Django的Tastypie通用關係

我有一個這樣的Django項目的一些型號：Django的Tastypie通用關係

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')

我想創建一個類似與Tasypie API來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()

當我想嘗試查看工作資源結果，links屬性始終是一個空數組。爲什麼我無法建立2資源之間的關係？

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

來源

2014-08-31 Murat Çorlu

它有點棘手，因爲有與ContentTypes飛來飛去的2路關係。我想這會有所幫助：

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))

來源

2014-09-03 07:13:48 frail

Django的Tastypie通用關係

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