Python Program To Open File Manager

In this article, we will demonstrate a Python program that opens the file manager to access any folder.

Python Code:

“`python
import os
import subprocess

folder_path = input(“Enter the path of the folder you want to open: “)

if os.path.exists(folder_path):
if os.name == ‘nt’:
os.startfile(folder_path)
elif os.name == ‘posix’:
subprocess.Popen([‘xdg-open’, folder_path])
else:
print(“Unsupported operating system.”)
else:
print(“Folder does not exist.”)

“`

Explanation:

In this Python program, we first import the necessary modules – ‘os’ and ‘subprocess’. We then prompt the user to enter the path of the folder they want to open.

We check if the folder exists using the ‘os.path.exists()’ function. If it does, then we check the operating system using the ‘os.name’ attribute. Depending on the operating system, we either use ‘os.startfile()’ for Windows or ‘subprocess.Popen()’ for POSIX systems to open the file manager and access the specified folder.

If the folder path provided by the user does not exist, we display an error message indicating that the folder does not exist.

Conclusion:

This Python program demonstrates how to open the file manager to access any folder specified by the user. It is a simple yet useful script that can be used to quickly navigate to a specific folder on your system.