Question

对于我的应用程序，我需要创建一个带有相同实体的点的链接。目前我是这样做的。

association_table = db.Table('main_connectedpoints', Base.metadata,
                             db.Column('point_id', db.Integer, db.ForeignKey('main_placepoint.id')),
                             db.Column('connected_point_id', db.Integer, db.ForeignKey('main_placepoint.id')))


class PlacePoint(db.Model):
    __tablename__ = 'main_placepoint'

    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String())
    lat = db.Column(db.Float())
    lon = db.Column(db.Float())
    address = db.Column(db.String())
    organizations = db.relationship('Organization', backref='main_placepoint',
                                    lazy='dynamic')
    related_points = db.relationship("PlacePoint",
                                     secondary=lambda: association_table)

    def __init__(self, dict_):
        self.name = dict_['name']
        self.lat = dict_['lat']
        self.lon = dict_['lon']
        self.address = dict_['address']

    def __repr__(self):
        return "Place %r" % self.address

    def as_dict(self):
        return {"name": self.name, "lat": self.lat, "lon": self.lon, "address": self.address}

但是我有以下错误：NoReferencedTableError：与列关联的外键＆＃39; main_connectedpoints.point_id＆＃39;找不到表＆＃39; main_placepoint＆＃39;用于生成目标列＆＃39; id＆＃39;

的外键

我认为这是因为PlacePoint类没有实例化。有什么想法来解决它吗？

SQL Alchemy与当前表有很多很多关系

0 个答案: