[python] Choosing a file in Python with simple Dialog

I would like to get file path as input in my Python console application.

Currently I can only ask for full path as an input in the console.

Is there a way to trigger a simple user interface where users can select file instead of typing the full path?

I obtained much better results with wxPython than tkinter, as suggested in this answer to a later duplicate question:

https://stackoverflow.com/a/9319832

The wxPython version produced the file dialog that looked the same as the open file dialog from just about any other application on my OpenSUSE Tumbleweed installation with the xfce desktop, whereas tkinter produced something cramped and hard to read with an unfamiliar side-scrolling interface.

Python 3.x version of Etaoin's answer for completeness:

from tkinter.filedialog import askopenfilename
filename = askopenfilename()

With EasyGui:

import easygui
print(easygui.fileopenbox())

To install:

pip install easygui

Demo:

import easygui
easygui.egdemo()

In Python 2 use the tkFileDialog module.

import tkFileDialog

tkFileDialog.askopenfilename()

In Python 3 use the tkinter.filedialog module.

import tkinter.filedialog

tkinter.filedialog.askopenfilename()

Another option to consider is Zenity: http://freecode.com/projects/zenity.

I had a situation where I was developing a Python server application (no GUI component) and hence didn't want to introduce a dependency on any python GUI toolkits, but I wanted some of my debug scripts to be parameterized by input files and wanted to visually prompt the user for a file if they didn't specify one on the command line. Zenity was a perfect fit. To achieve this, invoke "zenity --file-selection" using the subprocess module and capture the stdout. Of course this solution isn't Python-specific.

Zenity supports multiple platforms and happened to already be installed on our dev servers so it facilitated our debugging/development without introducing an unwanted dependency.