sqlalchemy 中的多对多自引用关系

发布于 2024-10-01 20:41:26 字数 905 浏览 0 评论 0原文

我试图在 sqlalchemy 中建立自引用多对多关系(这意味着 Line 可以有许多父行和许多子行),如下所示:

Base = declarative_base()

class Association(Base):
 __tablename__ = 'association'

 prev_id = Column(Integer, ForeignKey('line.id'), primary_key=True)                            
 next_id = Column(Integer, ForeignKey('line.id'), primary_key=True)


class Line(Base):
 __tablename__ = 'line'

 id = Column(Integer, primary_key = True)
 text = Column(Text)
 condition = Column(Text)
 action = Column(Text)

 next_lines = relationship(Association, backref="prev_lines")



class Root(Base):
 __tablename__ = 'root'

 name = Column(String, primary_key = True)
 start_line_id = Column(Integer, ForeignKey('line.id'))

 start_line = relationship('Line')

但出现以下错误: sqlalchemy.exc.ArgumentError:无法确定父/之间的连接条件 关系 Line.next_lines 上的子表。指定“primaryjoin”表达式 名词如果存在“Secondary”,则还需要“SecondaryJoin”。

你知道我该如何解决这个问题吗?

I'm trying to make a self-referential many-to-many relationship (it means that Line can have many parent lines and many child lines) in sqlalchemy like this:

Base = declarative_base()

class Association(Base):
 __tablename__ = 'association'

 prev_id = Column(Integer, ForeignKey('line.id'), primary_key=True)                            
 next_id = Column(Integer, ForeignKey('line.id'), primary_key=True)


class Line(Base):
 __tablename__ = 'line'

 id = Column(Integer, primary_key = True)
 text = Column(Text)
 condition = Column(Text)
 action = Column(Text)

 next_lines = relationship(Association, backref="prev_lines")



class Root(Base):
 __tablename__ = 'root'

 name = Column(String, primary_key = True)
 start_line_id = Column(Integer, ForeignKey('line.id'))

 start_line = relationship('Line')

But I get the following error:
sqlalchemy.exc.ArgumentError: Could not determine join condition between parent/
child tables on relationship Line.next_lines. Specify a 'primaryjoin' expressio
n. If 'secondary' is present, 'secondaryjoin' is needed as well.

Do you know how I could remedy this?

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(1

生寂 2024-10-08 20:41:26

您应该只需要:

prev_lines = relationship(
    Association,
    backref="next_lines",
    primaryjoin=id==Association.prev_id)

由于这指定了 next_lines 反向引用,因此不需要有 next_lines 关系。

您还可以使用关系的 remote_side 参数来执行此操作:http://www.sqlalchemy.org/trac/browser/examples/adjacency_list/adjacency_list.py

You should just need:

prev_lines = relationship(
    Association,
    backref="next_lines",
    primaryjoin=id==Association.prev_id)

Since this specifies the next_lines back reference there is no need to have a next_lines relationship.

You can also do this using the remote_side parameter to a relationship: http://www.sqlalchemy.org/trac/browser/examples/adjacency_list/adjacency_list.py

~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文