3

我的 api 上有一个始终返回登录用户的资源。该资源是只读的。我希望列表 uri 充当详细 uri，并删除详细 url。

因此，/api/v1/user/将返回登录的用户，并且任何其他 url 都会失败。这就是我为实现这一目标所做的：

class UserResource(ModelResource):
    class Meta:
        queryset = User.objects.all()
        fields = ['email', 'name']
        authentication = MultiAuthentication(SessionAuthentication(), BasicAuthentication())
        authorization = Authorization()
        list_allowed_methods = []
        detail_allowed_methods = ['get']

    def base_urls(self):
        '''
        The list endpoint behaves as the list endpoint.
        '''
        return [
            url(r"^(?P<resource_name>%s)%s$" % (self._meta.resource_name, trailing_slash()), self.wrap_view('dispatch_detail'), name="api_dispatch_detail"),
            url(r"^(?P<resource_name>%s)/schema%s$" % (self._meta.resource_name, trailing_slash()), self.wrap_view('get_schema'), name="api_get_schema")
        ]

    def obj_get(self, bundle, **kwargs):
        '''
        Always returns the logged in user.
        '''
        return bundle.request.user

    def get_resource_uri(self, bundle_or_obj=None, url_name='api_dispatch_detail'):
        bundle_or_obj = None
        try:
            return self._build_reverse_url(url_name, kwargs=self.resource_uri_kwargs(bundle_or_obj))
        except NoReverseMatch:
            return ''

我使用base_urls()而不是prepend_urls()因为我想删除其他网址。

它工作正常，但是当我点击/api/v1/url 时，我收到了这个错误：

Traceback:
File "/home/vagrant/workspace/expenses/venv/local/lib/python2.7/site-packages/Django-1.5-py2.7.egg/django/core/handlers/base.py" in get_response
  115.                         response = callback(request, *callback_args, **callback_kwargs)
File "/home/vagrant/workspace/expenses/venv/local/lib/python2.7/site-packages/django_tastypie-0.9.15-py2.7.egg/tastypie/api.py" in wrapper
  80.                 return getattr(self, view)(request, *args, **kwargs)
File "/home/vagrant/workspace/expenses/venv/local/lib/python2.7/site-packages/django_tastypie-0.9.15-py2.7.egg/tastypie/api.py" in top_level
  137.                     'resource_name': name,
File "/home/vagrant/workspace/expenses/venv/local/lib/python2.7/site-packages/django_tastypie-0.9.15-py2.7.egg/tastypie/api.py" in _build_reverse_url
  166.         return reverse(name, args=args, kwargs=kwargs)
File "/home/vagrant/workspace/expenses/venv/local/lib/python2.7/site-packages/Django-1.5-py2.7.egg/django/core/urlresolvers.py" in reverse
  496.     return iri_to_uri(resolver._reverse_with_prefix(view, prefix, *args, **kwargs))
File "/home/vagrant/workspace/expenses/venv/local/lib/python2.7/site-packages/Django-1.5-py2.7.egg/django/core/urlresolvers.py" in _reverse_with_prefix
  416.                 "arguments '%s' not found." % (lookup_view_s, args, kwargs))

Exception Type: NoReverseMatch at /api/v1/
Exception Value: Reverse for 'api_dispatch_list' with arguments '()' and keyword arguments '{'api_name': u'v1', 'resource_name': 'user'}' not found.

它试图到达丢失的列表端点。我该如何摆脱这个？

谢谢。

感谢 Rudy 的指导，我得到了以下结果：

class UserResource(ModelResource):
    class Meta:
        queryset = User.objects.all()
        fields = ['email', 'name']
        authentication = MultiAuthentication(SessionAuthentication(), BasicAuthentication())
        authorization = Authorization()
        list_allowed_methods = []
        detail_allowed_methods = ['get']

    def dispatch_list(self, request, **kwargs):
        return self.dispatch_detail(request, **kwargs)

    def obj_get(self, bundle, **kwargs):
        '''
        Always returns the logged in user.
        '''
        return bundle.request.user

    def get_resource_uri(self, bundle_or_obj=None, url_name='api_dispatch_list'):
        bundle_or_obj = None
        try:
            return self._build_reverse_url(url_name, kwargs=self.resource_uri_kwargs(bundle_or_obj))
        except NoReverseMatch:
            return ''
4

1 回答 1

4

您应该使用自定义授权类来阻止列表端点并优雅地引发错误，而不是仅仅删除所有 URL，这样它仍然可以很好地与 Tastypie 配合使用。

class UserObjectsOnlyAuthorization(Authorization):
def read_list(self, object_list, bundle):
    raise Unauthorized("Sorry, no list reads.")

def read_detail(self, object_list, bundle):
    # Is the requested object the user?
    return bundle.obj == bundle.request.user

def create_list(self, object_list, bundle):
    raise Unauthorized("Sorry, no creates.")

def create_detail(self, object_list, bundle):
    raise Unauthorized("Sorry, no creates.")

def update_list(self, object_list, bundle):
    raise Unauthorized("Sorry, no updates.")

def update_detail(self, object_list, bundle):
    raise Unauthorized("Sorry, no updates.")

def delete_list(self, object_list, bundle):
    # Sorry user, no deletes for you!
    raise Unauthorized("Sorry, no deletes.")

def delete_detail(self, object_list, bundle):
    raise Unauthorized("Sorry, no deletes.")

编辑：

如果您想强制此 API 始终为“详细信息”请求，那么您可以覆盖 Tastypie 的内置函数。基本上，如果您在 URL 中指定一个 ID，那么美味派将其路由为 _detail 请求，如果您不指定，则将其路由为 _list 请求。如果您覆盖检测到这一点的调度函数，您可以将对该资源的所有请求更改为 _detail 并指定查找用户的主键。这可能有点hacky，但会完成你想要的：

def dispatch(self, request_type, request, **kwargs):
    # Force this to be a single User object
    return super(UserResource, self).dispatch('detail', request, **kwargs)

def get_detail(self, request, **kwargs):
    # Place the authenticated user's id in the get detail request
    kwargs['id'] = request.user.pk
    return super(UserResource, self).get_detail(request, **kwargs)
于 2013-05-17T22:09:30.823 回答