[RESOLVED] how to get path location

I used common dialog, How I can get the path location of my file?

Code:

---


dlgOpen.Filter = "Tin Files (*.tin)|*.tin"
dlgOpen.ShowOpen
InputFileTIN = dlgOpen.FileName
MsgBox dlgOpen.FileName

---

There's a few ways, ie using Split with "\" as the delimiter, APIs and:

vb Code:

Private Sub Form_Load()
   MsgBox GetDirectoryFromPath("E:\VideoPaper\Videos\Atomic Kitten - Whole Again.mpg")
   MsgBox GetFilenameFromPath("E:\VideoPaper\Videos\Atomic Kitten - Whole Again.mpg")
End Sub
 
Public Function GetDirectoryFromPath(FullPath As String) As String
'Returns "E:\VideoPaper\Videos\".
   GetDirectoryFromPath = Left(FullPath, InStrRev(FullPath, "\"))
End Function
 
Public Function GetFilenameFromPath(FullPath As String) As String
'Returns "Atomic Kitten - Whole Again.mpg".
   GetFilenameFromPath = Right(FullPath, Len(FullPath) - InStrRev(FullPath, "\"))
End Function

I use common dialog I got this messagebox "C:\Project\D.tin"

dlgOpen.FileName

i want to get this C:\Project\

Using the code above:

vb Code:

Option Explicit
 
Private Sub Form_Load()
   Dim strFN As String
   
   CommonDialog.Filter = "All Files (*.*)|*.*"
   CommonDialog.ShowOpen
   strFN = CommonDialog.FileName
   MsgBox CommonDialog.FileName
 
   MsgBox GetDirectoryFromPath(CommonDialog.FileName)
End Sub
 
Public Function GetDirectoryFromPath(FullPath As String) As String
   GetDirectoryFromPath = Left(FullPath, InStrRev(FullPath, "\"))
End Function