问题:不可JSON序列化

我有以下代码序列化查询集;

def render_to_response(self, context, **response_kwargs):

    return HttpResponse(json.simplejson.dumps(list(self.get_queryset())),
                        mimetype="application/json")

以下是我的 get_querset()

[{'product': <Product: hederello ()>, u'_id': u'9802', u'_source': {u'code': u'23981', u'facilities': [{u'facility': {u'name': {u'fr': u'G\xe9n\xe9ral', u'en': u'General'}, u'value': {u'fr': [u'bar', u'r\xe9ception ouverte 24h/24', u'chambres non-fumeurs', u'chambres familiales',.........]}]

我需要序列化。但是它说无法序列化<Product: hederello ()>。因为列表由Django对象和字典组成。有任何想法吗 ?

I have the following code for serializing the queryset;

def render_to_response(self, context, **response_kwargs):

    return HttpResponse(json.simplejson.dumps(list(self.get_queryset())),
                        mimetype="application/json")

And following is my get_querset()

[{'product': <Product: hederello ()>, u'_id': u'9802', u'_source': {u'code': u'23981', u'facilities': [{u'facility': {u'name': {u'fr': u'G\xe9n\xe9ral', u'en': u'General'}, u'value': {u'fr': [u'bar', u'r\xe9ception ouverte 24h/24', u'chambres non-fumeurs', u'chambres familiales',.........]}]

Which I need to serialize. But it says not able to serialize the <Product: hederello ()>. Because list composed of both django objects and dicts. Any ideas ?


回答 0

simplejson并且json不能很好地与Django对象配合使用。

Django的内置序列化器只能序列化由django对象填充的查询集:

data = serializers.serialize('json', self.get_queryset())
return HttpResponse(data, content_type="application/json")

就您而言,self.get_queryset()其中包含django对象和dict的混合。

一种选择是摆脱中的模型实例,self.get_queryset()并使用dict将其替换为model_to_dict

from django.forms.models import model_to_dict

data = self.get_queryset()

for item in data:
   item['product'] = model_to_dict(item['product'])

return HttpResponse(json.simplejson.dumps(data), mimetype="application/json")

希望能有所帮助。

simplejson and json don’t work with django objects well.

Django’s built-in serializers can only serialize querysets filled with django objects:

data = serializers.serialize('json', self.get_queryset())
return HttpResponse(data, content_type="application/json")

In your case, self.get_queryset() contains a mix of django objects and dicts inside.

One option is to get rid of model instances in the self.get_queryset() and replace them with dicts using model_to_dict:

from django.forms.models import model_to_dict

data = self.get_queryset()

for item in data:
   item['product'] = model_to_dict(item['product'])

return HttpResponse(json.simplejson.dumps(data), mimetype="application/json")

Hope that helps.


回答 1

最简单的方法是使用JsonResponse

对于查询集,您应传递该查询集的的列表values,如下所示:

from django.http import JsonResponse

queryset = YourModel.objects.filter(some__filter="some value").values()
return JsonResponse({"models_to_return": list(queryset)})

The easiest way is to use a JsonResponse.

For a queryset, you should pass a list of the the values for that queryset, like so:

from django.http import JsonResponse

queryset = YourModel.objects.filter(some__filter="some value").values()
return JsonResponse({"models_to_return": list(queryset)})

回答 2

我发现可以使用“ .values”方法相当简单地完成此操作,该方法还提供了命名字段:

result_list = list(my_queryset.values('first_named_field', 'second_named_field'))
return HttpResponse(json.dumps(result_list))

必须使用“列表”来获取可迭代的数据,因为“值查询集”类型仅当作为可迭代的拾取时才是字典。

文档:https : //docs.djangoproject.com/en/1.7/ref/models/querysets/#values

I found that this can be done rather simple using the “.values” method, which also gives named fields:

result_list = list(my_queryset.values('first_named_field', 'second_named_field'))
return HttpResponse(json.dumps(result_list))

“list” must be used to get data as iterable, since the “value queryset” type is only a dict if picked up as an iterable.

Documentation: https://docs.djangoproject.com/en/1.7/ref/models/querysets/#values


回答 3

从1.9版本开始,更轻松和官方的获取json的方式

from django.http import JsonResponse
from django.forms.models import model_to_dict


return JsonResponse(  model_to_dict(modelinstance) )

From version 1.9 Easier and official way of getting json

from django.http import JsonResponse
from django.forms.models import model_to_dict


return JsonResponse(  model_to_dict(modelinstance) )

回答 4

我们的js程序员要求我向她返回确切的JSON格式数据,而不是json编码的字符串。

下面是解决方案(这将返回一个可以在浏览器中直接使用/查看的对象)

import json
from xxx.models import alert
from django.core import serializers

def test(request):
    alert_list = alert.objects.all()

    tmpJson = serializers.serialize("json",alert_list)
    tmpObj = json.loads(tmpJson)

    return HttpResponse(json.dumps(tmpObj))

Our js-programmer asked me to return the exact JSON format data instead of a json-encoded string to her.

Below is the solution.(This will return an object that can be used/viewed straightly in the browser)

import json
from xxx.models import alert
from django.core import serializers

def test(request):
    alert_list = alert.objects.all()

    tmpJson = serializers.serialize("json",alert_list)
    tmpObj = json.loads(tmpJson)

    return HttpResponse(json.dumps(tmpObj))

回答 5

首先,我在模型中添加了to_dict方法;

def to_dict(self):
    return {"name": self.woo, "title": self.foo}

然后我有这个;

class DjangoJSONEncoder(JSONEncoder):

    def default(self, obj):
        if isinstance(obj, models.Model):
            return obj.to_dict()
        return JSONEncoder.default(self, obj)


dumps = curry(dumps, cls=DjangoJSONEncoder)

最后使用此类来序列化我的查询集。

def render_to_response(self, context, **response_kwargs):
    return HttpResponse(dumps(self.get_queryset()))

这个效果很好

First I added a to_dict method to my model ;

def to_dict(self):
    return {"name": self.woo, "title": self.foo}

Then I have this;

class DjangoJSONEncoder(JSONEncoder):

    def default(self, obj):
        if isinstance(obj, models.Model):
            return obj.to_dict()
        return JSONEncoder.default(self, obj)


dumps = curry(dumps, cls=DjangoJSONEncoder)

and at last use this class to serialize my queryset.

def render_to_response(self, context, **response_kwargs):
    return HttpResponse(dumps(self.get_queryset()))

This works quite well


声明:本站所有文章,如无特殊说明或标注,均为本站原创发布。任何个人或组织,在未征得本站同意时,禁止复制、盗用、采集、发布本站内容到任何网站、书籍等各类媒体平台。如若本站内容侵犯了原著者的合法权益,可联系我们进行处理。