Checking if a website is up via 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

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

The HTTPConnection object from the httplib module in the standard library will probably do the trick for you. BTW, if you start doing anything advanced with HTTP in Python, be sure to check out httplib2; it's a great library.

If server if down, on python 2.7 x86 windows urllib have no timeout and program go to dead lock. So use urllib2

import urllib2
import socket

def check_url( url, timeout=5 ):
    try:
        return urllib2.urlopen(url,timeout=timeout).getcode() == 200
    except urllib2.URLError as e:
        return False
    except socket.timeout as e:
        print False


print check_url("http://google.fr")  #True 
print check_url("http://notexist.kc") #False     

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

In my opinion, caisah's answer misses an important part of your question, namely dealing with the server being offline.

Still, using requests is my favorite option, albeit as such:

import requests

try:
    requests.get(url)
except requests.exceptions.ConnectionError:
    print(f"URL {url} not reachable")

I use requests for this, then it is easy and clean. Instead of print function you can define and call new function (notify via email etc.). Try-except block is essential, because if host is unreachable then it will rise a lot of exceptions so you need to catch them all.

import requests

URL = "https://api.github.com"

try:
    response = requests.head(URL)
except Exception as e:
    print(f"NOT OK: {str(e)}")
else:
    if response.status_code == 200:
        print("OK")
    else:
        print(f"NOT OK: HTTP response code {response.status_code}")

If by up, you simply mean "the server is serving", then you could use cURL, and if you get a response than it's up.

I can't give you specific advice because I'm not a python programmer, however here is a link to pycurl http://pycurl.sourceforge.net/.

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

my 2 cents

def getResponseCode(url):
conn = urllib.request.urlopen(url)
return conn.getcode()

if getResponseCode(url) != 200:
    print('Wrong URL')
else:
    print('Good URL')