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 s

Solution 1:

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

Solution 2:

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

Solution 3:

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.

Solution 4:

import httplib
import socket
import re

defis_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:
        returnFalseelse:
        returnTruedefis_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)):
            returnTrueexcept StandardError:
        returnNone

Solution 5:

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