How to select a directory and store the location using tkinter in Python

I am creating a GUI with a browse button which I only want to return the path. I’ve been looking at solutions using code like below.

Tkinter.Button(subframe, text = "Browse", command = self.loadtemplate, width = 10).pack()

   def loadtemplate(self): 
        filename = tkFileDialog.askopenfilename(filetypes = (("Template files", "*.tplate")
                                                             ,("HTML files", "*.html;*.htm")
                                                             ,("All files", "*.*") ))
        if filename: 
            try: 
                self.settings["template"].set(filename)
            except: 
                tkMessageBox.showerror("Open Source File", "Failed to read file n'%s'"%filename)

However I know Tkinter has a built in askopenfilename which is a super easy one line of code for opening files. Is there some way to modify this to return the directory instead of a file? Is there a smaller option than the larger chunk of code I posted?

Answers:

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Method 1

It appears that tkFileDialog.askdirectory should work. documentation

Method 2

This code may be helpful for you.

from tkinter import filedialog
from tkinter import *
root = Tk()
root.withdraw()
folder_selected = filedialog.askdirectory()

Method 3

Go with this code
First, select the directory for creating a new file

import tkinter as tk
from tkinter import filedialog

root = tk.Tk()
root.withdraw()
# file_path = filedialog.askopenfilename()
file_path = filedialog.askdirectory()

new_file = input("Name filen")
open_file = open(f"{file_path}%s.py" % new_file, 'w')

in my case

i created (ok.py) file in ppppp directory

path is: PS C:UsersdemoDesktoppppppok.py

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0