如果同一个外键有 2 个字段，如何声明一对多
我是 python(sqlalchemy) 的新手，我正在学习使用 pylons 和 sqlalchemy 构建网站。
当我声明模型之间的关系时，我遇到了问题。我已经尝试了几个小时，但失败了。但我认为这应该是一个基本问题。
我有两个类：User和Article，用户可以创建文章，并修改其他人的文章（如wiki）。 因此，用户创建了文章并编辑了文章。
class Article(Base):
__tablename__ = 'articles'
id = Column(Integer, primary_key=True)
title = ...
user_id = Column(Integer, ForeignKey('users.id'))
editor_id = Column(Integer, ForeignKey('users.id'))
# relations
user = relationship('User', backref='articles') # -> has error
class User(Base):
__tablename__ = "users"
id = Column(Integer, primary_key=True)
name = Column(String(20))
def __init__(self):
pass
但显示错误：
InvalidRequestError: One or more mappers failed to compile. Exception was probably suppressed within a hasattr() call. Message was: Could not determine join condition between parent/child tables on relationship Article.user. Specify a 'primaryjoin' expression. If this is a many-to-many relationship, 'secondaryjoin' is needed as well.
我尝试将 primaryjoin 添加到该行（'有错误'），但不知道它应该是什么。我尝试了一些代码，但没有一个有效。
先感谢您！
I'm new to python(sqlalchemy), and I'm learning to build web site with pylons and sqlalchemy.
I have a problem when I declare the relationship between models. I've tried it several hours, but failed. But I think it should be a basic question.
I have two classes: User and Article, user can create articles, and modified the other people's article(like wiki).
So a user has created-articles and edited-articles.
class Article(Base):
__tablename__ = 'articles'
id = Column(Integer, primary_key=True)
title = ...
user_id = Column(Integer, ForeignKey('users.id'))
editor_id = Column(Integer, ForeignKey('users.id'))
# relations
user = relationship('User', backref='articles') # -> has error
class User(Base):
__tablename__ = "users"
id = Column(Integer, primary_key=True)
name = Column(String(20))
def __init__(self):
pass
But there is an error displayed:
InvalidRequestError: One or more mappers failed to compile. Exception was probably suppressed within a hasattr() call. Message was: Could not determine join condition between parent/child tables on relationship Article.user. Specify a 'primaryjoin' expression. If this is a many-to-many relationship, 'secondaryjoin' is needed as well.
I tried to add primaryjoin to the line('has error'), but don't know what it should be. I tried some codes, but none works.
Thank you in advance!
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啊，那是显而易见的。
Article 类有两个对 User 的引用，user_id 和 editor_id，因此 SQLA 不知道对您的关系使用其中哪一个。只需使用显式的primaryjoin：
Ah, thats obvious one.
Article class has two references to User, user_id and editor_id, so SQLA does not know which one of them to use for your relation. Just use explicit primaryjoin: