如何获取文件名而不是文件的整个文件路径？ [英] How do you get just the filename rather than the entire file path of an open file?

问题描述

换句话说，在调用 Application.GetOpenFileName（）方法后，是否需要进行一些字符串处理？

解决方案

我正在使用这些函数进行文件名处理。最后一个是您需要的那个。

 公共函数FilePathOf（ByVal as As String）As String 
 Dim pos As Integer 
 
 pos = InStrRev（s，\）
如果pos = 0然后
 FilePathOf =
 Else 
 FilePathOf =左$（s，pos）
结束如果
结束函数
 
公共函数FileNameOf（ByVal s As String）As String 
 Dim pos1 As Integer，pos2 As Integer 
 
 pos1 = InStrRev（s，\）+ 1 
 pos2 = InStrRev（s，。）
如果pos2 = Len（s）则pos2 = pos2 + 1 
如果pos2 = 0则pos2 = Len（s）+ 1 
 FileNameOf = Mid $（s，pos1，pos2  -  pos1）
结束函数
 
公共函数FileExtOf（ByVal s As String）As String 
 Dim pos As Integer 
 
 pos = InStrRev（s，。）
如果pos = 0然后
 FileExtOf =
 Else 
 FileExtOf = Mid $（s，pos + 1）
如果
结束函数
 
公共函数FileNameExt （ByVal s As String）As String 
 FileNameExtOf = Mid $（s，InStrRev（s，\）+ 1）
结束函数
  / pre> 
In other words, would I need to do some string processing after invoking the Application.GetOpenFileName() Method?
解决方案
I am using these functions for filename processing. The last one is the one you need here.
Public Function FilePathOf(ByVal s As String) As String
    Dim pos As Integer

    pos = InStrRev(s, "\")
    If pos = 0 Then
        FilePathOf = ""
    Else
        FilePathOf = Left$(s, pos)
    End If
End Function

Public Function FileNameOf(ByVal s As String) As String
    Dim pos1 As Integer, pos2 As Integer

    pos1 = InStrRev(s, "\") + 1
    pos2 = InStrRev(s, ".")
    If pos2 = Len(s) Then pos2 = pos2 + 1
    If pos2 = 0 Then pos2 = Len(s) + 1
    FileNameOf = Mid$(s, pos1, pos2 - pos1)
End Function

Public Function FileExtOf(ByVal s As String) As String
    Dim pos As Integer

    pos = InStrRev(s, ".")
    If pos = 0 Then
        FileExtOf = ""
    Else
        FileExtOf = Mid$(s, pos + 1)
    End If
End Function

Public Function FileNameExtOf(ByVal s As String) As String
    FileNameExtOf = Mid$(s, InStrRev(s, "\") + 1)
End Function


                        
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