blocks|key|1365746|text|似乎你不能序列化一个实例，你必须序列化一个对象的QuerySet。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1365747|from+django.core+import+serializers
from+models+import+*

def+getUser(request):
++++return+HttpResponse(json(Users.objects.filter(id=88)))|code-block|syntax|javascript|1365748|我已经用完了django的svn发行版，所以在早期版本中可能没有这个版本。|offset|length|style|CODE|1365749|entityMap^0|0|0|D|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

It doesn't seem you can serialize an instance, you'd have to serialize a QuerySet of one object.

<pre><code>from django.core import serializers
from models import *

def getUser(request):
 return HttpResponse(json(Users.objects.filter(id=88)))
</code></pre>

I run out of the <code>svn</code> release of django, so this may not be in earlier versions.

blocks|key|1365836|text|听起来您所询问的内容涉及到序列化Django模型实例的数据结构以实现互操作性。其他的帖子是正确的:如果您希望序列化的表单与可以通过Django的api查询数据库的python应用程序一起使用，那么您会希望使用一个对象来序列化一个查询集。另一方面，如果您需要一种在不接触数据库或不使用Django的情况下在其他地方重新膨胀模型实例的方法，那么您需要做一些工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1365837|下面是我要做的：|1365838|首先，我使用demjson进行转换。它碰巧是我最先发现的，但可能不是最好的。我的实现依赖于它的一个特性，但其他转换器也应该有类似的方法。|offset|length|style|CODE|1365839|其次，在可能需要序列化的所有模型上实现json_equivalent方法。对于demjson来说，这是一个神奇的方法，但无论您选择哪种实现，都可能需要考虑它。其思想是返回一个可直接转换为json的对象(即数组或字典)。如果您真的想自动执行此操作：|1365840|def+json_equivalent(self):
++++dictionary+=+{}
++++for+field+in+self._meta.get_all_field_names()
++++++++dictionary[field]+=+self.__getattribute__(field)
++++return+dictionary|code-block|syntax|javascript|1365841|除非您有一个完全扁平的数据结构(没有ForeignKeys，数据库中只有数字和字符串等)，否则这对您没有帮助。否则，您应该认真考虑实现此方法的正确方式。|1365842|第三，调用demjson.JSON.encode(instance)，你就得到了你想要的东西。|1365843|entityMap|0|LINK|mutability|MUTABLE|url|http://deron.meranda.us/python/demjson/^0|0|0|6|7|6|7|0|0|J|F|13|7|2L|4|0|0|I|B|0|5|T|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|14|8|@$F|15|G|16|H|I]]|9|@$F|17|G|18|1|19]]|A|$]]|$1|J|3|K|5|6|7|1A|8|@$F|1B|G|1C|H|I]|$F|1D|G|1E|H|I]|$F|1F|G|1G|H|I]]|9|@]|A|$]]|$1|L|3|M|5|N|7|1H|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1I|8|@$F|1J|G|1K|H|I]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1L|8|@$F|1M|G|1N|H|I]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

It sounds like what you're asking about involves serializing the data structure of a Django model instance for interoperability. The other posters are correct: if you wanted the serialized form to be used with a python application that can query the database via Django's api, then you would wan to serialize a queryset with one object. If, on the other hand, what you need is a way to re-inflate the model instance somewhere else without touching the database or without using Django, then you have a little bit of work to do.

Here's what I do:

First, I use <a href="http://deron.meranda.us/python/demjson/" rel="noreferrer"><code>demjson</code></a> for the conversion. It happened to be what I found first, but it might not be the best. My implementation depends on one of its features, but there should be similar ways with other converters.

Second, implement a <code>json_equivalent</code> method on all models that you might need serialized. This is a magic method for <code>demjson</code>, but it's probably something you're going to want to think about no matter what implementation you choose. The idea is that you return an object that is directly convertible to <code>json</code> (i.e. an array or dictionary). If you really want to do this automatically:

<pre><code>def json_equivalent(self):
 dictionary = {}
 for field in self._meta.get_all_field_names()
 dictionary[field] = self.__getattribute__(field)
 return dictionary
</code></pre>

This will not be helpful to you unless you have a completely flat data structure (no <code>ForeignKeys</code>, only numbers and strings in the database, etc.). Otherwise, you should seriously think about the right way to implement this method.

Third, call <code>demjson.JSON.encode(instance)</code> and you have what you want.

blocks|key|1581050|text|ville+=+UneVille.objects.get(nom='lihlihlihlih')
....
blablablab
.......

return+HttpResponse(simplejson.dumps(ville.__dict__))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1581051|我返回我的实例的判决|unstyled|1581052|因此它返回类似{'field1':value，"field2":value，....}|1581053|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>ville = UneVille.objects.get(nom='lihlihlihlih')
....
blablablab
.......

return HttpResponse(simplejson.dumps(ville.__dict__))
</code></pre>

I return the dict of my instance

so it return something like {'field1':value,"field2":value,....}

blocks|key|1581130|text|我通过向我的模型添加一个序列化方法解决了这个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1581131|def+toJSON(self):
++++import+simplejson
++++return+simplejson.dumps(dict([(attr,+getattr(self,+attr))+for+attr+in+[f.name+for+f+in+self._meta.fields]]))|code-block|syntax|javascript|1581132|对于那些不喜欢一行程序的人，下面是详细的等价物：|1581133|def+toJSON(self):
++++fields+=+[]
++++for+field+in+self._meta.fields:
++++++++fields.append(field.name)

++++d+=+{}
++++for+attr+in+fields:
++++++++d[attr]+=+getattr(self,+attr)

++++import+simplejson
++++return+simplejson.dumps(d)|1581134|_meta.fields是模型字段的有序列表，可以从实例和模型本身进行访问。|offset|length|style|CODE|1581135|entityMap^0|0|0|0|0|0|C|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

I solved this problem by adding a serialization method to my model:

<pre><code>def toJSON(self):
 import simplejson
 return simplejson.dumps(dict([(attr, getattr(self, attr)) for attr in [f.name for f in self._meta.fields]]))
</code></pre>

Here's the verbose equivalent for those averse to one-liners:

<pre><code>def toJSON(self):
 fields = []
 for field in self._meta.fields:
 fields.append(field.name)

 d = {}
 for attr in fields:
 d[attr] = getattr(self, attr)

 import simplejson
 return simplejson.dumps(d)
</code></pre>

<code>_meta.fields</code> is an ordered list of model fields which can be accessed from instances and from the model itself.

blocks|key|1366086|text|您可以轻松地使用列表来包装所需的对象，这就是django序列化程序正确序列化它所需的全部内容，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366087|from+django.core+import+serializers

#+assuming+obj+is+a+model+instance
serialized_obj+=+serializers.serialize('json',+[+obj,+])|code-block|syntax|javascript|1366088|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can easily use a list to wrap the required object and that's all what django serializers need to correctly serialize it, eg.:

<pre><code>from django.core import serializers

# assuming obj is a model instance
serialized_obj = serializers.serialize('json', [ obj, ])
</code></pre>

blocks|key|1581271|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1581272|def+ins2dic(obj):
++++SubDic+=+obj.__dict__
++++del+SubDic['id']
++++del+SubDic['_state']
return+SubDic|code-block|syntax|javascript|1581273|或者排除任何你不想要的东西。|1581274|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

how about this way:

<pre><code>def ins2dic(obj):
 SubDic = obj.__dict__
 del SubDic['id']
 del SubDic['_state']
return SubDic
</code></pre>

or exclude anything you don't want.

blocks|key|1581337|text|要避免数组包装，请在返回响应之前将其删除：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1581338|import+json
from+django.core+import+serializers

def+getObject(request,+id):
++++obj+=+MyModel.objects.get(pk=id)
++++data+=+serializers.serialize('json',+[obj,])
++++struct+=+json.loads(data)
++++data+=+json.dumps(struct[0])
++++return+HttpResponse(data,+mimetype='application/json')|code-block|syntax|javascript|1581339|我也发现了这篇关于这个话题的有趣的帖子：|1581340|http://timsaylor.com/convert-django-model-instances-to-dictionaries|offset|length|1581341|它使用django.forms.models.model_to_dict，这看起来是完成这项工作的完美工具。|1581342|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|1V|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|X|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Y|8|@]|9|@$K|Z|L|10|1|11]]|A|$]]|$1|M|3|N|5|6|7|12|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|J]]]]

To avoid the array wrapper, remove it before you return the response:

<pre><code>import json
from django.core import serializers

def getObject(request, id):
 obj = MyModel.objects.get(pk=id)
 data = serializers.serialize('json', [obj,])
 struct = json.loads(data)
 data = json.dumps(struct[0])
 return HttpResponse(data, mimetype='application/json')
</code></pre>

I found this interesting post on the subject too:

<a href="http://timsaylor.com/convert-django-model-instances-to-dictionaries">http://timsaylor.com/convert-django-model-instances-to-dictionaries</a>

It uses django.forms.models.model_to_dict, which looks like the perfect tool for the job.

blocks|key|1366247|text|下面是我的解决方案，它允许您轻松地定制JSON以及组织相关记录|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366248|首先在模型上实现一个方法。我称它为json，但你可以随心所欲地叫它，例如：|offset|length|style|CODE|1366249|class+Car(Model):
++++...
++++def+json(self):
++++++++return+{
++++++++++++'manufacturer':+self.manufacturer.name,
++++++++++++'model':+self.model,
++++++++++++'colors':+[color.json+for+color+in+self.colors.all()],
++++++++}|code-block|syntax|javascript|1366250|然后在视图中我这样做：|1366251|data+=+[car.json+for+car+in+Car.objects.all()]
return+HttpResponse(json.dumps(data),+content_type='application/json;+charset=UTF-8',+status=status)|1366252|entityMap^0|0|H|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Here's my solution for this, which allows you to easily customize the JSON as well as organize related records

Firstly implement a method on the model. I call is <code>json</code> but you can call it whatever you like, e.g.:

<pre><code>class Car(Model):
 ...
 def json(self):
 return {
 'manufacturer': self.manufacturer.name,
 'model': self.model,
 'colors': [color.json for color in self.colors.all()],
 }
</code></pre>

Then in the view I do:

<pre><code>data = [car.json for car in Car.objects.all()]
return HttpResponse(json.dumps(data), content_type='application/json; charset=UTF-8', status=status)
</code></pre>

blocks|key|1581438|text|要序列化和反序列化，请使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1581439|from+django.core+import+serializers

serial+=+serializers.serialize("json",+[obj])
...
#+.next()+pulls+the+first+object+out+of+the+generator
#+.object+retrieves+django+object+the+object+from+the+DeserializedObject
obj+=+next(serializers.deserialize("json",+serial)).object|code-block|syntax|javascript|1581440|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

To serialize and deserialze, use the following:

<pre><code>from django.core import serializers

serial = serializers.serialize("json", [obj])
...
# .next() pulls the first object out of the generator
# .object retrieves django object the object from the DeserializedObject
obj = next(serializers.deserialize("json", serial)).object
</code></pre>

blocks|key|1366311|text|使用列表，它会解决问题|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366312|Step1：|1366313|+result=YOUR_MODELE_NAME.objects.values('PROP1','PROP2').all();|code-block|syntax|javascript|1366314|Step2：|1366315|+result=list(result)++#after+getting+data+from+model+convert+result+to+list|1366316|Step3：|1366317|+return+HttpResponse(json.dumps(result),+content_type+=+"application/json")|1366318|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Use list, it will solve problem

Step1:

<pre><code> result=YOUR_MODELE_NAME.objects.values('PROP1','PROP2').all();
</code></pre>

Step2:

<pre><code> result=list(result) #after getting data from model convert result to list
</code></pre>

Step3:

<pre><code> return HttpResponse(json.dumps(result), content_type = "application/json")
</code></pre>

blocks|key|1366383|text|这个问题有一个很好的答案，我很惊讶它没有被提及。只需几行代码，您就可以处理日期、模型和其他所有内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366384|制作一个可以处理模型的自定义编码器：|1366385|from+django.forms+import+model_to_dict
from+django.core.serializers.json+import+DjangoJSONEncoder
from+django.db.models+import+Model

class+ExtendedEncoder(DjangoJSONEncoder):

++++def+default(self,+o):

++++++++if+isinstance(o,+Model):
++++++++++++return+model_to_dict(o)

++++++++return+super().default(o)|code-block|syntax|javascript|1366386|现在，在使用json.dumps时使用它|1366387|json.dumps(data,+cls=ExtendedEncoder)|1366388|现在，模型、日期和一切都可以序列化，而不必在数组中或序列化和非序列化。您拥有的任何自定义内容都可以添加到default方法中。|offset|length|style|CODE|1366389|你甚至可以这样使用Django的原生JsonResponse：|1366390|from+django.http+import+JsonResponse

JsonResponse(data,+encoder=ExtendedEncoder)|1366391|entityMap^0|0|0|0|0|0|1G|7|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|13|8|@$O|14|P|15|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

There is a good answer for this and I'm surprised it hasn't been mentioned. With a few lines you can handle dates, models, and everything else.
Make a custom encoder that can handle models:
<pre><code>from django.forms import model_to_dict
from django.core.serializers.json import DjangoJSONEncoder
from django.db.models import Model

class ExtendedEncoder(DjangoJSONEncoder):

 def default(self, o):

 if isinstance(o, Model):
 return model_to_dict(o)

 return super().default(o)
</code></pre>
Now use it when you use json.dumps
<pre><code>json.dumps(data, cls=ExtendedEncoder)
</code></pre>
Now models, dates and everything can be serialized and it doesn't have to be in an array or serialized and unserialized. Anything you have that is custom can just be added to the <code>default</code> method.
You can even use Django's native JsonResponse this way:
<pre><code>from django.http import JsonResponse

JsonResponse(data, encoder=ExtendedEncoder)
</code></pre>

blocks|key|1366431|text|与我对框架的期望相比，所有这些答案都有点老生常谈，如果你使用的是rest框架，我认为这是目前为止最简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1366432|rep+=+YourSerializerClass().to_representation(your_instance)
json.dumps(rep)|code-block|syntax|javascript|1366433|这将直接使用序列化程序，考虑到您在序列化程序上定义的字段以及任何关联等。|1366434|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

All of these answers were a little hacky compared to what I would expect from a framework, the simplest method, I think by far, if you are using the rest framework:

<pre><code>rep = YourSerializerClass().to_representation(your_instance)
json.dumps(rep)
</code></pre>

This uses the Serializer directly, respecting the fields you've defined on it, as well as any associations, etc.

blocks|key|1581653|text|这是一个项目，它可以序列化(现在基于JSON)模型中的所有数据，并自动将它们放到特定的目录中，然后它可以随时反序列化它……我亲自使用这个脚本序列化了数千条记录，然后将所有记录加载回另一个数据库，而不会丢失任何数据。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1581654|任何对开源项目感兴趣的人都可以贡献这个项目，并为其添加更多功能。|1581655|serializer_deserializer_model|offset|length|1581656|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/rainyman2012/model_serializer_deserializer/blob/master/serializer_deserializer.py^0|0|0|0|T|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@$F|S|G|T|1|U]]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

This is a project that it can serialize(JSON base now) all data in your model and put them to a specific directory automatically and then it can deserialize it whenever you want... I've personally serialized thousand records with this script and then load all of them back to another database without any losing data.
Anyone that would be interested in opensource projects can contribute this project and add more feature to it.
<a href="https://github.com/rainyman2012/model_serializer_deserializer/blob/master/serializer_deserializer.py" rel="nofollow noreferrer">serializer_deserializer_model</a>

There is a lot of documentation on how to serialize a Model QuerySet but how do you just serialize to JSON the fields of a Model Instance?

How do you serialize a model instance in Django?

数据结构

数据库

Django

JSON

有很多关于如何序列化Model的文档，但是如何才能将QuerySet实例的字段序列化呢？

问如何在Django中序列化模型实例？
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