检查网站是否通过 Python 运行

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/.

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.

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

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

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

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))

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)

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

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

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")

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')

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}")