函数！
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
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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？？
[解决办法]
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
[解决办法]
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
[解决办法]
cdate( "2007/02/12 ")
用 字符串
[解决办法]
weekday( #2007-2-12#,vbMonday)=1
