Select filename using FileDialog - vba

Question

I am trying to get a fullpath and filename using the file dialog from MS Access VBA.
What I am trying to do is to open the file Dialog on button click by calling this function. This function should return the fullpath and filename that was selected from the filedialog.
I commented the loop part because I only want to select single file.
This function is returning an error Error: 0 after I select a file
So far this is my code.
Anyone can figure out what's wrong?
Thanks
Public Function SelectTheFile() As String
On Error GoTo SelectTheFile_ErrorHandler
Dim fDialog As Office.FileDialog
Dim varFile As Variant
Set fDialog = Application.FileDialog(msoFileDialogFilePicker)
With fDialog
.AllowMultiSelect = False
.Title = "Please select one file"
If .Show = True Then
'For Each varFile In .SelectedItems
'SelectTheFile = varFile
'Debug.Print SelectTheFile
'Next
SelectTheFile = .SelectedItems(1)
Debug.Print SelectTheFile
Else
Debug.Print "Cancel"
End If
End With
SelectTheFile_ErrorHandler:
Set fd = Nothing
MsgBox "Error " & Err & ": " & Error(Err)
End Function

Answer 1

your code always reach the SelectTheFile_ErrorHandler: section whatever the file dialog result
you must exit the function before that section
Public Function SelectTheFile() As String
On Error GoTo SelectTheFile_ErrorHandler
Dim fDialog As Office.FileDialog
Dim varFile As Variant
Set fDialog = Application.FileDialog(msoFileDialogFilePicker)
With fDialog
.AllowMultiSelect = False
.Title = "Please select one file"
If .Show = True Then
'For Each varFile In .SelectedItems
'SelectTheFile = varFile
'Debug.Print SelectTheFile
'Next
SelectTheFile = .SelectedItems(1)
Debug.Print SelectTheFile
Else
Debug.Print "Cancel"
End If
End With
Exit Function '<==== exit here, otherwise code goes on to following section
SelectTheFile_ErrorHandler:
Set fDialog = Nothing
MsgBox "Error " & Err & ": " & Error(Err)
End Function

Answer 2

Try just using this:
Application.GetOpenFilename
That works for me and saves the full file path without actually opening the file. Much simpler unless I am missing what you're trying to do. Read more here in the docs: https://msdn.microsoft.com/en-us/library/office/ff834966.aspx