Sqlalchemy返回关系与使用marshmallow的嵌套联接

目前，我有一组表，它们之间建立了明确的关系。但是，我需要一个与答案表和问题表有关系的表（应用程序），以便只显示带有标记为活动的问题的答案。我试图通过应用程序模型来实现这一点，以便在棉花糖中它可以返回到应用程序内部。我还尝试在不生成N+1查询的情况下执行此操作

桌子

application_question_table = db.Table(
    'application_questions',
    db.Column(
        'application_id',
        db.Integer,
        db.ForeignKey(
            'application.id',
            name="application_questions_application_id_fkey",
            ondelete="CASCADE"
        )
    ),
    db.Column(
        'question_id',
        db.Integer,
    db.ForeignKey(
            'question.id', 
            name="application_questions_question_id_fkey",
            ondelete="CASCADE"
        )
    )
)

class Application(Base):
    id = db.Column(db.Integer, primary_key=True)
    answers = db.relationship('Answer', back_populates="application", cascade=‘delete)
    active_answers = db.relationship("should only show answerswhere question is active")
    questions = db.relationship('Question', lazy="dynamic", secondary=application_question_table)


class Answer(Base):
    id = db.Column(db.Integer, primary_key=True)
    question_id = db.Column(db.Integer, db.ForeignKey('question.id', ondelete="SET NULL", name="answer_question_id_fkey"))
    question = db.relationship('Question', back_populates=“answers")


class Question(Base):
    id = db.Column(db.Integer, primary_key=True)
    answers = db.relationship("Answer", back_populates="question")
    applications = db.relationship('Application',lazy="dynamic", secondary=application_question_table)
    active = db.Column(db.Boolean, default=True)

模式-请注意，GetVisible问题本质上就是我希望发生的事情。但我不喜欢这样做，因为这将是我找到的每个应用程序的附加查询

class ApplicationSchema(ModelSchema):
    visible_answers = fields.Method('get_visible_answers')
    class Meta:
        model = Application

    def get_visible_answers(self, application):
        schema = ApplicationAnswerSchema(many=True)
        answers = Answer.query.join(
            Answer.application,
            Answer.question
        ).filter(
            Application.id == application.id,
            Question.active == True,
        ).all()
        return schema.dump(answers).data


class ApplicationAnswerSchema(ModelSchema):

    class Meta:
        model = Answer

我的预期行为是当我做类似的事情时

applications = Application.query.options(joinedload(Application.active_answers)).all()
schema = ApplicationSchema(many=True)
results = schema.dump(applications)

只有一个查询，我可以输入每个应用程序并获得可见的答案

我得到的实际结果是为每个应用程序发出一个额外的查询