Django的Tastypie通用关系

我有一个这样的Django项目的一些型号：Django的Tastypie通用关系

class Link(BaseModel, BeginEndModel): 
    entity0_content_type = models.ForeignKey(ContentType, related_name='link_from') 
    entity0_object_id = models.PositiveIntegerField() 
    entity0_content_object = generic.GenericForeignKey('entity0_content_type', 'entity0_object_id') 

    entity1_content_type = models.ForeignKey(ContentType, related_name='link_to') 
    entity1_object_id = models.PositiveIntegerField() 
    entity1_content_object = generic.GenericForeignKey('entity1_content_type', 'entity1_object_id') 

    link_type = models.ForeignKey(LinkType) 

class Work(BaseModel, SluggedModel): 
    """ Eser """ 
    name = models.CharField(max_length=255) 
    links = generic.GenericRelation('Link', content_type_field='entity0_content_type', object_id_field='entity0_object_id')

我想创建一个类似与Tasypie API来WorkResource：

from tastypie.resources import ModelResource, ALL, ALL_WITH_RELATIONS 
from tastypie import fields, utils 
from tastypie.contrib.contenttypes.fields import GenericForeignKeyField 
from tastypie.authentication import Authentication, SessionAuthentication 
from tastypie.authorization import DjangoAuthorization, Authorization 
from models import Link, LinkType, LinkPhrase 
from models import Work 

.... 

class WorkResource(BaseModelResource): 
    links = fields.ToManyField('musiclibrary.api.LinkResource', 'links_set') 

    class Meta: 
     queryset = Work.objects.all() 
     always_return_data = True 
     filtering = { 
      'slug': ALL, 
      'name': ['contains', 'exact'] 
     } 

class LinkResource(ModelResource): 
    entity0_content_object = GenericForeignKeyField({ 
     Work: WorkResource, 
     Artist: ArtistResource 
    }, 'entity0_content_object') 
    entity1_content_object = GenericForeignKeyField({ 
     Work: WorkResource, 
     Artist: ArtistResource 
    }, 'entity1_content_object') 

    link_type = fields.ForeignKey(LinkTypeResource, 'link_type', full=True, null=True) 

    class Meta: 
     queryset = Link.objects.all()

当我想尝试查看工作资源结果，links属性始终是一个空数组。为什么我无法建立2资源之间的关系？

注：我使用Django 1.6.5，django-tastypie 0.11.1。我简化了我上面的models.py和api.py示例。如果需要，我可以分享我的完整代码。

来源

2014-08-31 Murat Çorlu

它有点棘手，因为有与ContentTypes飞来飞去的2路关系。我想这会有所帮助：

class WorkResource(BaseModelResource): 
    links = fields.ToManyField('musiclibrary.api.LinkResource', attribute=lambda bundle: Link.objects.filter(entity0_content_type=ContentType.objects.get_for_model(bundle.obj), entity0_object_id=bundle.obj.id))

来源

2014-09-03 07:13:48 frail

Django的Tastypie通用关系

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