Checking if a website is up via Python

By using python, how can I check if a website is up? From what I read, I need to check the "HTTP HEAD" and see status code "200 OK", but how to do so ?

Cheers

Related

• How do you send a HEAD HTTP request in Python?

You could try to do this with getcode() from urllib

import urllib.request

print(urllib.request.urlopen("https://www.stackoverflow.com").getcode())

200

For Python 2, use

print urllib.urlopen("http://www.stackoverflow.com").getcode()

200

I think the easiest way to do it is by using Requests module.

import requests

def url_ok(url):
    r = requests.head(url)
    return r.status_code == 200

You can use httplib

import httplib
conn = httplib.HTTPConnection("www.python.org")
conn.request("HEAD", "/")
r1 = conn.getresponse()
print r1.status, r1.reason

prints

200 OK

Of course, only if www.python.org is up.

import httplib
import socket
import re

def is_website_online(host):
    """ This function checks to see if a host name has a DNS entry by checking
        for socket info. If the website gets something in return, 
        we know it's available to DNS.
    """
    try:
        socket.gethostbyname(host)
    except socket.gaierror:
        return False
    else:
        return True


def is_page_available(host, path="/"):
    """ This function retreives the status code of a website by requesting
        HEAD data from the host. This means that it only requests the headers.
        If the host cannot be reached or something else goes wrong, it returns
        False.
    """
    try:
        conn = httplib.HTTPConnection(host)
        conn.request("HEAD", path)
        if re.match("^[23]\d\d$", str(conn.getresponse().status)):
            return True
    except StandardError:
        return None

from urllib.request import Request, urlopen
from urllib.error import URLError, HTTPError
req = Request("http://stackoverflow.com")
try:
    response = urlopen(req)
except HTTPError as e:
    print('The server couldn\'t fulfill the request.')
    print('Error code: ', e.code)
except URLError as e:
    print('We failed to reach a server.')
    print('Reason: ', e.reason)
else:
    print ('Website is working fine')

Works on Python 3