Django Tastypie 通用关系答案

【问题标题】：Django Tastypie Generic RelationDjango Tastypie 通用关系
【发布时间】：2014-08-31 22:21:28
【问题描述】：

我在 Django 项目中有一些类似的模型：

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')

我想像这样使用 Tasypie Api 创建一个 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时，links 属性始终是一个空数组。为什么我无法在 2 个资源之间建立关系？

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

【问题讨论】：

标签： python django django-models tastypie

【解决方案1】：

这有点棘手，因为 ContentType 有 2 种方式的关系。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))

【讨论】：