尝试在 sqlalchemy orm 查询中加入两个表，出现错误

Question

尝试使用此命令连接下表：

Subscription.query.filter(    return Subscription.query.filter(Subscription.watch_id == id).join(User).filter_by(watch_id=id)

我得到这个错误：

sqlalchemy.exc.InvalidRequestError: Could not find a FROM clause to join from.  Tried joining to <class 'app.user.model.User'>, but got: Can't find any foreign key relationships between 'wm_subscription' and 'user'.

基本上我的最终目标是获取一个查询，该查询获取共享 watch_id 的用户列表。不确定模型或查询是否正确。有谁知道怎么回事？

Database = declarative_base(cls=DbBase)
class Subscription(Database):
    __tablename__ = 'wm_subscription'
    subscription_id = UniqueIdPk()
    watch_id = UniqueIdRefNotNull(index=True)
    user_id = UniqueIdRefNotNull(ForeignKey('User.user_id'), index=True)

    subscription_watch = relationship('Watch',
                                      primaryjoin='Subscription.watch_id == Watch.watch_id',
                                      foreign_keys='Watch.watch_id',
                                      uselist=True)

    subscription_user = relationship('User',
                                     primaryjoin='Subscription.watch_id == User.user_id',
                                     foreign_keys='User.user_id',
                                     uselist=True,
                                     backref='user')

class User(Database, UserMixin):
    __tablename__ = 'user'

    user_id = UniqueIdPk()

    # Google sub ID - unique to user https://developers.google.com/identity/protocols/OpenIDConnect
    google_id = Column(String(length=50))

    # override email mixin for unique index
    email = Email(unique=True)

    first_name = Name()
    last_name = Name()

    def get_id(self):
        return self.user_id

Answer 1

这是正确的查询：

Subscription.query.filter(Subscription.watch_id == id).join(User)