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?
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
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
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
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.
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
import requests
url = "https://www.google.com"
page = requests.get(url)
print (page.status_code)
>> 200