CODEWITHSUNDEEP
pythonuser-interfacetkinter
RPiAwesomeness
RPiAwesomeness

Reputation: 5159

How to get a string from the tkinter filedialog in Python 3?

I'm attempting to use the tkinter filedialog to get the user's choice of a file in my Python 3.4 program.

Previously, I was trying to use the Gtk FileChooserDialog, but I keep running into wall after wall getting it to work (here's my question about that.) So, I (attempted to) switched over to tkinter and use the filedialog.

Here's the code I'm using for the GUI:

import tkinter
from tkinter import filedialog

root = tkinter.Tk()
root.withdraw()

path = filedialog.askopenfile()

print(type(path)) # <- Not actually in the code, but I've included it to show the type

It works perfectly, except for the fact that it returns an <class '_io.TextIOWrapper'> object instead of a string, like I expected/need it to.

Calling str() on that doesn't work, and neither does using the io module function getvalue().

Does anyone know how I could get the chosen file path as a string from the filedialog.askopenfile() function?

Upvotes: 2

Views: 5462

Answers (1)

chdorr
chdorr

Reputation: 298

I'm sure there are several ways, but what about getting path.name? This should be a string.

print("type(path):", type(path))
# <class '_io.TextIOWrapper'>

print("path:", path)
# <_io.TextIOWrapper name='/some/path/file.txt' mode='r' encoding='UTF-8'>

print("path.name:", path.name)
# /some/path/file.txt

print("type(path.name):", type(path.name))
# <class 'str'>

Note that askopenfile opens and returns the file in read mode by default. If you just want the filename and plan on opening it yourself later, try using askopenfilename instead. See this link for more:

First you have to decide if you want to open a file or just want to get a filename in order to open the file on your own. In the first case you should use tkFileDialog.askopenfile() in the latter case tkFileDialog.askopenfilename().

Upvotes: 11

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