Checking if a website is up via Python

Question

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 ?

Answer 1

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

Answer 2

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.

Answer 3

Here's my solution using PycURL and validators

import pycurl, validators


def url_exists(url):
    """
    Check if the given URL really exists
    :param url: str
    :return: bool
    """
    if validators.url(url):
        c = pycurl.Curl()
        c.setopt(pycurl.NOBODY, True)
        c.setopt(pycurl.FOLLOWLOCATION, False)
        c.setopt(pycurl.CONNECTTIMEOUT, 10)
        c.setopt(pycurl.TIMEOUT, 10)
        c.setopt(pycurl.COOKIEFILE, '')
        c.setopt(pycurl.URL, url)
        try:
            c.perform()
            response_code = c.getinfo(pycurl.RESPONSE_CODE)
            c.close()
            return True if response_code < 400 else False
        except pycurl.error as err:
            errno, errstr = err
            raise OSError('An error occurred: {}'.format(errstr))
    else:
        raise ValueError('"{}" is not a valid url'.format(url))

Answer 4

Requests and httplib2 are great options:

# Using requests.
import requests
request = requests.get(value)
if request.status_code == 200:
    return True
return False

# Using httplib2.
import httplib2

try:
    http = httplib2.Http()
    response = http.request(value, 'HEAD')

    if int(response[0]['status']) == 200:
        return True
except:
    pass
return False

If using Ansible, you can use the fetch_url function:

from ansible.module_utils.basic import AnsibleModule
from ansible.module_utils.urls import fetch_url

module = AnsibleModule(
    dict(),
    supports_check_mode=True)

try:
    response, info = fetch_url(module, url)
    if info['status'] == 200:
        return True

except Exception:
    pass

return False

Answer 5

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

>>> print urllib.urlopen("http://www.stackoverflow.com").getcode()
>>> 200

EDIT: For more modern python, i.e. python3, use:

import urllib.request
print(urllib.request.urlopen("http://www.stackoverflow.com").getcode())
>>> 200

Answer 6

You may use requests library to find if website is up i.e. status code as 200

import requests
url = "https://www.google.com"
page = requests.get(url)
print (page.status_code) 

>> 200