Python - 从 Google Alerts feeds 获取链接的重定向 url

Question

如果您将 google 提醒创建为 rss 提要(不会自动发送到您的电子邮件地址)，它包含如下链接:https://www.google.com/url?rct=j&sa=t&url=http://www.statesmanjournal.com/story/opinion/readers/2014/10/13/gmo-labels-encourage-people-make-choices/17171289/&ct=ga&cd=CAIyGjkyZjE1NGUzMGIwZjRkNGQ6Y29tOmVuOlVT&usg=AFQjCNHrCLmbml7baTXaqySagcuKHp-KHA .

这个链接显然是一个重定向(只要尝试一下，你就会在这里结束:http://www.statesmanjournal.com/story/opinion/readers/2014/10/13/gmo-labels-encourage-people-make-choices/17171289/)，但我无法用 Python 获得这个最终 url(除非删除 url 的开头，这是非常难看的).

到目前为止，我已经尝试使用包 urllib2、httplib2 和请求:

• urllib2.urlopen 和 geturl() 从返回值
• 使用 follow_all_redirects=True 和返回值中的“content-location”的 httplib2 请求
• requests.get 和返回值的历史记录

有人遇到过这个问题吗？谢谢!

Answer 1

Google 不会为您提供 HTTP 重定向；返回 200 OK 响应，而不是 30x 重定向:

>>> import requests
>>> url = 'https://www.google.com/url?rct=j&sa=t&url=http://www.statesmanjournal.com/story/opinion/readers/2014/10/13/gmo-labels-encourage-people-make-choices/17171289/&ct=ga&cd=CAIyGjkyZjE1NGUzMGIwZjRkNGQ6Y29tOmVuOlVT&usg=AFQjCNHrCLmbml7baTXaqySagcuKHp-KHA'
>>> response = requests.get(url)
>>> response.url
u'https://www.google.com/url?rct=j&sa=t&url=http://www.statesmanjournal.com/story/opinion/readers/2014/10/13/gmo-labels-encourage-people-make-choices/17171289/&ct=ga&cd=CAIyGjkyZjE1NGUzMGIwZjRkNGQ6Y29tOmVuOlVT&usg=AFQjCNHrCLmbml7baTXaqySagcuKHp-KHA'
>>> response.text
u'<script>window.googleJavaScriptRedirect=1</script><script>var m={navigateTo:function(b,a,d){if(b!=a&&b.google){if(b.google.r){b.google.r=0;b.location.href=d;a.location.replace("about:blank");}}else{a.location.replace(d);}}};m.navigateTo(window.parent,window,"http://www.statesmanjournal.com/story/opinion/readers/2014/10/13/gmo-labels-encourage-people-make-choices/17171289/");\n</script><noscript><META http-equiv="refresh" content="0;URL=\'http://www.statesmanjournal.com/story/opinion/readers/2014/10/13/gmo-labels-encourage-people-make-choices/17171289/\'"></noscript>'

响应是一段 HTML 和 JavaScript，您的浏览器会将其解释为加载新的 URL。您必须解析该响应以提取目标。

字符串拆分可以实现这一点:

>>> response.text.partition("URL='")[-1].rpartition("'\"")[0]
u'http://www.statesmanjournal.com/story/opinion/readers/2014/10/13/gmo-labels-encourage-people-make-choices/17171289/'

如果我们假设正文中的 URL 参数只是查询字符串中 url 参数的直接反射(reflect)，那么您也可以从那里提取它，我们甚至不必要求 Google 执行重定向:

try:
    from urllib.parse import parse_qs, urlsplit
except ImportError:
    # Python 2
    from urlparse import parse_qs, urlsplit

target = parse_qs(urlsplit(url).query)['url'][0]