["python","openfiledialog"]
Quick and easy file dialog in Python?

I have a simple script which parses a file and loads it's contents to a database. I don't need a UI, but right now I'm prompting the user for the file to parse using raw_input which is most unfriendly, especially because the user can't copy/paste the path. I would like a quick and easy way to present a file selection dialog to the user, they can select the file, and then it's loaded to the database. (In my use case, if they happened to chose the wrong file, it would fail parsing, and wouldn't be a problem even if it was loaded to the database.) import tkFileDialog file_path_string = tkFileDialog.askopenfilename() This code is close to what I want, but it leaves an annoying empty frame open (which isn't able to be closed, probably because I haven't registered a close event handler). I don't have to use tkInter, but since it's in the Python standard library it's a good candidate for quickest and easiest solution. Whats a quick and easy way to prompt for a file or filename in a script without any other UI?
Tkinter is the easiest way if you don't want to have any other dependencies. To show only the dialog without any other GUI elements, you have to hide the root window using the withdraw method: import tkinter as tk from tkinter import filedialog root = tk.Tk() root.withdraw() file_path = filedialog.askopenfilename() Python 2 variant: import Tkinter, tkFileDialog root = Tkinter.Tk() root.withdraw() file_path = tkFileDialog.askopenfilename()