Python请求lib在LinkedIn上不起作用
[英]Python requests lib doesn't work with LinkedIn
问题描述
Sample code: 
样例代码：
import requests
print requests.get("https://www.linkedin.com/")
I get: <Response [200]> 我得到： <Response [200]>
Simple curl request does work: 简单的curl请求确实起作用：
curl "https://www.linkedin.com/"
2 个解决方案
解决方案1
2 2018-07-31 15:21:48
If you get <Response [200]> that means that it worked properly. 如果得到<Response [200]> ，则表明它正常工作。 You should refer to the documentation for unpacking this Response object to get the data inside of it. 您应该参考文档来解压缩此Response对象以获取其中的数据。
Eg: 例如：
>>> r = requests.get('https://linkedin.com/')
>>> r.text
'<!DOCTYPE html> ...'
解决方案2
2 2018-07-31 15:32:56
The requests.get() function returns a Response object that contains attributes about the status_code, headers, and content: request.get（）函数返回一个Response对象，其中包含有关status_code，标头和内容的属性：
[In]: type(requests.get("https://www.linkedin.com/")
[Out]: <class 'requests.models.Response'> . [Out]: <class 'requests.models.Response'> 。
I would recommend saving the returned Response into a variable: 我建议将返回的响应保存到变量中：
response = requests.get("https://www.linkedin.com/")
Then you can access the contents of the Response using response.json() if it is a JSON file or response.text if it is an html page. 然后，您可以使用访问响应的内容response.json()如果它是一个JSON文件或response.text如果它是一个HTML页面。
In your use case, response.text should return the same thing as curl "https://www.linkedin.com/" . 在您的用例中， response.text应该返回与curl "https://www.linkedin.com/"相同的东西。
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