使用Django Rest Framework 3.2.2的可写嵌套序列化器和现有对象

时间:2015-08-20 16:06:08

标签: django django-rest-framework

考虑具有关联类别的Listing模型。我想通过对数据执行POST来为现有类别创建新的列表: {"title": "myapp", "category": {"name": "Business"}},其中title是应创建的列表的标题,Business是要用于此新列表的现有类别的名称。

当我尝试发出这样的请求并为此实例化ListingSerializer时,我收到错误,指示类别名称必须是唯一的 - 我不想创建新的类别,但是使用而现有的。我已经尝试将类别字段上的验证器设置为[],但这并没有改变行为。

我可以使用SlugRelatedField,但这会强制我的请求数据看起来更像{"title": "myapp", "category": "Business"},这不是我想要的。我尝试使用source的{​​{1}}参数来指定嵌套关系,但这不起作用:

SlugRelatedField

的产率:

category = serializers.SlugRelatedField(
        slug_field='category.name',
        queryset=models.Category.objects.all()
    )

models.py:

  "category": [
    "Object with name={'name': 'Business'} does not exist."
  ]

serializers.py:

import django.contrib.auth
from django.db import models
from django.conf import settings

class Profile(models.Model):
    display_name = models.CharField(max_length=255)
    user = models.OneToOneField(settings.AUTH_USER_MODEL)

class Category(models.Model):
    name = models.CharField(max_length=50, unique=True)
    description = models.CharField(max_length=200)

class Listing(models.Model):
    title = models.CharField(max_length=50, unique=True)
    category = models.ForeignKey(Category, related_name='listings', null=True)
    owners = models.ManyToManyField(
        Profile,
        related_name='owned_listings',
        db_table='profile_listing',
        blank=True
    )

views.py:

import logging
import django.contrib.auth
from rest_framework import serializers
import myapp.models as models

logger = logging.getLogger('mylogger')

class ShortUserSerializer(serializers.ModelSerializer):
    class Meta:
        model = django.contrib.auth.models.User
        fields = ('username', 'email')

class ProfileSerializer(serializers.ModelSerializer):
    user = ShortUserSerializer()
    class Meta:
        model = models.Profile
        fields = ('user', 'display_name')
        read_only = ('display_name',)

class CategorySerializer(serializers.ModelSerializer):
    class Meta:
        model = models.Category
        fields = ('name', 'description')
        read_only = ('description',)


class ListingSerializer(serializers.ModelSerializer):
    owners = ProfileSerializer(required=False, many=True)
    # TODO: how to indicate that this should look for an existing category?
    category = CategorySerializer(required=False, validators=[])

    class Meta:
        model = models.Listing
        depth = 2

    def validate(self, data):
        logger.info('inside ListingSerializer validate')
        return data

    def create(self, validated_data):
        logger.info('inside ListingSerializer.create')
        # not even getting this far...

完整示例:https://github.com/arw180/drf-example

3 个答案:

答案 0 :(得分:3)

CategorySerializer.create

上将update_or_create转换为name方法
class CategorySerializer(serializers.ModelSerializer):

    ...

    # update_or_create on `name`
    def create(self, validated_data):
        try:
            self.instance = Category.objects.get(name=validated_data['name'])
            self.instance = self.update(self.instance, validated_data)
            assert self.instance is not None, (
                '`update()` did not return an object instance.'
            )
            return self.instance
        except Category.DoesNotExist:
            return super(CategorySerializer, self).create(validated_data)

    ...

我建议您在需要创建自定义功能时查看DRF source

相关问题由DRF的创建者回答:django-rest-framework 3.0 create or update in nested serializer

修改

所以我仍然处于DRF 2心态,自动处理嵌套的可写字段。您可以在此处阅读此主题:http://www.django-rest-framework.org/topics/3.0-announcement/

我已经测试了以下代码并且它可以运行:

class CategorySerializer(serializers.ModelSerializer):
    class Meta:
        ...
        extra_kwargs = {
            'name': {'validators': []},
            'description': {'required': False},
        }


class ListingSerializer(serializers.ModelSerializer):

    ...

    def update_or_create_category(self, validated_data):
        data = validated_data.pop('category', None)
        if not data:
            return None

        category, created = models.Category.objects.update_or_create(
            name=data.pop('name'), defaults=data)

        validated_data['category'] = category

    def create(self, validated_data):
        self.update_or_create_category(validated_data)
        return super(ListingSerializer, self).create(validated_data)

    def update(self, instance, validated_data):
        self.update_or_create_category(validated_data)
        return super(ListingSerializer, self).update(instance, validated_data)

修改

使用SlugRelatedField的正确方法是这样的,以防您想知道:

class ListingSerializer(serializers.ModelSerializer):

    ...

    # slug_field should be 'name', i.e. the name of the field on the related model
    category = serializers.SlugRelatedField(slug_field='name',
        queryset=models.Category.objects.all())

    ...

答案 1 :(得分:3)

这不太理想,但我确实找到了解决问题的解决方案(我等待接受它作为答案,希望别人能做得更好)。分为两部分:

首先,在初始化partial=Truehttp://www.django-rest-framework.org/api-guide/serializers/#partial-updates)时使用ListingSerializer参数。然后使用序列化程序的validate方法获取与输入数据对应的实际模型实例。

其次,明确删除nameCategorySerializer字段的验证程序。这尤其糟糕,因为它影响的不仅仅是ListingSerializer

退出任何一块将导致在实例化序列化器时抛出验证错误。

对views.py的修改:

class ListingViewSet(viewsets.ModelViewSet):
    queryset = models.Listing.objects.all()
    serializer_class = serializers.ListingSerializer

    def create(self, request):
        serializer = serializers.ListingSerializer(data=request.data,
            context={'request': request}, partial=True)
        if not serializer.is_valid():
            logger.error('%s' % serializer.errors)
            return Response(serializer.errors,
                  status=status.HTTP_400_BAD_REQUEST)

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

对serializers.py的修改:

class CategorySerializer(serializers.ModelSerializer):
    class Meta:
        model = models.Category
        fields = ('name', 'description')
        read_only = ('description',)
        # also need to explicitly remove validators for `name` field
        extra_kwargs = {
            'name': {
                'validators': []
            }
        }

class ListingSerializer(serializers.ModelSerializer):
    owners = ProfileSerializer(required=False, many=True)
    category = CategorySerializer(required=False)
    class Meta:
        model = models.Listing
        depth = 2

    def validate(self, data):
        # manually get the Category instance from the input data
        data['category'] = models.Category.objects.get(name=data['category']['name'])
        return data

    def create(self, validated_data):
        title = validated_data['title']

        listing = models.Listing(title=validated_data['title'],
            category=validated_data['category'])

        listing.save()

        if 'owners' in validated_data:
            logger.debug('owners: %s' % validated_data['owners'])
            for owner in validated_data['owners']:
                print ('adding owner: %s' % owner)
                listing.owners.add(owner)
        return listing

我会稍等一下接受这个作为答案以防有人能提出更好的解决方案(比如如何使source参数与SlugRelatedField正常工作) - 我有如果您想进行实验,请使用上面https://github.com/arw180/drf-example解决方案的工作示例。我也很想听到关于extra_kwargs中为什么CategorySerializer这些东西是必要的评论 - 为什么不像这样实例化category = CategorySerializer(required=False, validators=[])就足够了(ListingSerializer )?更新:我认为这不起作用,因为独特的验证器是从数据库约束中自动添加的,无论在此处设置任何显式验证器都会运行,如本答案中所述:http://iswwwup.com/t/3bf20dfabe1f/python-order-of-serializer-validation-in-django-rest-framework.html

答案 2 :(得分:1)

我有类似的问题:我需要检查嵌套序列化程序(CategorySerializer)是否存在,如果是,则使用它,如果不存在 - 从嵌套序列化程序(ListingSerializer)创建它。只有当我没有对嵌套序列化器中的字段使用自定义验证时,@ demux的解决方案才对我有用(如果此实例存在,我将通过嵌套序列化程序检查该字段)。因此,我将create()方法添加到嵌套序列化程序,并将update_or_create_category()create()update() ListingSerializer添加到class CategorySerializer(serializers.ModelSerializer): class Meta: model = Category ... def create(self, validated_data): if Category.objects.filter(name=self.validated_data['name']).exists(): raise serializers.ValidationError("This category name already exists") return Category.objects.create(**validated_data)

Mysecret.objects.filter(creator=request.user).delete()
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