我正在寻找一种从URL(例如200、404等)获取HTTP响应代码的快速方法。我不确定要使用哪个库。
Answers:
使用精彩的请求库进行更新。请注意,我们使用的是HEAD请求,它比完整的GET或POST请求发生得更快。
import requests
try:
r = requests.head("https://stackoverflow.com")
print(r.status_code)
# prints the int of the status code. Find more at httpstatusrappers.com :)
except requests.ConnectionError:
print("failed to connect")
requests提供403了链接。
这是httplib替代使用的解决方案。
import httplib
def get_status_code(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
None instead.
"""
try:
conn = httplib.HTTPConnection(host)
conn.request("HEAD", path)
return conn.getresponse().status
except StandardError:
return None
print get_status_code("stackoverflow.com") # prints 200
print get_status_code("stackoverflow.com", "/nonexistant") # prints 404
except至少将限制限制在StandardError这样的范围内,以免导致错误捕获诸如此类的东西KeyboardInterrupt。
curl -I http://www.amazon.com/。
您应该使用urllib2,如下所示:
import urllib2
for url in ["http://entrian.com/", "http://entrian.com/does-not-exist/"]:
try:
connection = urllib2.urlopen(url)
print connection.getcode()
connection.close()
except urllib2.HTTPError, e:
print e.getcode()
# Prints:
# 200 [from the try block]
# 404 [from the except block]
http://entrian.com/为http://entrian.com/blog,即使结果涉及重定向到http://entrian.com/blog/(注意末尾的斜杠),结果200也将是正确的。
这是一个httplib行为类似于urllib2的解决方案。您可以给它一个URL,它就可以工作。无需费心将URL拆分为主机名和路径。该功能已经做到了。
import httplib
import socket
def get_link_status(url):
"""
Gets the HTTP status of the url or returns an error associated with it. Always returns a string.
"""
https=False
url=re.sub(r'(.*)#.*$',r'\1',url)
url=url.split('/',3)
if len(url) > 3:
path='/'+url[3]
else:
path='/'
if url[0] == 'http:':
port=80
elif url[0] == 'https:':
port=443
https=True
if ':' in url[2]:
host=url[2].split(':')[0]
port=url[2].split(':')[1]
else:
host=url[2]
try:
headers={'User-Agent':'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:26.0) Gecko/20100101 Firefox/26.0',
'Host':host
}
if https:
conn=httplib.HTTPSConnection(host=host,port=port,timeout=10)
else:
conn=httplib.HTTPConnection(host=host,port=port,timeout=10)
conn.request(method="HEAD",url=path,headers=headers)
response=str(conn.getresponse().status)
conn.close()
except socket.gaierror,e:
response="Socket Error (%d): %s" % (e[0],e[1])
except StandardError,e:
if hasattr(e,'getcode') and len(e.getcode()) > 0:
response=str(e.getcode())
if hasattr(e, 'message') and len(e.message) > 0:
response=str(e.message)
elif hasattr(e, 'msg') and len(e.msg) > 0:
response=str(e.msg)
elif type('') == type(e):
response=e
else:
response="Exception occurred without a good error message. Manually check the URL to see the status. If it is believed this URL is 100% good then file a issue for a potential bug."
return response