Question

class Team(db.Model):
    __tablename__ = "Teams"
    id = db.Column(db.Integer, primary_key=True)
    flashscore_id = db.Column(db.String(255), nullable=False, unique=True)
    name = db.Column(db.String(255), nullable=False)

    leagues = db.relationship("League",
                    secondary=league_teams_table)

    standings = db.relationship('Standing', backref='teams', cascade="all,delete")
    fixtures = db.relationship('Fixture', backref='teams', cascade="all,delete")

related_fixtures_table = db.Table('RelatedFixtures',
    db.Column('fixture_id', db.Integer, db.ForeignKey('Fixtures.id')),
    db.Column('related_fixture_id', db.Integer, db.ForeignKey('Fixtures.id')))

class Fixture(db.Model):
    __tablename__ = "Fixtures"
    id = db.Column(db.Integer, primary_key=True)

    home_id = db.Column(db.Integer, db.ForeignKey('Teams.id'),
        nullable=False)
    away_id = db.Column(db.Integer, db.ForeignKey('Teams.id'),
        nullable=False)

    home_ref = db.relationship("Teams", backref="fixture", uselist=False, foreign_keys=[home_id])
    away_ref = db.relationship("Teams", backref="fixture", uselist=False, foreign_keys=[away_id])

    flashscore_id = db.Column(db.String(255), nullable=False, unique=True)
    total_goals = db.Column(db.Integer, nullable=False)

    total_fh_goals = db.Column(db.Integer, nullable=False, default=0)
    total_sh_goals = db.Column(db.Integer, nullable=False, default=0)

    total_home_goals = db.Column(db.Integer, nullable=False)
    total_away_goals = db.Column(db.Integer, nullable=False)

    total_home_fh_goals = db.Column(db.Integer, nullable=False, default=0)
    total_home_sh_goals = db.Column(db.Integer, nullable=False, default=0)

    total_away_fh_goals = db.Column(db.Integer, nullable=False, default=0)
    total_away_sh_goals = db.Column(db.Integer, nullable=False, default=0)

    related_fixtures = db.relationship("Fixture",
                    secondary=related_fixtures_table)

我编写了以上代码，试图定义Fixture和Team之间的多个关系。当我运行应用程序并执行操作时，出现以下错误：

sqlalchemy.exc.AmbiguousForeignKeysError: Could not determine join condition between parent/child tables on relationship Team.fixtures - there are multiple foreign key paths linking the tables.  Specify the 'foreign_keys' argument, providing a list of those columns which should be counted as containing a foreign key reference to the parent table.

我尝试通过添加以下两行来解决此问题：

home_ref = db.relationship("Teams", backref="fixture", uselist=False, foreign_keys=[home_id])
    away_ref = db.relationship("Teams", backref="fixture", uselist=False, foreign_keys=[away_id])

但是我仍然遇到相同的错误。有人可以帮我吗？

Answer 1

您需要在public static String get(String value) { String key = null; try { FileInputStream in = new FileInputStream(file); props.load(in); key = props.getProperty(value); in.close(); } catch (Exception fnf) { System.out.println(fnf); } return key; }类上配置fixtures关系，因为这就是错误告诉您要做的事情：Team。

但是，您无法使用Could not determine join condition ... on relationship Team.fixtures解决这种情况；期望有一个单个（可能是复合）外键，并且您有两个不同的外键关系。您必须创建两个单独的关系，一个用于本队是主队的固定装置，另一个用于foreign_keys关系。

如果您希望away返回一个通过一个或两个外键引用团队的所有Team.fixtures行的列表，则需要创建一个自定义Fixture条件，该条件要匹配您的primaryjoin与Team.id列中的home_id 或（您可能想向查询中添加away_id条件以避免重复的结果用于通过两个外键连接到您的团队的灯具。

无法确定父/子表之间的连接条件，设置了foreign_keys

1 个答案: