通过Python检查网站是否已启动

Question

通过使用python,我如何检查网站是否已启动？从我读到的,我需要检查"HTTP HEAD"并查看状态代码"200 OK",但该怎么做？

干杯

有关

• 如何在Python中发送HEAD HTTP请求？

Answer 1

您可以尝试使用要做到这一点getcode()从urllib的

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

编辑:对于更现代的python,即python3使用:

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

Answer 2

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

Answer 3

你可以使用httplib

import httplib
conn = httplib.HTTPConnection("www.python.org")
conn.request("HEAD", "/")
r1 = conn.getresponse()
print r1.status, r1.reason

版画

200 OK

当然,只有这样www.python.org.

Answer 4

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

Answer 5

我使用requests来做到这一点，然后它既简单又干净。您可以定义和调用新函数（通过电子邮件等通知），而不是打印函数。Try- except块是必不可少的，因为如果主机无法访问，那么它将引发大量异常，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}")

Answer 6

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')

适用于Python 3

Answer 7

您可以使用requests库来查找网站是否已启动，status code即200

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

>> 200