Question

我如何获得文件名（没有路径和扩展名）
喜欢＆＃34; MyFileName＆＃34;
从以下完整路径？
C：\ A_B \ C.D \ E_ \ F0123456789 \ G变\ MyFileName.txt

Answer 1

InStrRev会在字符串中找到最后一个字符。搜索\并将其拆分

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)

Answer 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

Answer 3

如果它是您可以访问的真实文件，则可以使用Dir

sFileOnly = Dir(sPathAndFile)

如果它不是真实文件或您无法访问它，则会返回一个空字符串。

Answer 4

Set regEx = New RegExp
regEx.Pattern = ".*\\"
regEx.IgnoreCase = True
filename = regEx.Replace(fullpath, "")

Answer 5

    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

Excel VBA：正则表达式 - 获取文件名

5 个答案: