如何序列化 Django MPTT 系列并保持分层？答案

【问题标题】：How to serialize a Django MPTT family and keep it hierarchical?如何序列化 Django MPTT 系列并保持分层？
【发布时间】：2018-05-25 13:57:31
【问题描述】：

我试图为this (Google cache) 寻找解决方案

我只能想出这样的解决方案：

import json

from mptt.utils import tree_item_iterator
from rest_framework import generics
from rest_framework.response import Response

from .models import Category

def tree_family_items_to_json(instances):
    data = ''
    channel = '"{}"'.format(instances[0].channel.slug)
    for category, structure in tree_item_iterator(instances):
        if structure['new_level']:
        data += '{'
        else:
            data += '],'
            data += '"channel": {}'.format(channel)
            data += '},{'
        data += '"slug": "{}",'.format(category.slug)
        data += '"name": "{}",'.format(category.name)
        data += '"subcategories": ['
        for level in structure['closed_levels']:
            data += '],'
            data += '"channel": {}'.format(channel)
            data += '}'

    return json.loads(data)

class CategoryFamily(generics.RetrieveAPIView):
    lookup_field = 'slug'
    queryset = Category.objects.all()

    def retrieve(self, request, *args, **kwargs):
        instances = self.get_object().get_family()
        json_data = tree_family_items_to_json(instances)
        return Response(json_data)

关键是我使用了来自 mptt 的tree_item_iterator，现在我正在寻找更花哨的东西。

它满足了一段时间的需要。但现在确定要多久。

有什么想法吗？

【问题讨论】：

标签： python django django-rest-framework django-mptt mptt

【解决方案1】：

这是一种在rest api中具有树结构的方法：

# serializers.py
class CategoryTreeSerializer(ModelSerializer):
    children = SerializerMethodField(source='get_children')
    class Meta:
        fields = ('children',)  # add here rest of the fields from model 

    def get_children(self, obj):
        children = self.context['children'].get(obj.id, [])
        serializer = CategoryTreeSerializer(children, many=True, context=self.context)
        return serializer.data


 # views.py
 class CategoryViewSet(viewsets.ModelViewSet):

    queryset = Category.objects.all()
    serializer_class = CategoryTreeSerializer

    @detail_route()
    def tree(self, request, pk=None):
        """
        Detail route of an category that returns it's descendants in a tree structure.
        """
        category = self.get_object()
        descendants = category.get_descendants() # add here any select_related/prefetch_related fields to improve api performance

        children_dict = defaultdict(list)
        for descendant in descendants:
            children_dict[descendant.get_parent().pk].append(descendant)

        context = self.get_serializer_context()
        context['children'] = children_dict
        serializer = CategoryTreeSerializer(category, context=context)

        return Response(serializer.data)

在我的情况下，您会获得一个新端点（取决于您的 url），它将是这样的：category/<category_pk>/tree，您可以在其中获得指定类别的树结构。

这个想法是获取所有后代并为每个父代填充 children_dict，这将传递给序列化器的上下文以避免多次查询。

【讨论】：

看起来我可以像你说的那样做一些事情。我几乎到了那里。现在我正在加倍努力。
我希望我能给你更多的信任，我很惊讶这个页面没有被那么多人访问，因为这是一个常见问题的非常合法的答案！

【解决方案2】：

您可以使用djangorestframework-recursive 更方便。

views.py

from rest_framework import viewsets, generics
from yourapp.serializers import CategorySerializer
from yourapp.models import Category

class CategoryViewSet(viewsets.ModelViewSet):
    queryset = Category.objects.root_nodes()
    serializer_class = CategorySerializer

序列化器.py

from rest_framework import serializers
from yourapp.models import Category
from rest_framework_recursive.fields import RecursiveField

class CategorySerializer(serializers.ModelSerializer):
    children = RecursiveField(many=True)

    class Meta:
        model = Category
        fields = ['id', 'name', 'children']

【讨论】：