脱水期间的 TastyPie 序列化答案

【问题标题】：TastyPie Serializing during dehydrate脱水期间的 TastyPie 序列化
【发布时间】：2016-07-11 08:18:30
【问题描述】：

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/，它应该返回带有属性响应的问题。虽然它们没有被序列化。

{
"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" 制作成json 数组而不是字符串？

编辑： 在 raphv 的帮助下，我在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
    }]
}

【问题讨论】：

Python 中编码的 JSON 始终是字符串。要将其转换为 array，您必须对其进行解码。 json.loads 这样做。

标签： python django api serialization tastypie

【解决方案1】：

您应该创建一个单独的ResponseResource 并在 api.py 中链接两者。

full=True 参数使 API 返回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()
        ...

【讨论】：