Question

我有GFK的页面模型。

class Page(models.Model):
    title = models.CharField(max_length=200)
    content_type = models.ForeignKey(ContentType,null=True,blank=True)
    object_id = models.CharField(max_length=255,null=True,blank=True)
    content_object = generic.GenericForeignKey('content_type', 'object_id')

和

class TextContent(models.Model):
    content = models.TextField(null=True, blank=True)
    pages = generic.GenericRelation(Page)

我做了Page.objects.get（pk = 1）.content_object，我明白了。

请帮我看一下REST中锚定对象的链接（或输出到JSON）。

class PageResource(ModelResource):
    content_object = fields.?????

    class Meta:
        queryset = Page.objects.all()
        resource_name = 'page'

怎么做对了？

谢谢！

的Vitaliy

Answer 1

目前在tastypie中使用泛型关系并不容易。在tastypie github page提交了一些补丁，但截至本文撰写时尚未合并。

最简单的方法是，定义contenttype资源并将其用于具有通用关系的资源。有点像：

class ContentTypeResource(ModelResource):
    class Meta:
        queryset = ContentType.objects.all()
        resource_name = "contrib/contenttype"
        fields = ['model']
        detail_allowed_methods = ['get',]
        list_allowed_methods = ['get']

class PageResource(ModelResource):
    content_object = fields.ToOneField('myresources.ContentTypeResource', 'content_object')


    class Meta:
        queryset = Page.objects.all()
        resource_name = 'page'

希望这有帮助。

Answer 2

“myresources”是包含ContentTypeResource的应用程序。如果它与您的其他资源位于同一个应用程序中，则无需限定它。在下面的代码中删除。

“contrib / contenttype”是资源的名称。设置您自己的名称是可选的。如果您没有指定，Tastypie将为您创建一个。我已在下面的更新代码中将其删除。

fields = ['model']部分限制此资源所代表的模型中的可访问字段。如果你看一下Django代码中ContentType模型的定义，你会发现它有一个名为'model'的字段。

我认为最初的答案是其字段名称混淆了。您正在尝试为content_type创建新资源，并将其挂钩到模型中的content_type外键。上面的代码对此进行了排序。

class ContentTypeResource(ModelResource):
    class Meta:
        queryset = ContentType.objects.all()
        fields = ['model']
        detail_allowed_methods = ['get',]
        list_allowed_methods = ['get']

class PageResource(ModelResource):
    content_type = fields.ToOneField('ContentTypeResource', 'content_type')

    class Meta:
        queryset = Page.objects.all()
        resource_name = 'page'

您还需要在urls.py中注册ContentTypeResource，就像您拥有所有其他资源一样：

from myapp.api import ContentTypeResource

v1_api = Api(api_name='v1')
v1_api.register(ContentTypeResource())

“myapp”位再次是包含ContentTypeResource的api代码的应用程序。

我希望这可以解决问题。我只是让它自己工作......

Answer 3

看起来这个月被正式添加到Tastypie，请查看此处的示例。

https://github.com/toastdriven/django-tastypie/blob/master/docs/content_types.rst

Answer 4

我们破解了代码！

class ContentTypeResource(ModelResource):

    class Meta:
        queryset = ContentType.objects.all()
        resource_name = 'content_type'
        allowed_methods = ['get',]

class PageObjectResource(ModelResource):

    content_object = fields.CharField()

    content_type = fields.ToOneField(
        ContentTypeResource,
        attribute = 'content_type',
        full=True)

    class Meta:
        queryset = models.PageObject.objects.all()
        resource_name = 'page_object'
        allowed_methods = ['get',]

    def dehydrate_content_object(self, bundle):
        for resource in api._registry.values():
            if resource._meta.object_class == bundle.obj.content_object.__class__:
                return resource.full_dehydrate(resource.build_bundle(obj=bundle.obj.content_object, request=bundle.request)).data
        return ''

结果如下：

"page_objects": [
{
"content_object": {
"id": "186",
"look_stills": [
{
"_image": "/static/media/uploads/looks/DSC_0903_PR_MEDIUM_QUALITY_RGB_FA.jpg",
"aspect": "front",
"id": "186",
"look_still_icons": [
{
"colour_code": "58",
"enabled": true,
"id": "186",
"in_stock_only": true,
"look_product": {
"colour_code": "58",
"enabled": true,
"id": "186",
"resource_uri": "/api/look_product/186/",
"style_code": "420215"
},
"resource_uri": "/api/look_still_icon/186/",
"x_coord": 76,
"y_coord": 5
}
],
"ordering": 1,
"resource_uri": "/api/look_still/186/"
}
],
"resource_uri": "/api/look_still_set/186/",
"slug": ""
},
"content_type": {
"app_label": "looks_beta",
"id": "97",
"model": "lookstillset",
"name": "look still set",
"resource_uri": "/api/content_type/97/"
},
"id": "2",
"object_id": 186,
"resource_uri": "/api/page_object/2/"
}
],
"page_order": 3,
"page_template": "look_still",
"resource_uri": "/api/page/2/",
"slug": "",
"spread_number": 2,
"title": ""
},

Answer 5

这为您提供了content_object字段作为嵌套对象。这很简单，很有效，并且（不幸的是）技术可以实现高效率。

class PageResource(ModelResource):

    def full_dehydrate(self, bundle):
        new_bundle = super(PageResource, self).full_dehydrate(bundle)
        new_bundle.data['content_object'] = get_serializable(bundle.obj.content_object)
        return new_bundle

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


def get_serializable(model):

    data = {'type': model.__class__.__name__}
    for field in model._meta.fields:
        data[field.name] = getattr(model, field.name)
    return data

Answer 6

我们设法获取内容对象的uri，如果它有相应的ModelResource：

class ContentTypeResource(ModelResource):

    class Meta:
        queryset = ContentType.objects.all()
        resource_name = 'content_type'
        allowed_methods = ['get',]

class PageObjectResource(ModelResource):

    content_object_uri = fields.CharField()

    content_type = fields.ToOneField(
        ContentTypeResource,
        attribute = 'content_type',
        full=True)

    class Meta:
        queryset = models.PageObject.objects.all()
        resource_name = 'page_object'
        allowed_methods = ['get',]

    def dehydrate_content_object_uri(self, bundle):
        for resource in api._registry.values():
            if resource._meta.object_class == bundle.obj.content_object.__class__:
                return resource.get_resource_uri(bundle.obj.content_object)
        return ''

Answer 7

事实上，正如马里奥所建议的，他们已经加入了对此的支持。既然花了很长时间才弄清楚我认为这可能会帮助一些人。以下是使用Django内置注释模型的示例，其中我得到与注释对象的注释的反向关系：

将此添加到注释附加到的模型：

class CmntedObject(models.Model):
    comments = generic.GenericRelation(Comment,
                           content_type_field='content_type',
                           object_id_field='object_pk')

并且资源看起来像这样：

class UserResource(ModelResource):
    what ever you need here....

class CmntedObjectResource(ModelResource):
    comments = fields.ToManyField('path.to.api.CmntedObjectResource', 'comments', full=True, null=True)
    class Meta:
        queryset = CmntedObject.objects.all()
        resource_name = 'cmntedobject'
        allowed_methods = ['get', 'post', 'delete']
        authorization = DjangoAuthorization()

class CommentResource(ModelResource):
    user = fields.ToOneField('path.to.api.UserResource', 'user', full=True)
    content_type_id = fields.CharField(attribute = 'content_type_id')
    site_id = fields.CharField(attribute = 'site_id')
    content_object = GenericForeignKeyField({
                       CmntedObject: CmntedObjectResource, #shown above
                       OtherCmntedObject: OtherCmntedObjectResource, #optional
                    }, 'content_object', null=True)

    class Meta:
        queryset = Comment.objects.all()
        resource_name = 'cmnt'
        allowed_methods = ['get', 'post', 'delete']
        authorization = DjangoAuthorization()

    def obj_create(self, bundle, **kwargs):
        #here we get the current authenticated user as the comment user.
        bundle = super(CmntResource, self).obj_create(bundle, user=bundle.request.user)
        return bundle

Django tastypie和GenericForeignKey

7 个答案: