TastyPie 在脱水过程中序列化

Question

所以我有一个 QuestionResource:

class QuestionResourse(ModelResource):
def dehydrate(self, bundle):
    bundle.data['responses'] = Responses.objects.filter(question_id=bundle.data['id'])
    return bundle
class Meta:
    resource_name='question'
    queryset = Questions.objects.all()
    allowed_methods = ['get', 'post']

如果 url 类似于 https://domain.com/api/v1/question/，它应该 return 附有属性响应的问题。虽然他们没有被连载。

{
"date": "2015-10-03T16:53:22",
"id": "1",
"question": "Where is my mind?",
"resource_uri": "/api/v1/question/1/",
"responses": "[<Responses: Responses object>, <Responses: Responses object>, <Responses: Responses object>, <Responses: Responses object>, <Responses: Responses object>]",
"totalresponses": 5
}

如何序列化 <Responses: Responses object>？另外，如何将 "responses" 变成 json 数组而不是字符串？

编辑： 在 raphv 的帮助下，我在我的资源中使用了这段代码：

class ResponseResourse(ModelResource):
    class Meta:
        resource_name='response'
        queryset = Responses.objects.all()
        allowed_methods = ['get', 'post']
class QuestionResourse(ModelResource):
    responses = fields.ToManyField(ResponseResourse, attribute=lambda bundle: Responses.objects.filter(question_id = bundle.obj.id), full=True)
    class Meta:
        resource_name='question'
        queryset = Questions.objects.all()
        allowed_methods = ['get', 'post']

生产：

{
"date": "2015-10-03T16:53:22",
"id": "1",
"question": "Where is my mind?",
"resource_uri": "/api/v1/question/1/",
"responses": [
    {
        "id": "54",
        "resource_uri": "/api/v1/response/54/",
        "response": "ooooooo oooooo",
    },
    {
        "id": "60",
        "resource_uri": "/api/v1/response/60/",
        "response": "uhh, test",
        "votes": 0
    }]
}

Answer 1

您应该在 api.py.

中创建一个单独的 ResponseResource 和 link

full=True 参数使 API return 成为每个响应的完整表示

from tastypie import resources, fields

class ResponseResource(resources.ModelResource):
    class Meta:
        resource_name = 'response'
        queryset = Responses.objects.all()
        ...

class QuestionResource(resources.ModelResource):
    responses = fields.ToManyField(ResponseResource, 'responses', full=True)
    class Meta:
        resource_name='question'
        queryset = Questions.objects.all()
        ...

TastyPie 在脱水过程中序列化

TastyPie Serializing during dehydrate

python

api

django

serialization

tastypie