1

如何 从以下完整路径中 获取
像“MyFileName”这样 的文件名（没有路径和扩展名）？ C:\A_B\CD\E_\F0123456789\G\MyFileName.txt

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5 回答 5

2

InStrRev将查找字符串中最后一次出现的字符。在那里搜索\并拆分它

FullFileName="C:\A_B\C.D\E_\F0123456789\G\MyFileName.txt"
FileName=mid(FullFileName,instrrev(FullFileName,"\")+1)

现在取消扩展

FileNameWithoutExt=left(FileName,instrrev(FileName,".")-1)
于 2012-09-10T15:29:23.053 回答
2 
Public Function GetFileNameWithoutExt(ByVal fullPath As String) As String
Dim fileName As String
Dim fileNameWithoutExt As String

Dim lastSlash As Integer
Dim positionOfDot As Integer

lastSlash = InStrRev(fullPath, "\")
fileName = Mid(fullPath, lastSlash + 1)

positionOfDot = InStr(1, fileName, ".")
fileNameWithoutExt = Mid(fileName, 1, positionOfDot - 1)

GetFileNameWithoutExt = fileNameWithoutExt
End Function

使用即时窗口

?GetFileNameWithoutExt("C:\A_B\C.D\E_\F0123456789\G\MyFileName.txt")

编辑：另一种方法

Public Function GetFileNameWithoutExt2(ByVal fullPath As String) As String
Dim fileName As String
Dim splittedData
Dim fileNameWithoutExt As String

splittedData = Split(fullPath, "\")
fileName = splittedData(UBound(splittedData))

fileNameWithoutExt = Split(fileName, ".")(0)

GetFileNameWithoutExt2 = fileNameWithoutExt
End Function
于 2012-09-10T15:30:52.517 回答
1

If it's a real file that you have access to, you can use Dir

sFileOnly = Dir(sPathAndFile)

If it's not a real file or you don't have access to it, this will return an empty string.

于 2012-09-10T18:30:23.700 回答
0 
    Sub Test()
        Dim fileNameOnly As String
        fileNameOnly = Left$(Split("C:\A_B\C.D\E_\F0123456789\G\MyFileName.txt", "\")(UBound(Split("C:\A_B\C.D\E_\F0123456789\G\MyFileName.txt", "\"))), InStrRev(Split("C:\A_B\C.D\E_\F0123456789\G\MyFileName.txt", "\")(UBound(Split("C:\A_B\C.D\E_\F0123456789\G\MyFileName.txt", "\"))), ".") - 1)
        Debug.Print Strtf    
    End Sub
于 2012-09-10T15:39:37.840 回答
0 
Set regEx = New RegExp
regEx.Pattern = ".*\\"
regEx.IgnoreCase = True
filename = regEx.Replace(fullpath, "")
于 2012-09-10T15:22:09.777 回答