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

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/.

Hi this class can do speed and up test for your web page with this class:

 from urllib.request import urlopen
 from socket import socket
 import time


 def tcp_test(server_info):
     cpos = server_info.find(':')
     try:
         sock = socket()
         sock.connect((server_info[:cpos], int(server_info[cpos+1:])))
         sock.close
         return True
     except Exception as e:
         return False


 def http_test(server_info):
     try:
         # TODO : we can use this data after to find sub urls up or down    results
         startTime = time.time()
         data = urlopen(server_info).read()
         endTime = time.time()
         speed = endTime - startTime
         return {'status' : 'up', 'speed' : str(speed)}
     except Exception as e:
         return {'status' : 'down', 'speed' : str(-1)}


 def server_test(test_type, server_info):
     if test_type.lower() == 'tcp':
         return tcp_test(server_info)
     elif test_type.lower() == 'http':
         return http_test(server_info)

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')

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))