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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

11 Answers
11

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

getcode()

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


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

python3

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


urlopen.getcode

getcode

req = urllib.request.Request(url, headers = headers) resp = urllib.request.urlopen(req)

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


url = "http://foo.example.org/"

False

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.

www.python.org

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]dd$", str(conn.getresponse().status)):
return True
except StandardError:
return None


is_website_online

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.

HTTPConnection

httplib

httplib2

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

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/.

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


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)


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))


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

requests

status code

200

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

>> 200


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