Python открыть папку в проводнике windows

Trilarion

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Kirill TitovKirill Titov

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A nicer and safer solution (only in Windows unfortunately) is os.startfile().

When it’s given a folder instead of a file, it will open Explorer.

Im aware that i do not completely answer the question since its not selecting a file, but using subprocess is always kind of a bad idea (for security reasons) and this solution may help other people.

answered Mar 7, 2018 at 19:40

As explorer could be overridden it would be a little safer to point to the executable directly. (just had to be schooled on this too)

And while you’re at it: use Python 3s current subprocess API: run()

import os
import subprocess
FILEBROWSER_PATH = os.path.join(os.getenv('WINDIR'), 'explorer.exe')

def explore(path):
    # explorer would choke on forward slashes
    path = os.path.normpath(path)

    if os.path.isdir(path):
        subprocess.run([FILEBROWSER_PATH, path])
    elif os.path.isfile(path):
        subprocess.run([FILEBROWSER_PATH, '/select,', path])

answered Jun 21, 2018 at 9:56

For some reason, on windows 7 it always opens the users Path, for me following worked out:

import subprocess
subprocess.call("explorer C:\temp\yourpath", shell=True)

answered Dec 2, 2014 at 13:51

Alternatively, you could use the fileopenbox module of EasyGUI to open the file explorer for the user to click through and then select a file (returning the full filepath).

import easygui
file = easygui.fileopenbox()

answered Oct 18, 2018 at 19:48

For anyone wondering how to use a variable in place of a direct file path. The code below will open explorer and highlight the file specified.

import subprocess
subprocess.Popen(f'explorer /select,{variableHere}')

The code below will just open the specified folder in explorer without highlighting any specific file.

import subprocess
subprocess.Popen(f'explorer "{variableHere}"')

Ive only tested on windows

answered Dec 15, 2020 at 15:57

asked Jul 8, 2011 at 22:41

I am surprised no one has mentioned using xdg-open for *nix which will work for both files and folders:

import os
import platform
import subprocess

def open_file(path):
    if platform.system() == "Windows":
        os.startfile(path)
    elif platform.system() == "Darwin":
        subprocess.Popen(["open", path])
    else:
        subprocess.Popen(["xdg-open", path])

answered Apr 24, 2013 at 23:36

You can use subprocess.

import subprocess
import sys

if sys.platform == 'darwin':
    def openFolder(path):
        subprocess.check_call(['open', '--', path])
elif sys.platform == 'linux2':
    def openFolder(path):
        subprocess.check_call(['xdg-open', '--', path])
elif sys.platform == 'win32':
    def openFolder(path):
        subprocess.check_call(['explorer', path])

answered Jul 8, 2011 at 22:46

The following works on Macintosh.

import webbrowser
webbrowser.open('file:///Users/test/test_folder')

On GNU/Linux, use the absolute path of the folder. (Make sure the folder exists)

import webbrowser
webbrowser.open('/home/test/test_folder')

As pointed out in the other answer, it works on Windows, too.

answered Sep 13, 2012 at 13:21

I think you may have to detect the operating system, and then launch the relevant file explorer accordingly.

This could come in userful for OSX’s Finder: Python «show in finder»

(The below only works for windows unfortunately)

import webbrowser as wb
wb.open('C:/path/to/folder')

This works on Windows. I assume it would work across other platforms. Can anyone confirm? Confirmed windows only

Reblochon MasqueReblochon Masque

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asked Jul 8, 2011 at 22:41

I am surprised no one has mentioned using xdg-open for *nix which will work for both files and folders:

import os
import platform
import subprocess

def open_file(path):
    if platform.system() == "Windows":
        os.startfile(path)
    elif platform.system() == "Darwin":
        subprocess.Popen(["open", path])
    else:
        subprocess.Popen(["xdg-open", path])

answered Apr 24, 2013 at 23:36

You can use subprocess.

import subprocess
import sys

if sys.platform == 'darwin':
    def openFolder(path):
        subprocess.check_call(['open', '--', path])
elif sys.platform == 'linux2':
    def openFolder(path):
        subprocess.check_call(['xdg-open', '--', path])
elif sys.platform == 'win32':
    def openFolder(path):
        subprocess.check_call(['explorer', path])

answered Jul 8, 2011 at 22:46

The following works on Macintosh.

import webbrowser
webbrowser.open('file:///Users/test/test_folder')

On GNU/Linux, use the absolute path of the folder. (Make sure the folder exists)

import webbrowser
webbrowser.open('/home/test/test_folder')

As pointed out in the other answer, it works on Windows, too.

answered Sep 13, 2012 at 13:21

I think you may have to detect the operating system, and then launch the relevant file explorer accordingly.

This could come in userful for OSX’s Finder: Python «show in finder»

(The below only works for windows unfortunately)

import webbrowser as wb
wb.open('C:/path/to/folder')

This works on Windows. I assume it would work across other platforms. Can anyone confirm? Confirmed windows only

Reblochon MasqueReblochon Masque

34.6k10 gold badges53 silver badges76 bronze badges

необходимо из скрипта открыть папку..так чтоб ее содержимое отображалось в стандартном виндовом проводнике..

Т.е. нужно вызвать из скрипта проводник — explorer “c:program Files”

import subprocess
subprocess.Popen('explorer "%s"'%(PATH))

Например http://python.su/forum/topic/21507/