[英]Multiple foreign keys in one to many relationship - SqlAlchemy
我是 SQL Alchemy 的新手,我需要帮助来实现以下关系:
我有两个表Trends和ClosestTrends ,我想声明两个一对多关系:
在 SQL 中,它将是:
ALTER TABLE "closest_trends" ADD FOREIGN KEY ("id_trend_ref") REFERENCES "trends" ("id") ON DELETE CASCADE;
ALTER TABLE "closest_trends" ADD FOREIGN KEY ("id_trend_close") REFERENCES "trends" ("id") ON DELETE CASCADE;
我尝试了以下实现:
class Trends(Base):
__tablename__ = "trends"
__table_args__ = (
UniqueConstraint(
"name", "id_region", "language_iso", name="name_id_region_language"
),
)
id = Column(Integer, primary_key=True, index=True, unique=True)
.
.
.
closest_trends = relationship("ClosestTrends", backref="Trends")
def __str__(self):
return "Trends"
class ClosestTrends(Base):
__tablename__ = "closest_trends"
__table_args__ = (
UniqueConstraint(
"id_trend_ref", "id_trend_close", name="id_trend_ref_id_trend_close"
),
)
id = Column(Integer, primary_key=True, index=True, unique=True)
.
.
.
id_trend_ref = Column(
Integer, ForeignKey("trends.id", ondelete="CASCADE"), nullable=False
)
id_trend_close = Column(
Integer, ForeignKey("trends.id", ondelete="CASCADE"), nullable=False
)
def __str__(self):
return "ClosestTrends"
我不工作,我收到以下错误:
sqlalchemy.exc.AmbiguousForeignKeysError:无法确定关系 Trends.closest_trends 上父/子表之间的连接条件 - 有多个外键路径链接表。 指定“foreign_keys”参数,提供应被视为包含对父表的外键引用的列的列表。
有谁知道如何解决这个问题?
非常感谢
您的解决方案写在官方网站上。
https://docs.sqlalchemy.org/en/13/orm/join_conditions.html#handling-multiple-join-paths
根据官方网站,我是这样实现的。
class Trends(Base):
__tablename__ = "trends"
__table_args__ = (
UniqueConstraint(
"name", "id_region", "language_iso", name="name_id_region_language"
),
)
id = Column(Integer, primary_key=True, index=True, unique=True)
.
.
.
# closest_trends = relationship("ClosestTrends", backref="Trends")
def __str__(self):
return "Trends"
class ClosestTrends(Base):
__tablename__ = "closest_trends"
__table_args__ = (
UniqueConstraint(
"id_trend_ref", "id_trend_close", name="id_trend_ref_id_trend_close"
),
)
id = Column(Integer, primary_key=True, index=True, unique=True)
.
.
.
id_trend_ref = Column(
Integer, ForeignKey("trends.id", ondelete="CASCADE"), nullable=False
)
id_trend_close = Column(
Integer, ForeignKey("trends.id", ondelete="CASCADE"), nullable=False
)
trend_ref = relationship("Trends", foreign_keys=[id_trend_ref])
trend_close = relationship("Trends", foreign_keys=[id_trend_close])
def __str__(self):
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.