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[英]Django rest-framework filter models by the name of other models connected with foreign key
[英]Django rest-framework howto serialize a object/foreign key
我的Django Rest Framework有問題。 我有以下代碼:
serializers.py
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,結果是一個錯誤的請求錯誤(400),因為is_valid()方法返回False。 沒關系。 字段“項目”不在數據對象中序列化。
但是,如何在View或Model方法中插入此值? 我不想在參數中傳遞此值(項目),因為它已經存在於URL中:
url = u'http://localhost:8080/text/4/'
怎么樣
data = JSONParser().parse(request)
data.update({'pk': project_id})
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