GetOpenFileName

I hope your macro code is only showing the openfile dialog ( using GetOpen FileName), do you want to hardcode the filename here instead of returning the selected value from browse dialog, then i think it will solve your problem, hope my understanding of your requirement is correct, let me know otherwise,

Sub OpenFile()
    Dim fileName
    
    fileName = Application.GetOpenFilename("Comma Separated Values (*.csv),*.csv")
    
    If fileName <> "False" Then

      //
    End If
End Sub