通过使用python，我如何检查网站是否已启动？根据我的阅读，我需要检查“HTTP HEAD”并查看状态代码“200 OK”，但该怎么做？
干杯
您可以尝试使用
urllib中的
getcode() 来执行此操作
import urllib.request
print(urllib.request.urlopen("https://www.stackoverflow.com").getcode())
200
对于 Python 2，请使用
print urllib.urlopen("http://www.stackoverflow.com").getcode()
200
我认为最简单的方法是使用 Requests 模块。
import requests
def url_ok(url):
r = requests.head(url)
return r.status_code == 200
您可以使用httplib
import httplib
conn = httplib.HTTPConnection("www.python.org")
conn.request("HEAD", "/")
r1 = conn.getresponse()
print r1.status, r1.reason
打印
200 OK
当然，前提是
www.python.orgfrom 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')
适用于Python 3
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
我使用requests来做到这一点，那么它既简单又干净。 您可以定义和调用新函数（通过电子邮件等通知），而不是 print 函数。 Try- except 块是必不可少的，因为如果主机无法访问，那么它将引发很多异常，因此您需要捕获所有异常。
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}")
您可以使用
requestsstatus code200import requests
url = "https://www.google.com"
page = requests.get(url)
print (page.status_code)
>> 200
HTTPConnection 模块中的
httplib对象可能会为您解决问题。顺便说一句，如果您开始在 Python 中使用 HTTP 进行任何高级操作，请务必查看
httplib2；这是一个很棒的图书馆。
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
caisah的回答遗漏了你问题的一个重要部分，即处理服务器离线的问题。
不过，使用requests
是我最喜欢的选择，尽管如此：
import requests
try:
requests.get(url)
except requests.exceptions.ConnectionError:
print(f"URL {url} not reachable")
我无法给你具体的建议，因为我不是Python程序员，但是这里有一个pycurl的链接
http://pycurl.sourceforge.net/.
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)
# 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
如果使用
Ansible，您可以使用fetch_url函数：
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
def getResponseCode(url):
conn = urllib.request.urlopen(url)
return conn.getcode()
if getResponseCode(url) != 200:
print('Wrong URL')
else:
print('Good URL')
PycURL 和 validators 的解决方案
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))
this link 中找到了更好的解决方案，它使用 https://www.isitdownrightnow.com 检查您的域，该解决方案使用世界上多个分布式服务器请求您的域。所以你可以使用这个代码：
domain = 'your domain ex:a.com'
print(f"checking '{domain}' is up?")
isitdown_url = f'https://www.isitdownrightnow.com/check.php?domain={domain}'
r = requests.get(isitdown_url)
r_text = (r.text.lower().replace('</div>', ' '))
status = (re.compile(f'{domain} is (.*) (?:it is not|and reachable)').search(r_text).group(1).split(' ')[0])
if status == 'down':
raise ValueError(f"'{domain}' is down in all the world!please request later!")