Dim dialog As OpenFileDialog = New OpenFileDialog
If dialog.ShowDialog() = Windows.Forms.DialogResult.OK Then
Dim filename As String
filename = dialog.FileName
Dim results() As String
results = filename.Split("\")
filename = results(results.Length - 1)
filename = filename.Substring(0, filename.LastIndexOf("."))
MessageBox.Show(filename)
End If
dialog.Dispose()
system.io.path.getfilename(FilePath)
用祥基vb怎样获得程序本身的文件乎宴滑名和路径?万分感谢 App.Path 路径 App.EXEName 文件名岁腊
filename.split()