在urllib2.urlopen上更改用户代理

Question

如何使用urllib2.urlopen上默认用户代理以外的用户代理下载网页？

Answer 1

I answered a similar question a couple weeks ago.

There is example code in that question, but basically you can do something like this: (Note the capitalization of User-Agent as of RFC 2616 , section 14.43.)

opener = urllib2.build_opener()
opener.addheaders = [('User-Agent', 'Mozilla/5.0')]
response = opener.open('http://www.stackoverflow.com')

Answer 2

headers = { 'User-Agent' : 'Mozilla/5.0' }
req = urllib2.Request('www.example.com', None, headers)
html = urllib2.urlopen(req).read()

Or, a bit shorter:

req = urllib2.Request('www.example.com', headers={ 'User-Agent': 'Mozilla/5.0' })
html = urllib2.urlopen(req).read()

Answer 3

Setting the User-Agent from everyone's favorite Dive Into Python .

The short story: You can use Request.add_header to do this.

You can also pass the headers as a dictionary when creating the Request itself, as the docs note :

headers should be a dictionary, and will be treated as if add_header() was called with each key and value as arguments. This is often used to “spoof” the User-Agent header, which is used by a browser to identify itself – some HTTP servers only allow requests coming from common browsers as opposed to scripts. For example, Mozilla Firefox may identify itself as "Mozilla/5.0 (X11; U; Linux i686) Gecko/20071127 Firefox/2.0.0.11" , while urllib2 's default user agent string is "Python-urllib/2.6" (on Python 2.6).

Answer 4

For python 3, urllib is split into 3 modules...

import urllib.request
req = urllib.request.Request(url="http://localhost/", headers={'User-Agent':' Mozilla/5.0 (Windows NT 6.1; WOW64; rv:12.0) Gecko/20100101 Firefox/12.0'})
handler = urllib.request.urlopen(req)

Answer 5

All these should work in theory, but (with Python 2.7.2 on Windows at least) any time you send a custom User-agent header, urllib2 doesn't send that header. If you don't try to send a User-agent header, it sends the default Python / urllib2

None of these methods seem to work for adding User-agent but they work for other headers:

opener = urllib2.build_opener(proxy)
opener.addheaders = {'User-agent':'Custom user agent'}
urllib2.install_opener(opener)

request = urllib2.Request(url, headers={'User-agent':'Custom user agent'})

request.headers['User-agent'] = 'Custom user agent'

request.add_header('User-agent', 'Custom user agent')

Answer 6

For urllib you can use:

from urllib import FancyURLopener

class MyOpener(FancyURLopener, object):
    version = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; it; rv:1.8.1.11) Gecko/20071127 Firefox/2.0.0.11'

myopener = MyOpener()
myopener.retrieve('https://www.google.com/search?q=test', 'useragent.html')

Answer 7

Another solution in urllib2 and Python 2.7:

req = urllib2.Request('http://www.example.com/')
req.add_unredirected_header('User-Agent', 'Custom User-Agent')
urllib2.urlopen(req)

Answer 8

Try this :

html_source_code = requests.get("http://www.example.com/",
                   headers={'User-Agent':'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/44.0.2403.107 Safari/537.36',
                            'Upgrade-Insecure-Requests': '1',
                            'x-runtime': '148ms'}, 
                   allow_redirects=True).content

Answer 9

there are two properties of urllib.URLopener() namely:
addheaders = [('User-Agent', 'Python-urllib/1.17'), ('Accept', '*/*')] and
version = 'Python-urllib/1.17' .
To fool the website you need to changes both of these values to an accepted User-Agent. for eg
Chrome browser : 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/33.0.1750.149 Safari/537.36'
Google bot : 'Googlebot/2.1'
like this

import urllib
page_extractor=urllib.URLopener()  
page_extractor.addheaders = [('User-Agent', 'Googlebot/2.1'), ('Accept', '*/*')]  
page_extractor.version = 'Googlebot/2.1'
page_extractor.retrieve(<url>, <file_path>)

changing just one property does not work because the website marks it as a suspicious request.