ORM查詢多對多對一，Flasksql煉金術

我有三個flask-sqlalchemy-models。Books-管理員的唯一條目：

class Book(db.Model):
    id = db.Column(db.Integer, primary_key=True, autoincrement=True)
    book_name = db.Column(db.String(20), unique=True, nullable=False)
    def __repr__(self):
        return self.book_name

Bookscomp-用戶條目，與以上相關：

class Bookscomp(db.Model):
    id = db.Column(db.Integer, primary_key=True, autoincrement=True)
    book = db.Column(db.Integer, db.ForeignKey('book.id'))

Company-用戶，與以上相關：

class Company(db.Model):
    id = db.Column(db.Integer, primary_key=True, autoincrement=True)
    name = db.Column(db.String(100), nullable=False)
    books = db.relationship('Bookscomp', secondary=companybook, lazy='dynamic',
        backref=db.backref('company'))

companybook = db.Table('companybook',
    db.Column('companyid', db.Integer, db.ForeignKey('company.id'), primary_key=True),
    db.Column('bookid', db.Integer, db.ForeignKey('bookscomp.id'), primary_key=True),
                                )

問題：我試圖通過Company+Bookscomp從Book模型中獲取book_name。所以一個公司有很多書，每本書都參考了一般的圖書信息。像這樣嘗試：

company = OrganizatCompanyion.query.filter_by(id=comp.id).first()
    books = company.books.all()
for item in books:
        print(item.book.book_name)
        #AttributeError: 'int' object has no attribute 
        print(item.book)
        #Gives book id from book model, but I need name

為什么我不能直接在上面的代碼片段中獲得book_name？如何才能做到這一點呢？