Django 2 -403禁止错误：无法从views.py发出获取请求

Question

我想从views.py中的功能启用GET请求并读取数据。 我的功能在views.py中，这是在POST请求后触发的。

@csrf_exempt
def paysuccess(request):
    #process requests
    URL ="http://example/apiusername=111390&pwd=123&circlecode=2&format=json"
    #make get request
    request=urllib.request.Request(URL) 
    response = urllib.request.urlopen(request)
    #read response 

追溯显示csrf.py文件中的问题。

Traceback (most recent call last):
  File "/usr/lib/python3.4/site-packages/django/core/handlers/exception.py", line 35, in inner
    response = get_response(request)
  File "/usr/lib/python3.4/site-packages/django/core/handlers/base.py", line 128, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "/usr/lib/python3.4/site-packages/django/core/handlers/base.py", line 126, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "/usr/lib/python3.4/site-packages/django/views/decorators/csrf.py", line 54, in wrapped_view
    return view_func(*args, **kwargs)
  File "/home/rhtry/redirect/redirect/views.py", line 174, in paysuccess
    response = urllib.request.urlopen(request)
  File "/usr/lib/python3.4/urllib/request.py", line 161, in urlopen
    return opener.open(url, data, timeout)
  File "/usr/lib/python3.4/urllib/request.py", line 470, in open
    response = meth(req, response)
  File "/usr/lib/python3.4/urllib/request.py", line 580, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python3.4/urllib/request.py", line 508, in error
    return self._call_chain(*args)
  File "/usr/lib/python3.4/urllib/request.py", line 442, in _call_chain
    result = func(*args)
  File "/usr/lib/python3.4/urllib/request.py", line 588, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

Answer 1

您可能需要设置用户代理标题，以使站点认为您正在使用浏览器。 代码如下：

@csrf_exempt
def paysuccess(request):

    URL ="http://example/apiusername=111390&pwd=123&circlecode=2&format=json"

    request=urllib.request.Request(URL, headers={'User-Agent': 'Mozilla/5.0'}) 
    response = urllib.request.urlopen(request)