14

我有字符串“ololo123”。我需要获取第一位数字的位置 - 1. 如何设置搜索掩码？

4

6 回答 6

11

这是一种轻量级且快速的方法，可避免添加正则表达式/引用，因此如果这是一个优势，则有助于降低开销和可移植性。

Public Function GetNumLoc(xValue As String) As Integer

For GetNumLoc = 1 To Len(xValue)
    If Mid(xValue, GetNumLoc, 1) Like "#" Then Exit Function
Next

GetNumLoc = 0

End Function
于 2015-08-06T19:09:10.863 回答
9

Something like this should do the trick for you:

Public Function GetPositionOfFirstNumericCharacter(ByVal s As String) As Integer
    For i = 1 To Len(s)
        Dim currentCharacter As String
        currentCharacter = Mid(s, i, 1)
        If IsNumeric(currentCharacter) = True Then
            GetPositionOfFirstNumericCharacter = i
            Exit Function
        End If
    Next i
End Function

You can then call it like this:

Dim iPosition as Integer
iPosition = GetPositionOfFirstNumericCharacter("ololo123")
于 2010-08-23T13:10:02.230 回答
2

我实际上有这个功能：

Public Function GetNumericPosition(ByVal s As String) As Integer
    Dim result As Integer
    Dim i As Integer
    Dim ii As Integer

    result = -1
    ii = Len(s)
    For i = 1 To ii
        If IsNumeric(Mid$(s, i, 1)) Then
            result = i
            Exit For
        End If
    Next
    GetNumericPosition = result
End Function
于 2010-08-23T14:19:46.087 回答
2

不确定您的环境，但这在 Excel 2010 中有效

'Added reference for Microsoft VBScript Regular Expressions 5.5

Const myString As String = "ololo123"
Dim regex As New RegExp
Dim regmatch As MatchCollection

regex.Pattern = "\d"
Set regmatch = regex.Execute(myString)
MsgBox (regmatch.Item(0).FirstIndex)   ' Outputs 5
于 2010-08-23T14:01:29.577 回答
1

你可以试试正则表达式，然后你会遇到两个问题。我的 VBAfu 达不到标准，但我会试一试：

Function FirstDigit(strData As String) As Integer
    Dim RE As Object REMatches As Object

    Set RE = CreateObject("vbscript.regexp")
    With RE
        .Pattern = "[0-9]"
    End With

    Set REMatches = RE.Execute(strData)
    FirstDigit = REMatches(0).FirstIndex
End Function

然后你只需用FirstDigit("ololo123").

于 2010-08-23T13:27:59.197 回答
0

If speed is an issue, this will run a bit faster than Robs (noi Rob):

Public Sub Example()
    Const myString As String = "ololo123"
    Dim position As Long
    position = GetFirstNumeric(myString)
    If position > 0 Then
        MsgBox "Found numeric at postion " & position & "."
    Else
        MsgBox "Numeric not found."
    End If
End Sub

Public Function GetFirstNumeric(ByVal value As String) As Long
    Dim i As Long
    Dim bytValue() As Byte
    Dim lngRtnVal As Long
    bytValue = value
    For i = 0 To UBound(bytValue) Step 2
        Select Case bytValue(i)
            Case vbKey0 To vbKey9
                If bytValue(i + 1) = 0 Then
                    lngRtnVal = (i \ 2) + 1
                    Exit For
                End If
        End Select
    Next
    GetFirstNumeric = lngRtnVal
End Function
于 2010-08-23T13:14:36.747 回答