关于graphene-django突变文档的问题

Question

虽然我已经能够使我的应用程序正常工作，但我还是担心以正确的方式做事。因此，有些东西我“不明白”：

在文档here 中，QuestionMutation 类中有一个question 属性。这实际上是什么意思？它说它定义了响应，我不明白这是什么意思。

文档中的代码：

class QuestionType(DjangoObjectType):
    class Meta:
        model = Question


class QuestionMutation(graphene.Mutation):
    class Arguments:
        # The input arguments for this mutation
        text = graphene.String(required=True)
        id = graphene.ID()

    # The class attributes define the response of the mutation
    question = graphene.Field(QuestionType)

    @classmethod
    def mutate(cls, root, info, text, id):
        question = Question.objects.get(pk=id)
        question.text = text
        question.save()
        # Notice we return an instance of this mutation
        return QuestionMutation(question=question)

在我的代码中，我已经能够做到这一点：

我的类型：

class UserType(DjangoObjectType):

    class Meta:
        model = User
        fields = (
            'id',
            'username',
            'password',
            'email',
            'first_name',
            'last_name',
            'is_active',
            'group_ids',
        )

    full_name = graphene.String()           # Python property
    full_identification = graphene.String() # Python property

我的突变：

class UpdateUser(graphene.Mutation):
    # --------> I could comment these and no problem. Why ? <--------
    # id = graphene.ID()
    # username = graphene.String()
    # email = graphene.String()
    # first_name = graphene.String()
    # last_name = graphene.String()
    # is_active = graphene.Boolean()

    class Arguments:
        id = graphene.ID()
        username = graphene.String()
        email = graphene.String()
        first_name = graphene.String()
        last_name = graphene.String()
        is_active = graphene.Boolean()

    class Meta:
        output = UserType

    @login_required
    def mutate(self, info, **kwargs):
        user = get_object_or_404(User, id=kwargs['id'])
        for attr, value in kwargs.items():
            setattr(user, attr, value)
        user.save()
        return user
        # I'm not returning explicit UserType, is it a problem ?
        # I actually do it with the Meta class. I guess it is the same for this ?

它可以工作，而我没有为响应属性指定任何内容。我什至不返回同样的东西。 如果我做错了，有人可以解释一下吗？

你可以在这里看到，如果我称之为突变：

mutation {
  updateUser (
    id: 42
    username: "updated_user"
    firstName: "updated"
    lastName: "user"
    email: "updated.user@test.com"
  ) {
    id
    username
    firstName
    lastName
    email
  }
}

即使没有返回值，我也能得到这个答案：

{
  "data": {
    "updateUser": {
      "id": "42",
      "username": "updated_user",
      "firstName": "updated",
      "lastName": "user",
      "email": "updated.user@test.com"
    }
  }
}

Answer 1

我还问了HERE这个问题，看来有更多关于这些配置的信息THERE。

最后，由于我设置了ouput = UserType 和model = User，我的UserType 和User 类是duck typed，即使没有定义返回字段（问题中评论的那些），它也像一个魅力一样工作)。

换句话说，这个代码示例似乎是执行此操作的正确方法，官方文档并没有这样做，但它也打算像这样工作（您可以在@987654324 中阅读@)。

class UserType(DjangoObjectType):

    class Meta:
        model = User
        fields = (
            'id',
            'username',
            'password',
            'email',
            'first_name',
            'last_name',
            'is_active',
            'group_ids',
        )

    full_name = graphene.String()           # Python property
    full_identification = graphene.String() # Python property

class UpdateUser(graphene.Mutation):
        # We can omit return fields since UserType is defined as output ('output = UserType')
        # and UserType has already defined fields in its class.

    class Arguments:
        id = graphene.ID()
        username = graphene.String()
        email = graphene.String()
        first_name = graphene.String()
        last_name = graphene.String()
        is_active = graphene.Boolean()

    class Meta:
        output = UserType

    @login_required
    def mutate(self, info, **kwargs):
        user = get_object_or_404(User, id=kwargs['id'])
        for attr, value in kwargs.items():
            setattr(user, attr, value)
        user.save()
        return user
        # Returning a User can be done since it is linked to the UserType and we
        # said we return a UserType with 'output = UserType'