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
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()
You can use easygui:
import easygui path = easygui.fileopenbox()
easygui, you can use
pip3 install easygui
It is a single pure Python module (
easygui.py) that uses
Try with wxPython:
import wx def get_path(wildcard): app = wx.App(None) style = wx.FD_OPEN | wx.FD_FILE_MUST_EXIST dialog = wx.FileDialog(None, 'Open', wildcard=wildcard, style=style) if dialog.ShowModal() == wx.ID_OK: path = dialog.GetPath() else: path = None dialog.Destroy() return path print get_path('*.txt')
If you don’t need the UI or expect the program to run in a CLI, you could parse the filepath as an argument. This would allow you to use the autocomplete feature of your CLI to quickly find the file you need.
This would probably only be handy if the script is non-interactive besides the filepath input.
pywin32 provides access to the
GetOpenFileName win32 function. From the example
import win32gui, win32con, os filter='Python Scripts *.py;*.pyw;*.pys Text files *.txt ' customfilter='Other file types *.* ' fname, customfilter, flags=win32gui.GetOpenFileNameW( InitialDir=os.environ['temp'], Flags=win32con.OFN_ALLOWMULTISELECT|win32con.OFN_EXPLORER, File='somefilename', DefExt='py', Title='GetOpenFileNameW', Filter=filter, CustomFilter=customfilter, FilterIndex=0) print 'open file names:', repr(fname) print 'filter used:', repr(customfilter) print 'Flags:', flags for k,v in win32con.__dict__.items(): if k.startswith('OFN_') and flags & v: print 't'+k
Using tkinter (python 2) or Tkinter (python 3) it’s indeed possible to display file open dialog (See other answers here). Please notice however that user interface of that dialog is outdated and does not corresponds to newer file open dialogs available in Windows 10.
Moreover – if you’re looking on way to embedd python support into your own application – you will find out soon that tkinter library is not open source code and even more – it is commercial library.
(For example search for “activetcl pricing” will lead you to this web page: https://reviews.financesonline.com/p/activetcl/)
So tkinter library will cost money for any application wanting to embedd python.
I by myself managed to find pythonnet library:
Using following command it’s possible to install pythonnet:
pip3 install pythonnet
And here you can find out working example for using open file dialog:
Let me copy an example also here:
import sys import ctypes co_initialize = ctypes.windll.ole32.CoInitialize # Force STA mode co_initialize(None) import clr clr.AddReference('System.Windows.Forms') from System.Windows.Forms import OpenFileDialog file_dialog = OpenFileDialog() ret = file_dialog.ShowDialog() if ret != 1: print("Cancelled") sys.exit() print(file_dialog.FileName)
If you also miss more complex user interface – see Demo folder
in pythonnet git.
I’m not sure about portability to other OS’s, haven’t tried, but .net 5 is planned to be ported to multiple OS’s (Search “.net 5 platforms”, https://devblogs.microsoft.com/dotnet/introducing-net-5/ ) – so this technology is also future proof.
Check out EasyGUI, a very easy to use module that should do the job – http://easygui.sourceforge.net/
You would use the fileopenbox function detailed on this api documentation page – https://easygui.readthedocs.io/en/latest/api.html
Python 3.8 32bit
Easy way to use FileDialog to select File
from tkinter import filedialog import tkinter root = tkinter.Tk() root.withdraw() file_path = filedialog.askopenfilename() print(file_path)