[excel] How to get selected path and name of the file opened with file dialog?

I need the path name and file name of the file that is opened with File Dialog. I want to show this information with a hyperlink in my worksheet.

With this code I have the file path:

Sub GetFilePath()

Set myFile = Application.FileDialog(msoFileDialogOpen)
With myFile
    .Title = "Choose File"
    .AllowMultiSelect = False
    If .Show <> -1 Then
        Exit Sub
    End If
FileSelected = .SelectedItems(1)
End With

ActiveSheet.Range("A1") = FileSelected
End Sub

I'm still looking for a way to get the filename.

I think this is the simplest way to get to what you want.

Credit to JMK's answer for the first part, and the hyperlink part was adapted from http://msdn.microsoft.com/en-us/library/office/ff822490(v=office.15).aspx

'Gets the entire path to the file including the filename using the open file dialog
Dim filename As String
filename = Application.GetOpenFilename

'Adds a hyperlink to cell b5 in the currently active sheet
With ActiveSheet
 .Hyperlinks.Add Anchor:=.Range("b5"), _
 Address:=filename, _
 ScreenTip:="The screenTIP", _
 TextToDisplay:=filename
End With

After searching different websites looking for a solution as to how to separate the full path from the file name once the full one-piece information has been obtained from the Open File Dialog, and seeing how "complex" the solutions given were for an Excel newcomer like me, I wondered if there could be a simpler solution. So I started to work on it on my own and I came to this possibility. (I have no idea if somebody got the same idea before. Being so simple, if somebody has, I excuse myself.)

Dim fPath As String
Dim fName As String
Dim fdString As String

fdString = (the OpenFileDialog.FileName)

'Get just the path by finding the last "\" in the string from the end of it
 fPath = Left(fdString, InStrRev(fdString, "\"))

'Get just the file name by finding the last "\" in the string from the end of it
 fName = Mid(fdString, InStrRev(fdString, "\") + 1)

'Just to check the result
 Msgbox "File path: " & vbLF & fPath & vbLF & vblF & "File name: " & vbLF & fName

AND THAT'S IT!!! Just give it a try, and let me know how it goes...

The below command is enough to get the path of the file from a dialog box -

my_FileName = Application.GetOpenFilename("Excel Files (*.tsv), *.txt")

You can get any part of the file path using the FileSystemObject. GetFileName(filepath) gives you what you want.

Modified code below:

Sub GetFilePath()
Dim objFSO as New FileSystemObject

Set myFile = Application.FileDialog(msoFileDialogOpen)
With myFile
.Title = "Choose File"
.AllowMultiSelect = False
If .Show <> -1 Then
Exit Sub
End If
FileSelected = .SelectedItems(1)
End With

ActiveSheet.Range("A1") = FileSelected 'The file path
ActiveSheet.Range("A2") = objFSO.GetFileName(FileSelected) 'The file name
End Sub

I think you want this:

Dim filename As String
filename = Application.GetOpenFilename

Dim cell As Range
cell = Application.Range("A1")
cell.Value = filename

Sub GetFilePath()

Set myFile = Application.FileDialog(msoFileDialogOpen)

With myFile

.Title = "Choose File"

.AllowMultiSelect = False

If .Show <> -1 Then

Exit Sub

End If

FileSelected = Replace(.SelectedItems(1), .InitialFileName, "")

End With

ActiveSheet.Range("A1") = FileSelected

End Sub

FileNameOnly = Dir(.SelectedItems(1))

From office 2010, we won't be able to use the common dialog box control, so it's nice to use the Application object to get the desired results.

Here I got a text box and Command button - paste the following code under the command button click event, which will open the file dialog box and add the File name to the Text box.

Dim sFileName  As String

sFileName = Application.GetOpenFilename("MS Excel (*.xlsx), *.xls")

TextBox1.Text = sFileName

To extract only the filename from the path, you can do the following:

varFileName = Mid(fDialog.SelectedItems(1), InStrRev(fDialog.SelectedItems(1), "\") + 1, Len(fDialog.SelectedItems(1)))

I think this will do:

Dim filename As String
filename = Application.GetOpenFilename

The code starts file search from root colon, If I want to start search from a specific directory, to avoid going to that directory every time, where I should put one. I did it like

Sub GetFilePath()
FileSelected = "G:\Audits\A2010"
Set myFile = Application.FileDialog(msoFileDialogOpen)
With myFile
.Title = "Choose File"
.AllowMultiSelect = False
If .Show <> -1 Then
Exit Sub
End If
FileSelected = .SelectedItems(1)
End With

ActiveSheet.Range("C14") = FileSelected
End Sub

But it could not start reach from "G:\Audits\A2010"