在Python中，如何使用urllib来查看网站是404还是200？

Question

如何通过urllib获取标题的代码？如果URL是没有HTTP URL

Answer 1

的引用代码（）方法（由在python2.6的）返回与该响应发送的HTTP状态代码或无。

>>> a=urllib.urlopen('http://www.google.com/asdfsf') 
>>> a.getcode() 
404 
>>> a=urllib.urlopen('http://www.google.com/') 
>>> a.getcode() 
200 

Answer 2

您可以使用urllib2还有：

import urllib2 

req = urllib2.Request('http://www.python.org/fish.html') 
try: 
    resp = urllib2.urlopen(req) 
except urllib2.HTTPError as e: 
    if e.code == 404: 
     # do something... 
    else: 
     # ... 
except urllib2.URLError as e: 
    # Not an HTTP-specific error (e.g. connection refused) 
    # ... 
else: 
    # 200 
    body = resp.read() 

注意HTTPError是存储HTTP状态代码的URLError一个子类。

Answer 3

import urllib2 

try: 
    fileHandle = urllib2.urlopen('http://www.python.org/fish.html') 
    data = fileHandle.read() 
    fileHandle.close() 
except urllib2.URLError, e: 
    print 'you got an error with the code', e 

Answer 4

对于Python 3：

import urllib.request, urllib.error 

url = 'http://www.google.com/asdfsf' 
try: 
    conn = urllib.request.urlopen(url) 
except urllib.error.HTTPError as e: 
    # Return code error (e.g. 404, 501, ...) 
    # ... 
    print('HTTPError: {}'.format(e.code)) 
except urllib.error.URLError as e: 
    # Not an HTTP-specific error (e.g. connection refused) 
    # ... 
    print('URLError: {}'.format(e.reason)) 
else: 
    # 200 
    # ... 
    print('good')