Django Tastypie - 仅包含对象细节的资源

Question

在Django中，使用Tastypie，是否有配置资源的方法，使其仅显示对象详细信息？

我想要一个url /user，它返回认证用户的详细信息，而不是包含单个用户对象的列表。我不想使用/users/<id>来获取用户的详细信息。

这里是我的代码的相关部分：

from django.contrib.auth.models import User 
from tastypie.resources import ModelResource 

class UserResource(ModelResource): 

    class Meta: 
     queryset  = User.objects.all() 
     resource_name = 'user' 
     allowed_methods = ['get', 'put'] 
     serializer  = SERIALIZER  # Assume those are defined... 
     authentication = AUTHENTICATION # " 
     authorization = AUTHORIZATION # " 

    def apply_authorization_limits(self, request, object_list): 
     return object_list.filter(pk=request.user.pk) 

Answer 1

我能够通过使用下列资源相结合的方法

• override_urls
• apply_authorization_limits

做到这一点

例如用户资源

#Django 
from django.contrib.auth.models import User 
from django.conf.urls import url 

#Tasty 
from tastypie.resources import ModelResource 

class UserResource(ModelResource): 
    class Meta: 
     queryset = User.objects.all() 
     resource_name = 'users' 

     #Disallow list operations 
     list_allowed_methods = [] 
     detail_allowed_methods = ['get', 'put', 'patch'] 

     #Exclude some fields 
     excludes = ('first_name', 'is_active', 'is_staff', 'is_superuser', 'last_name', 'password',) 

    #Apply filter for the requesting user 
    def apply_authorization_limits(self, request, object_list): 
     return object_list.filter(pk=request.user.pk) 

    #Override urls such that GET:users/ is actually the user detail endpoint 
    def override_urls(self): 
     return [ 
      url(r"^(?P<resource_name>%s)/$" % self._meta.resource_name, self.wrap_view('dispatch_detail'), name="api_dispatch_detail"), 
     ] 

使用比获取资源的细节主键之外的东西更详细的Tastypie Cookbook