嵌套資源用於美食中的多對多關係。父母子女關係

Question

我有兩個資源，客戶和電話（我只是通過只包括幾個字段來簡化它）。不同的客戶可以擁有相同類型的電話。我寫了我的Modelresource類，並通過/ customer /和/ phone訪問API。我現在想做的是爲某個客戶獲取電話。 so/customer/1/phone/

這些是我的類的樣子。

Models.py

# Defines the phone Model 

class Phone(models.Model): 
    phone_id= models.AutoField(primary_key=True) 
    phone_type = models.CharField(max_length=100) 


# Defines the Customer Model 

class Customer(models.Model): 
    customer_id= models.AutoField(primary_key=True) 
    phone = models.ManyToManyField(Phone) 

Api.py

class PhoneResource(ModelResource): 
    class Meta: 
     queryset = Phone.objects.all() 
     allowed_methods = ['get'] 
     resource_name = 'phone' 

class CustomerResource(ModelResource): 
    phone = fields.ManyToManyField(PhoneResource, "phone") 

    class Meta: 
     queryset = Customer.objects.all() 
     allowed_methods = ['get', 'patch', 'put'] 
     resource_name = 'customer' 
     authentication = Authentication() 
     authorization = Authorization() 

    def prepend_urls(self): 
     return [ 
      url(r'^(?P<resource_name>%s)/(?P<pk>\w[\w/-]*)/phone%s$' % 
       (self._meta.resource_name, trailing_slash()), 
       self.wrap_view('get_customer_phone'), name='customer_phone'), 
     ] 

    def customer_phone(self, request, **kwargs): 
     # My Question is what goes in this function 
     # I want to get only the phones for the given customer, and exclude other phones that does not belong to them 

我已經調查http://django-tastypie.readthedocs.org/en/latest/cookbook.html#nested-resources

但它不工作。我一直在收回所有電話，而不僅僅是屬於某個客戶的電話。所以，如果約翰有一個android和一個ios，它應該返回這兩個列表，但如果約翰有android它應該只返回android。但是，我正在使用手機模型中的所有手機。

Answer 1

型號：

class Phone(models.Model): 
    phone_id= models.AutoField(primary_key=True) 
    phone_type = models.CharField(max_length=100) 


# Defines the Customer Model 

class Customer(models.Model): 
    customer_id= models.AutoField(primary_key=True) 
    phones = models.ManyToManyField(Phone, related_name='customers') 

API：

class PhoneResource(ModelResource): 
    # TODO: update path 
    customers = fields.ManyToManyField('path.to.CustomerResource', "customers") 

    class Meta: 
     queryset = Phone.objects.all() 
     allowed_methods = ['get'] 
     resource_name = 'phone' 

class CustomerResource(ModelResource): 
    phones = fields.ManyToManyField(PhoneResource, "phones") 

    class Meta: 
     queryset = Customer.objects.all() 
     allowed_methods = ['get', 'patch', 'put'] 
     resource_name = 'customer' 
     authentication = Authentication() 
     authorization = Authorization() 

    def prepend_urls(self): 
     return [ 
      url(r'^(?P<resource_name>%s)/(?P<pk>\w[\w/-]*)/phone%s$' % 
       (self._meta.resource_name, trailing_slash()), 
       self.wrap_view('get_customer_phone'), name='customer_phone'), 
     ] 

    def get_customer_phone(self, request, **kwargs): 
     try: 
      bundle = self.build_bundle(data={'pk': kwargs['pk']}, request=request) 
      obj = self.cached_obj_get(bundle=bundle, **self.remove_api_resource_names(kwargs)) 
     except ObjectDoesNotExist: 
      return HttpGone() 
     except MultipleObjectsReturned: 
      return HttpMultipleChoices("More than one resource is found at this URI.") 

     phone_resource = PhoneResource() 
     return phone_resource.get_list(request, customers=obj.pk) 

Answer 2

您確定需要單獨的prepend_urls嗎？你可以得到的電話清單爲每個客戶購買添加全=真到ManyToManyField參數：

class CustomerResource(ModelResource): 
    phone = fields.ManyToManyField(PhoneResource, "phone", full=True) 

    class Meta: 
     queryset = Customer.objects.all() 
     allowed_methods = ['get', 'patch', 'put'] 
     resource_name = 'customer' 
     authentication = Authentication() 
     authorization = Authorization()