ویب سائٹ کی حیثیت چیک کرنے کے لیے اسکرپٹ۔ ازگر

import requests

def check_website_status(url):
    try:
        response = requests.get(url)
        if response.status_code == 200:
            return "Website is up and running."
        else:
            return f"Website is down with status code {response.status_code}."
    except requests.ConnectionError:
        return "Could not connect to the website."

# Example usage:
url = "https://www.example.com"
status = check_website_status(url)
print(status)