For i = 1 To Len(strmessage)
If Asc(Mid(strmessage, i, 1)) >= 48 And Asc(Mid(strmessage, i, 1)) <= 57 Then
cue = cue & Mid(strmessage, i, 1)
End If
Next
End Function作者: huangqinyong 时间: 2007-8-13 16:38
俺试试作者: chenyingfengsx 时间: 2009-8-21 19:02
看看作者: Henry D. Sy 时间: 2009-8-21 19:09
可以判断是不是数字的位置作者: Henry D. Sy 时间: 2009-8-21 19:42
这样也可以
Public Function gStr(strFlds As String) As Long
Dim i As Integer
For i = 1 To Len(strFlds)
If IsNumeric(Mid(strFlds, i, 1)) Then
Exit For
End If
Next
gStr = Val(Mid(strFlds, i))
End Function