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json - 如何获取 https://stocks.exchange/api2/ticker

问题描述

import json, requests

def tick():
    r = requests.get('https://stocks.exchange/api2/ticker')
    return r.json()
print tick()

此代码输出，

Traceback (most recent call last):
File "C:\Users\Steven\Desktop\Auto\tradeogre\stocksexchange.py", line 6, in 
<module>
print tick()
File "C:\Users\Steven\Desktop\Auto\tradeogre\stocksexchange.py", line 4, in 
tick
r = requests.get('https://stocks.exchange/api2/ticker')
File "C:\Python27\lib\site-packages\requests\api.py", line 72, in get
return request('get', url, params=params, **kwargs)
File "C:\Python27\lib\site-packages\requests\api.py", line 58, in request
return session.request(method=method, url=url, **kwargs)
File "C:\Python27\lib\site-packages\requests\sessions.py", line 508, in 
request
resp = self.send(prep, **send_kwargs)
File "C:\Python27\lib\site-packages\requests\sessions.py", line 618, in send
r = adapter.send(request, **kwargs)
File "C:\Python27\lib\site-packages\requests\adapters.py", line 506, in send
raise SSLError(e, request=request)
SSLError: HTTPSConnectionPool(host='stocks.exchange', port=443): Max retries 
exceeded with url: /api2/ticker (Caused by SSLError(SSLEOFError(8, u'EOF 
occurred in violation of protocol (_ssl.c:661)'),))

在不违反协议的情况下我将如何做到这一点？我使用了 urllib、urllib2，并得到了类似的响应。如果我在浏览器中输入 url，它会显示我需要的数据。任何帮助将不胜感激。

标签： jsonpython-2.7python-requeststicker

解决方案

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