如何处理装饰器内的生成器代码?

时间:2016-10-14 20:29:10

标签: python django wsgi python-decorators

我尝试编写一个login_required装饰器,用于使用WSGI中间件验证对Django中受保护视图的请求。

这是我的代码:

def login_required(f, request_class=HTTPRequest):
    def _wrapper(*args, **kwargs):
        if not isinstance(args[0], request_class):
            req = request_class(environ=args[0])
        else:
            req = args[0]
        wsgi_app = WSGIController()
        settings = wsgi_app.settings                
        google_client = client.GoogleClient(
            settings.OAUTH2_CLIENT_ID,
            access_token=settings.OAUTH2_ACCESS_TOKEN,
            scope='email', 
            redirect_url=settings.OAUTH2_REDIRECT_URL,
            login_path="/session_login/")

         wsgi_app = google_client.wsgi_middleware(wsgi_app, \
            secret=settings.SECRET_KEY)
         def process_middleware(environ, start_response):
             return wsgi_app(environ, start_response)
         response = process_middleware(req.environ, f)
         return f(*args, **kwargs)
     return _wrapper

如何处理响应对象(生成器)以将用户重定向到oauth登录页面?

更新:如果我添加以下代码:

  

redirect = [response.next()中项目的项目

然后我收到以下错误:

  

TypeError:不可用类型:' list'

1 个答案:

答案 0 :(得分:0)

您可以在django/contrib/auth/decoractors.py

中引用login_required装饰器逻辑

当登录检查失败时,它会使用redirect_to_login函数重定向到登录页面。

from django.contrib.auth.views import redirect_to_login
return redirect_to_login(path, resolved_login_url, redirect_field_name)