为多表继承Django模型填充tastypie资源

时间：2012-08-12 17:12:37

标签： django django-models tastypie

鉴于以下代码，我想知道如何用每个真实记录数据填充RecordsResource：

models.py 

class Record(models.Model):
    content_type = models.ForeignKey(ContentType, editable=False, null=True)
    user = models.ForeignKey(User, related_name='records')
    issued = models.DateTimeField(auto_now_add=True)
    date = models.DateField()

    def save(self, *args, **kwargs):
        if not self.content_type:
            self.content_type = ContentType.objects.get_for_model(self.__class__)
        super(Record, self).save(*args, **kwargs)

    def as_leaf_class(self):
        model = self.content_type.model_class()
        if model == self.__class__:
            return self
        return model.objects.get(pk=self.id)


class Record1(Record):
    # some fields

# ...

class RecordN(Record):
    # some fields

api.py 

class BaseModelResource(ModelResource):
    class Meta(object):
        authentication = ApiKeyPlusWebAuthentication()
        authorization= Authorization()
        cache = SimpleCache()
        throttle = CacheDBThrottle(
            throttle_at=350,
            # 1 day
            expiration=86400
        )
        if settings.DEBUG:
            serializer = PrettyJSONSerializer()

    def obj_create(self, bundle, request=None, **kwargs):
        return super(BaseModelResource, self).obj_create(bundle, request, user=request.user)

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


class BaseRecordResource(BaseModelResource):
    class Meta(BaseModelResource.Meta):
        filtering = {
            'date': ALL
        }
        excludes = ['issued']

class RecordsResource(BaseRecordResource):
    class Meta(BaseRecordResource.Meta):
        resource_name = 'records'
        queryset = Record.objects.all()

class Record1Resource(BaseRecordResource):
    class Meta(BaseRecordResource.Meta):
        resource_name = 'record1'
        queryset = Record1.objects.all()

# ...

class RecordNResource(BaseRecordResource):
    class Meta(BaseRecordResource.Meta):
        resource_name = 'recordn'
        queryset = RecordN.objects.all()

3 个答案:

答案 0 :(得分：7)

好的，我刚解决了。我简化了代码。

鉴于以下代码......

models.py 

from django.db import models
from model_utils.managers import InheritanceManager


class Place(models.Model):
    name = models.CharField(max_length=50)
    address = models.CharField(max_length=80)

    # https://github.com/carljm/django-model-utils#inheritancemanager
    objects = InheritanceManager()


class Restaurant(Place):
    custom_field = models.BooleanField()


class Bar(Place):
    custom_field = models.BooleanField()

api.py 

from core.models import Place, Restaurant, Bar
# http://django-tastypie.readthedocs.org/en/latest/cookbook.html#pretty-printed-json-serialization
from core.utils import PrettyJSONSerializer
from tastypie.resources import ModelResource


class PlaceResource(ModelResource):
    class Meta:
        queryset = Place.objects.select_subclasses()
        resource_name = 'place'
        serializer = PrettyJSONSerializer()


class RestaurantResource(ModelResource):
    class Meta:
        queryset = Restaurant.objects.all()
        resource_name = 'restaurant'
        serializer = PrettyJSONSerializer()


class BarResource(ModelResource):
    class Meta:
        queryset = Bar.objects.all()
        resource_name = 'bar'
        serializer = PrettyJSONSerializer()

输出

http://localhost:8000/api/v1/bar/?format=json 

{
  "meta": {
    "limit": 20,
    "next": null,
    "offset": 0,
    "previous": null,
    "total_count": 1
  },
  "objects": [
    {
      "address": "dawdaw",
      "custom_field": true,
      "id": "1",
      "name": "dwdwad",
      "resource_uri": "/api/v1/bar/1/"
    }
  ]
}

行

http://localhost:8000/api/v1/restaurant/?format=json 

{
  "meta": {
    "limit": 20,
    "next": null,
    "offset": 0,
    "previous": null,
    "total_count": 1
  },
  "objects": [
    {
      "address": "nhnhnh",
      "custom_field": true,
      "id": "2",
      "name": "nhnhnh",
      "resource_uri": "/api/v1/restaurant/2/"
    }
  ]
}

行

http://localhost:8000/api/v1/place/?format=json 

{
  "meta": {
    "limit": 20,
    "next": null,
    "offset": 0,
    "previous": null,
    "total_count": 2
  },
  "objects": [
    {
      "address": "dawdaw",
      "id": "1",
      "name": "dwdwad",
      "resource_uri": "/api/v1/place/1/"
    },
    {
      "address": "nhnhnh",
      "id": "2",
      "name": "nhnhnh",
      "resource_uri": "/api/v1/place/2/"
    }
  ]
}

我想要实现的目标

{
  "meta": {
    "limit": 20,
    "next": null,
    "offset": 0,
    "previous": null,
    "total_count": 2
  },
  "objects": [
    {
      "address": "dawdaw",
      "custom_field": true,
      "id": "1",
      "name": "dwdwad",
      "resource_uri": "/api/v1/bar/1/"
    },
    {
      "address": "nhnhnh",
      "custom_field": true,
      "id": "2",
      "name": "nhnhnh",
      "resource_uri": "/api/v1/restaurant/2/"
    }
  ]
}

解决方案：

from core.models import Place, Restaurant, Bar
# http://django-tastypie.readthedocs.org/en/latest/cookbook.html#pretty-printed-json-serialization
from core.utils import PrettyJSONSerializer
from tastypie.resources import ModelResource

class RestaurantResource(ModelResource):
    class Meta:
        queryset = Restaurant.objects.all()
        resource_name = 'restaurant'
        serializer = PrettyJSONSerializer()


class BarResource(ModelResource):
    class Meta:
        queryset = Bar.objects.all()
        resource_name = 'bar'
        serializer = PrettyJSONSerializer()

class PlaceResource(ModelResource):
    class Meta:
        queryset = Place.objects.select_subclasses()
        resource_name = 'place'
        serializer = PrettyJSONSerializer()

    def dehydrate(self, bundle):
        # bundle.data['custom_field'] = "Whatever you want"
        if isinstance(bundle.obj, Restaurant):
            restaurant_res = RestaurantResource()
            rr_bundle = restaurant_res.build_bundle(obj=bundle.obj, request=bundle.request)
            bundle.data = restaurant_res.full_dehydrate(rr_bundle).data
        elif isinstance(bundle.obj, Bar):
            bar_res = BarResource()
            br_bundle = bar_res.build_bundle(obj=bundle.obj, request=bundle.request)
            bundle.data = bar_res.full_dehydrate(br_bundle).data
        return bundle

答案 1 :(得分：3)

在RecordsResource类中，您还需要添加模型字段（请参阅https://github.com/tomchristie/django-rest-framework/blob/master/djangorestframework/resources.py#L232-234）

class RecordsResource(BaseRecordResource):
    model = Record

    class Meta(BaseRecordResource.Meta):
        resource_name = 'records'
        queryset = Record.objects.all()

答案 2 :(得分：1)

从头开始解释：

Django中可以有three styles of inheritance。

  1. 通常，您只想使用父类来保存 您不希望为每个孩子输入的信息 模型。这个类不会被孤立地使用，所以 抽象的基类就是你所追求的。

  2. 如果你是现有模型的子类（可能是某些东西） 另一个应用程序完全）并希望每个模型都有自己的 数据库表，多表继承是可行的方法。

  3. 最后，如果你只想修改一个Python的Python级行为 模型，无需以任何方式更改模型字段，即可使用 代理模型。

    4. 这里的选择是多表继承

    多表继承 Django支持的第二种模型继承是当层次结构中的每个模型都是模型本身时。每个模型对应于自己的数据库表，可以单独查询和创建。继承关系引入子模型与其每个父模型之间的链接（通过自动创建的OneToOneField） Ref

    从Record转到Recordx 1 <= x <= n a_example_record = Record.objects,get(pk=3) Recordx，然后使用something like below检查if hasattr(a_example_record, 'record1'): # ... elif hasattr(a_example_record, 'record2'): # ... 的{​​{1}}类型

    TastyPie

    现在，我们知道如何从父级中获取子级，并且我们需要在其元中为queryset提供queryset，您需要编写一个自定义的Record {{1}}模型上的自定义管理器，它会记录您的所有记录（此处为Custom QuerySet and Manager without breaking DRY?），检查它的子类型并将其附加到查询集或列表中。您可以在此处阅读有关附加How to combine 2 or more querysets in a Django view?

    的信息
相关问题
最新问题