订购SQLAlchemy的关系，而不是通过主键

Question

我有一个金字塔应用程序中使用的SQLAlchemy以下模型：

class Issue(base_iop): 
    __tablename__ = 'issues' 
    issueid = Column(BIGINT, primary_key=True, name="ISSUEID") 
    issuenum = Column(VARCHAR(20), name="ISSUENUM") 
    status = Column(VARCHAR(32), name="STATUS") 
    datetime = Column(TIMESTAMP, name="ISSUETIME") 

class Related(base_iop): 
    __tablename__ = 'related' 
    relationid = Column(BIGINT, primary_key=True, name="ROWSTAMP") 
    parent_num = Column(VARCHAR(20), ForeignKey('issue.ISSUENUM'),name="RECORDKEY") 
    children_num = Column(VARCHAR(20), ForeignKey('issue.ISSUENUM'), name="RELATEDKEY") 
    issues = relationship(iop, foreign_keys=[child_num]) 

我能得到就好使用问题的相关问题：的issues属性相关表格：

for related in db.query(Issue).all()[0].issues: 
    print related.status 

但是，我没有找到解决方案来订购datetime属性的问题。如果Related已经datetime属性就是这样，它应该使用ORDER_BY的关系是非常简单的：

class Issue(base_iop): 
    __tablename__ = 'issues' 
    issueid = Column(BIGINT, primary_key=True, name="ISSUEID") 
    issuenum = Column(VARCHAR(20), name="ISSUENUM") 
    status = Column(VARCHAR(32), name="STATUS") 
    datetime = Column(TIMESTAMP, name="ISSUETIME") 
    children = relationship("Related", foreign_keys="[Related.parent_num]", backref="parent", order_by="[Related.datetime]") 

class Related(base_iop): 
    __tablename__ = 'related' 
    relationid = Column(BIGINT, primary_key=True, name="ROWSTAMP") 
    parent_num = Column(VARCHAR(20), ForeignKey('issue.ISSUENUM'),name="RECORDKEY") 
    children_num = Column(VARCHAR(20), ForeignKey('issue.ISSUENUM'), name="RELATEDKEY") 
    datetime = Column(TIMESTAMP, name="ISSUETIME") 
    issues = relationship(iop, foreign_keys=[child_num]) 

如何通过另一场责令相关问题，而不是主键就像是现在？

Answer 1

关系接受关键字参数order_by来处理关系的默认排序。 应该不会停，如：

issues = relationship(iop, foreign_keys=[child_num], order_by="Issue.datetime") 

注意backref函数具有相同的关键字，如果你需要处理的。

在这里，你可以阅读更多：http://docs.sqlalchemy.org/en/rel_1_0/orm/tutorial.html#building-a-relationship关于这个主题