Python Django rest框架如何序列化对象/外键
我对Django Rest框架有一个问题。 我有以下代码: 序列化程序.pyPython Django rest框架如何序列化对象/外键,python,django,django-rest-framework,Python,Django,Django Rest Framework,我对Django Rest框架有一个问题。 我有以下代码: 序列化程序.py class TextSerializer(serializers.ModelSerializer): class Meta: model = Text fields = ('title', 'project', 'content') def create(self, validated_data): return Text.objects.create(
class TextSerializer(serializers.ModelSerializer):
class Meta:
model = Text
fields = ('title', 'project', 'content')
def create(self, validated_data):
return Text.objects.create(**validated_data)
views.py
@csrf_exempt
def text_view_set(request, project_id):
project = get_object_or_404(Project, pk=project_id)
if request.method == 'POST':
data = JSONParser().parse(request)
serializer = TextSerializer(data=data)
# How to put project in serializer to is_valid() return True?
if serializer.is_valid():
serializer.save()
return JSONResponse(serializer.data, status=status.HTTP_201_CREATED)
return JSONResponse(serializer.errors, status=status.HTTP_400_BAD_REQUEST)
my_command.py
def handle(self, *args, **options):
text = u'Lorem ipsum dolor sit amet, deseruisse voluptatum est cu, ea elit tation delicatissimi per. Decore soleat pri at.'
url = u'http://localhost:8080/text/4/'
params = { 'title' : u'Hello World', 'content': text.encode('utf8'), }
req = requests.post(url, data=json.dumps(params), headers={"Content-Type": "application/json"})
因此,结果是错误的请求错误(400),因为is_valid()方法返回False。没关系。字段“project”未在数据对象中序列化
但是,如何在视图或模型方法中插入此值?
我不想在参数中传递此值(项目),因为它已在URL中:
url = u'http://localhost:8080/text/4/'
像……这样的怎么样
data = JSONParser().parse(request)
data.update({'pk': project_id})
为什么不将“url”映射到指向关系的HyperlinkedRelatedField(使用“源”)?太好了!这是我的工作<代码>数据.更新({'project':project_id})