Question

使用以下设计示例：

from django.db import models
from django_filters import FilterSet, OrderingFilter
from graphene import ObjectType, Schema, relay
from graphene_django import DjangoObjectType
from graphene_django.filter import DjangoFilterConnectionField

class Recipe(models.Model):
    name = models.CharField(max_length=50)
    ingredients = models.ManyToManyField('Ingredient', related_name='recipes')

class Ingredient(models.Model):
    name = models.CharField(max_length=50)

class RecipeFilter(FilterSet):

    order_by = OrderingFilter(fields=[('name', 'name')])

    class Meta:
        fields = {'name': ['icontains']}
        model = Recipe

class IngredientFilter(FilterSet):

    order_by = OrderingFilter(fields=[('name', 'name')])

    class Meta:
        fields = {'name': ['icontains']}
        model = Ingredient

class RecipeNode(DjangoObjectType):

    ingredients = DjangoFilterConnectionField(IngredientNode, filterset_class=IngredientFilter)

    class Meta:
        interfaces = [relay.Node]
        model = Recipe
        only_fields = ['name']

class IngredientNode(DjangoObjectType):

    recipes = DjangoFilterConnectionField(RecipeNode, filterset_class=RecipeFilter)

    class Meta:
        interfaces = [relay.Node]
        model = Ingredient
        only_fields = ['name']

class Queries(ObjectType):

    all_recipes = DjangoFilterConnectionField(RecipeNode, filterset_class=RecipeFilter)
    all_ingredients = DjangoFilterConnectionField(IngredientNode, filterset_class=IngredientFilter)

schema = Schema(query=Queries)

如何定义RecipeNode和IngredientNode之间的循环关系，以便我可以运行以下GraphQL查询：

{
  allRecipes(name_Icontains: "gg") {
    edges {
      node {
        name
        ingredients(name_Icontains: "gg") {
          edges {
            node {
              name
            }
          }
        }
      }
    }
  }
  allIngredients(name_Icontains: "gg") {
    edges {
      node {
        name
        recipes(name_Icontains: "gg") {
          edges {
            node {
              name
            }
          }
        }
      }
    }
  }
}

就目前而言，我无法从IngredientNode引用RecipeNode，因为它尚未定义。如果我尝试使用我在其他地方推荐过的lambda，我会收到AttributeError: 'function' object has no attribute '_meta'。

class IngredientNode(DjangoObjectType):

    recipes = DjangoFilterConnectionField(lambda: RecipeNode, filterset_class=RecipeFilter)

    class Meta:
        interfaces = [relay.Node]
        model = Ingredient
        only_fields = ['name']

如果我尝试在事后设置属性，我将无法从配方中查询ingredients。没有错误，Graphiql只是表现为从未定义ingredients。

class RecipeNode(DjangoObjectType):

    class Meta:
        interfaces = [relay.Node]
        model = Recipe
        only_fields = ['name']

class IngredientNode(DjangoObjectType):

    recipes = DjangoFilterConnectionField(RecipeNode, filterset_class=RecipeFilter)

    class Meta:
        interfaces = [relay.Node]
        model = Ingredient
        only_fields = ['name']

RecipeNode.ingredients = DjangoFilterConnectionField(IngredientNode, filterset_class=IngredientFilter)

我必须认为这是一个简单的解决方案，我只是没有看到。任何帮助，将不胜感激。谢谢！

Django 1.8.17，django-filter 0.15.3，graphene-django 1.2.0

Answer 1

对于后人，我们解决这个问题的方法是重新定义DjangoFilterConnectionField，以便需要filterset_class参数，并删除引用节点的Meta属性的代码。缺点是我们无法再利用filter_fields快捷方式。对我们来说，这不是问题，因为我们从一开始就使用过滤器。

整个最终解决方案/解决方法：

from django.db import models
from django_filters import FilterSet, OrderingFilter
from functools import partial
from graphene import ObjectType, Schema, relay
from graphene_django import DjangoObjectType, DjangoConnectionField
from graphene_django.filter.utils import get_filtering_args_from_filterset, get_filterset_class

class DjangoFilterConnectionField(DjangoConnectionField):

    def __init__(self, type, filterset_class, *args, **kwargs):

        self.filterset_class = get_filterset_class(filterset_class)
        self.filtering_args = get_filtering_args_from_filterset(self.filterset_class, type)
        kwargs.setdefault('args', {})
        kwargs['args'].update(self.filtering_args)
        super(DjangoFilterConnectionField, self).__init__(type, *args, **kwargs)

    @staticmethod
    def connection_resolver(resolver, connection, default_manager, filterset_class, filtering_args,
                            root, args, context, info):
        filter_kwargs = {k: v for k, v in args.items() if k in filtering_args}
        qs = default_manager.get_queryset()
        qs = filterset_class(data=filter_kwargs, queryset=qs).qs
        return DjangoConnectionField.connection_resolver(resolver, connection, qs, root, args, context, info)

    def get_resolver(self, parent_resolver):
        return partial(self.connection_resolver, parent_resolver, self.type, self.get_manager(),
                       self.filterset_class, self.filtering_args)

class Recipe(models.Model):

    name = models.CharField(max_length=50)
    ingredients = models.ManyToManyField('Ingredient', related_name='recipes')

class Ingredient(models.Model):

    name = models.CharField(max_length=50)

class RecipeFilter(FilterSet):

    order_by = OrderingFilter(fields=[('name', 'name')])

    class Meta:
        fields = {'name': ['icontains']}
        model = Recipe

class IngredientFilter(FilterSet):

    order_by = OrderingFilter(fields=[('name', 'name')])

    class Meta:
        fields = {'name': ['icontains']}
        model = Ingredient

class RecipeNode(DjangoObjectType):

    ingredients = DjangoFilterConnectionField(lambda: IngredientNode, filterset_class=IngredientFilter)

    class Meta:
        interfaces = [relay.Node]
        model = Recipe
        only_fields = ['name']

class IngredientNode(DjangoObjectType):

    recipes = DjangoFilterConnectionField(RecipeNode, filterset_class=RecipeFilter)

    class Meta:
        interfaces = [relay.Node]
        model = Ingredient
        only_fields = ['name']

class Queries(ObjectType):

    all_recipes = DjangoFilterConnectionField(RecipeNode, filterset_class=RecipeFilter)
    all_ingredients = DjangoFilterConnectionField(IngredientNode, filterset_class=IngredientFilter)

schema = Schema(query=Queries)

以这种方式重新定义DjangoFilterConnectionField允许我们使用lambda来引用尚未定义的节点。

使用石墨烯-django，是否可以定义两个节点之间的循环关系？

1 个答案: