launch a .exe file on startup

YOU.com answered on September 11, 2023 Popularity 6/10 Helpfulness 6/10

answer launch a .exe file on startup

launch a .exe file on startup

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import winreg
import os

def add_to_startup(file_path):
    # Open the Registry 'Run' key
    key = winreg.OpenKey(winreg.HKEY_CURRENT_USER, r"Software\Microsoft\Windows\CurrentVersion\Run", 0, winreg.KEY_ALL_ACCESS)

    # Add the .exe file path to the Registry startup key
    winreg.SetValueEx(key, "MyApp", 0, winreg.REG_SZ, file_path)

    # Close the Registry key
    winreg.CloseKey(key)

# Specify your .exe file path here
exe_file_path = r"C:\path\to\your\file.exe"

# Check if the .exe file exists
if os.path.exists(exe_file_path):
    add_to_startup(exe_file_path)
    print("Successfully added to startup!")
else:
    print("File does not exist.")

Popularity 6/10 Helpfulness 6/10 Language python

Source: Grepper

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Contributed on Sep 11 2023

YOU.com

0 Answers Avg Quality 2/10

launch a .exe file on startup

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