4

Tom Christie 帮助我在正确的方向上使用 REST 框架,但我现在遇到了另一个问题:

注意:这是使用 viewsets.ModelViewSet

在我的原始代码中,我可以通过在模型实例 xyz(它保存像“20x40x50”这样的坐标数据)上使用 zip() 和 split() 来返回坐标 JSON 数据。我调用了我自己的 toJSON() 函数来输出我需要的所有内容的 JSON 就绪输出。结果是这样的:

[
  {
   "id" : "4"
   "x" : "500",
   "Y" : "80",
   "z" : "150"
   "color" : "yellow"
  },
  ...
]

使用 REST Framework 序列化程序的问题是我只知道如何做 serializers.Field(source"xyz") 事情。我不知道如何将“x”“y”“z”作为单独的字段返回,而不是“xyz”作为一个大字段返回。

这是我的代码:

serializers.py:
---------------
class NoteSerializer(serializers.ModelSerializer):
    owner = serializers.Field(source='owner.username')
    firstname = serializers.Field(source='owner.first_name')
    lastname = serializers.Field(source='owner.last_name')

    x = ???
    y = ???
    z = ???

class Meta:
    model = Note
    fields = ('id','owner','firstname','lastname','text','color', 'x', 'y, 'z', 'time')

这是视图:

    views.py:
    ---------
    def list(self, request, format=None):
        if request.method == 'GET':
            queryset = Note.objects.filter(owner=request.user)
            serializer = NoteSerializer(queryset, many=True)
            if 'text' in request.GET:
                if self.is_numeric(request.GET['id']) and self.is_numeric(request.GET['x']) and self.is_numeric(request.GET['y']) and self.is_numeric(request.GET['z']):

                    serializer = NoteSerializer(data=request.QUERY_PARAMS)
                    intx = int(float(request.GET['x']))
                    inty = int(float(request.GET['y']))
                    intz = int(float(request.GET['z']))
                    serializer.object.xyz = str(intx) +'x'+ str(inty) +'x'+ str(intz)
                    serializer.save()

            return Response(serializer.data, status=status.HTTP_201_CREATED)

    def create(self, request, format=None):

        serializer = NoteSerializer(data=request.DATA)
        if serializer.is_valid():
            serializer.object.owner = request.user
            serializer.save()

            return Response(serializer.data, status=status.HTTP_201_CREATED)
        return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

这是我的模型:

from django.db import models
import datetime
import json
from django.utils import timezone
from django.core.urlresolvers import reverse
from django.core import serializers
from django.contrib.auth.models import User

class Note(models.Model):
owner = models.ForeignKey('auth.User', null=True)
text = models.CharField(max_length=500)
color = models.CharField(max_length=20)
xyz = models.CharField(max_length=20)
time = models.DateTimeField((u"Note Creation Date and Time"), auto_now_add=True,  blank=True)

def __unicode__(self):
    return unicode(self.owner)

非常感谢您的帮助!我是 Python/Django/REST 的新手。这似乎很有趣,但让我沮丧了好几天。

更新:

看来我无法通过带有 serializer.object.xyz 的 views.py 访问 xyz。它说同样的错误“Nonetype没有属性xyz”

serializer = NoteSerializer(data=request.QUERY_PARAMS)
intx = int(float(request.GET['x']))
inty = int(float(request.GET['y']))
intz = int(float(request.GET['z']))
serializer.object.xyz = str(intx) +'x'+ str(inty) +'x'+ str(intz)
serializer.save()
4

2 回答 2

2

您可以尝试使用get_serializer_context将 x,y,z 发送到序列化程序,例如...

from core import models

from rest_framework import generics
from rest_framework import serializers

class BloopModelSerializer(serializers.ModelSerializer):
    x_coord = serializers.SerializerMethodField('get_x')

    def get_x(self, instance):
        return self.context['x']

    class Meta:
        model = models.Bloop

class BloopAPIView(generics.ListCreateAPIView):
    serializer_class = BloopModelSerializer
    queryset = models.Bloop.objects.all()

    def get_serializer_context(self):
        context = super(BloopAPIView, self).get_serializer_context()
        # you have access to self.request here
        context.update({
            'x': 1111,
            'y': 2222
        })

        return context

这样您就不必再覆盖列表并创建函数了。

附带说明一下,您还可以尝试将您的坐标放入它自己的序列化程序中,并将它们分组到序列化程序中。rest_framework 可以嵌套序列化器,所以你可以添加一个CoordinatesSerializer类,你的模型序列化器看起来像这样

class CoordinateSerializer(serializers.Serializer):
    x = models.Field()
    y = models.Field()
    # ...

class BloopModelSerializer(serializers.ModelSerializer):
    coordinates = CoordinateSerializer('get_coords')

    def get_coords(self, instance):
        return self.context['coords']

    class Meta:
        model = models.Bloop
于 2013-09-23T02:14:02.983 回答
2

我对此的做法:

COORD = dict(x=0, y=1, z=2)

class CoordinateField(serializers.Field):
    def field_to_native(self, obj, field_name):
        # retrieve and split coordinates
        coor = obj.content.split('x')

        # get coordinate value depending on coordinate key (x,y,z)
        return int(coor[COORD[field_name]])

    def field_from_native(self, data, files, field_name, into):
        into['xyz'] = u'{x}x{y}x{z}'.format(**data)
        super(CoordinateField, self).field_from_native(data, files, field_name, into)

class BloopModelSerializer(serializers.ModelSerializer):
    x = CoordinateField()
    y = CoordinateField()
    z = CoordinateField()

    class Meta:
        model = Bloop

这就是我得到的结果:

{
    "x": 10,
    "y": 20,
    "z": 30,
    "content": "10x20x30"
},

编辑:

视图.py

class BloopList(generics.ListCreateAPIView):
    queryset = Bloop.objects.all()
    serializer_class = BloopModelSerializer

bloop_list = Bloop.as_view()

网址.py

url(r'^api/bloops/$', 'myapp.views.bloop_list', name='bloop-list'),

建议

您不应该使用 list GET 方法来更改/添加对象,DRF 具有内置类,使其非常容易并让您遵循正确的 REST 标准。

例如,您的列表方法使用 GET 参数获取请求数据,这是一个坏主意,每当您更新或添加新对象时,您应该使用 POST 或 PUT 在请求正文中提供数据。DRF 假设,这就是提供数据并处理所有事情的方式。

于 2013-09-23T21:55:06.957 回答