Sqlalchemy 从自参考多对多关系表中选择所需值时需要建议

我在flask SQLAlchemy中有以下自参考多对多模型，但无法返回SQL查询中提供的所有列值。 请有人建议：

如何在下面提供的代码中返回经理姓名

如果可以按特定的管理器（或）筛选输出行，则只能使用联接

以下是我的模型：

# Users will act as both users/employees and also manager
class Users(db.Model):
id = db.Column(db.Integer, primary_key=True)
username = db.Column(db.String(20), unique=True, nullable=False)
managers = db.relationship('Users',
            secondary = association_table, 
            primaryjoin=id == association_table.c.user_id, 
            secondaryjoin=id == association_table.c.manager_id, 
            backref=db.backref('employees', lazy='dynamic'))
user_subs = db.relationship('Submissions', backref='sub_created_user', lazy='dynamic')

def __repr__(self):
    return f"Name: {self.username}"

# association table to relate user-manager m:m relationship
association_table = db.Table('approvals', db.Model.metadata,
                                db.Column('user_id', db.Integer, db.ForeignKey('users.id')),
                                db.Column('manager_id', db.Integer, db.ForeignKey('users.id')))


# User submissions will be stored here
class Submissions(db.Model):
id = db.Column(db.Integer, primary_key=True)
submission_name = db.Column(db.String(10))
user_id = db.Column(db.Integer, db.ForeignKey('users.id'))

我想显示用户提交的所有内容以及他们的经理姓名。下面的代码为我提供了除经理姓名以外的所有信息-它作为

类返回

for s in Submissions.query.all():
    print(f"Submission: {s.submission_name} by User: {s.sub_created_user.username} whose Manager:{s.sub_created_user.managers}")

输出：

INFO sqlalchemy.engine.base.Engine ()
Submission: Sub1 by User: B whose Manager:[Name: A]
Submission: Sub2 by User: C whose Manager:[Name: B]
Submission: Sub3 by User: F whose Manager:[Name: E, Name: D]

所需的SQL：

感谢您的反馈
 即使在多次尝试之后，我也无法正确选择“managers”对象。所以我在数据库中创建了一个“用户”作为“管理者”的视图，并将其用作SQLAlchemy中的“管理者”模型，现在我能够准确地检索到我想要的信息。由于这不是确切的答案，我将其作为评论发布
select submissions.submission_name, users.username as user_name, managers.username as manager_name
from submissions join users
on submissions.user_id = users.id
join approvals
on users.id = approvals.user_id
join users as managers
on approvals.manager_id = managers.id
order by users.username, submission_name, managers.username