Python Flask SQLAlchemy-模型没有属性'；外键'；

我用Flask SQLalchemy创建了3个模型：用户、角色、用户角色

role.py：

class Role( ActiveRecord, db.Model ):

    __tablename__ = "roles"

    #   Schema
    id = db.Column( db.Integer, primary_key = True )
    name = db.Column( db.String( 24 ), unique = True )
    description = db.Column( db.String( 90 ) )

    users = db.relationship( "User", secondary = "UserRole", \
        backref = db.backref( "roles" ) )

user.py：

class User( db.Model, ActiveRecord ):

    __tablename__ = "users"

    #   Schema
    id = db.Column( db.Integer, primary_key = True )
    email = db.Column( db.String( 90 ), unique = True )
    password = db.Column( db.String( 64 ) )

    #   Relations
    roles = db.relationship( "Role", secondary = "UserRole", \
        backref = db.backref( "users" ) )

用户_role.py：

class UserRole( ActiveRecord, db.Model ):

    __tablename__ = "user_roles"

    #   Schema
    user_id = db.Column( db.Integer, db.ForeignKey( 'users.id' ), primary_key = True )
    role_id = db.Column( db.Integer, db.ForeignKey( 'roles.id' ), primary_key = True )

如果我尝试（在控制台中）通过

User.query.all（）

获取所有用户，我会获取

AttributeError:'NoneType'对象没有属性'all'

，如果我再次尝试，我会收到另一个错误消息：

sqlalchemy.exc.InvalidRequestError: One or more mappers failed to initialize - can't proceed with initialization of other mappers.  Original exception was: type object 'UserRole' has no attribute 'foreign_keys'

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谁能解释一下我到底做错了什么？我想几个月前我的代码运行正常，但最近我更新了SQLAlchemy、Flask和Flask SQLAlchemy，它停止了。这只是一个附带项目。

这里有点难，因为您使用的是一些未知的基类，如“ActiveRecord”等。然而，看起来“次要”论点是错误的：

class User( db.Model, ActiveRecord ):

    __tablename__ = "users"

    #   Schema
    id = db.Column( db.Integer, primary_key = True )
    email = db.Column( db.String( 90 ), unique = True )
    password = db.Column( db.String( 64 ) )

    #   Relations
    roles = db.relationship( "Role", secondary = "UserRole", \
        backref = db.backref( "users" ) )

次要

需要引用一个

表

对象，而不是一个映射类，如果按字符串名称，那么它将是

用户角色“

：

---

发布源代码的部分原因是为了让我们能够尝试重现我们这边的问题。但是，如果我们不知道ActiveRecord是什么，或者您是如何设置试图在其中运行命令的“控制台”，我们就无法做到这一点。有没有一种方法可以创建一个复制问题的脚本？更好的方法是避免硬代码，所以我更喜欢使用：

secondary=lambda:ModelClass.\uuuu table\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
    roles = db.relationship( "Role", secondary = "user_roles", \
        backref = db.backref( "users" ) )