Python: How to open a folder on Windows Explorer(Python 3.6.2, Windows 10)

Question

If I store the path that i want to open in a string called finalpath which looks something like this: "./2.8 Movies/English/Die Hard Series"

then how do i open this in Windows Explorer?(Windows 10)(Python 3.6.2)

P.S I know many people have asked this question but I did not find them clear. Please answer soon.

Luminiferous

Edit queue is full unfortunately. – Overshoe

Answer 1

I found a simple method.

import os
path = "C:/Users"
path = os.path.realpath(path)
os.startfile(path)

Luminiferous

Note that startfile is Windows only. – Schlock

True, but notice that the OP specifically tagged Windows and Windows-10. For platform independency you can consider this answer. – Overshoe

Answer 2

Other alternatives

import webbrowser, os
path="C:/Users"
webbrowser.open(os.path.realpath(path))

or with os alone

import os
os.system(f'start {os.path.realpath(path)}')

or subprocess

import subprocess,os
subprocess.Popen(f'explorer {os.path.realpath(path)}')

or

subprocess.run(['explorer', os.path.realpath(path)])

Agateware

webbrowser also works in Linux; unlike other methods. For subprocess, it will be open, xdg-open or explorer, depending on the OS. – Doorway

Already mentioned in this underrated answer – Overshoe

Answer 3

Cross platform:

import webbrowser


path = 'C:/Users'

webbrowser.open('file:///' + path)

Expansible

Note that on some platforms, trying to open a filename using this function, may work and start the operating system’s associated program. However, this is neither supported nor portable. – Hakenkreuz

Answer 4

Windows:

import os
path = r'C:\yourpath'
os.startfile(path)

This method is a simplified version of the approved answer.

Barque

Answer 5

If you want a GUI Try this

import os
from tkinter import *
from tkinter import filedialog

def open_directory():
    directory = filedialog.askdirectory()
    if directory: # if user didn't choose directory tkinter returns ''
        os.chdir(directory) # change current working directory
        for file_name in os.listdir(directory): # for every file in directory
            if os.path.isfile(os.path.join(directory, file_name)): # if it's file add file name to the list
                listbox.insert(END, file_name)

Button(root, text="Choose Directory", command=open_directory).pack()  # create a button which will call open_directory function

Outofdoors

Answer 6

import os

path = "C:\\Users"


def listdir(dir):
    filenames = os.listdir(dir)
    for files in filenames:
        print(files)


listdir(path)

ok here is another piece of cake it list all your files in a directory

Chickadee