requests.get返回403,而相同的URL在浏览器中工作

Question

我正在尝试使用rlsnet.ru上的搜索表单.这是我从源文件中提取的表单定义:

<form id="site_search_form" action="/search_result.htm" method="get">
    <input id="simplesearch_text_input" class="search__field" type="text" name="word" value="" autocomplete="off">
    <input type="hidden" name="path" value="/" id="path">
    <input type="hidden" name="enter_clicked" value="1">
    <input id="letters_id" type="hidden" name="letters" value="">
    <input type="submit" class="g-btn search__btn" value="?????" id="simplesearch_button">
    <div class="sf_suggestion">
        <ul style="display: none; z-index:1000; opacity:0.85;">
        </ul>
    </div>
    <div id="contentsf">
    </div>
</form>

这是我用来发送搜索请求的代码:

import requests
from urllib.parse import urlencode 

root = "http://www.rlsnet.ru/search_result.htm?"
response = requests.get(root + urlencode({"word": "????????????".encode('cp1251')})

每次我这样做,响应状态为403.当我在http://www.rlsnet.ru/search_result.htm?word=%D6%E5%F0%E5%E1%F0%EE%EB%E8%E7%E8%EDSafari/Chrome/Opera中输入相同的请求URL(即)时,它可以正常工作并返回预期的页面.我究竟做错了什么？谷歌搜索问题只带来了这个问题:为什么网址在浏览器中工作,但没有使用请求获取方法,这没什么用处.

Answer 1

嗯,这是因为默认User-Agent的requests是python-requests/2.13.0,和你的情况是网站不要像"非浏览器"流量,因此他们试图阻止此类流量.

>>> import requests
>>> session = requests.Session()
>>> session.headers
{'Connection': 'keep-alive', 'Accept-Encoding': 'gzip, deflate', 'Accept': '*/*', 'User-Agent': 'python-requests/2.13.0'}

您需要做的就是使请求看起来像来自浏览器,所以只需添加一个额外的header参数:

import requests

headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/56.0.2924.76 Safari/537.36'} # This is chrome, you can set whatever browser you like
response = requests.get('http://www.rlsnet.ru/search_result.htm?word=%D6%E5%F0%E5%E1%F0%EE%EB%E8%E7%E8%ED', headers=headers)

print response.status_code
print response.url

200 
http://www.rlsnet.ru/search_result.htm?word=%D6%E5%F0%E5%E1%F0%EE%EB%E8%E7%E8%ED