Question

我有两种模式：

class Report(Base):
    __tablename__ = 'report'
    id = Column(Integer, primary_key=True)

class ReportPhoto(Base):
    __tablename__ = 'report_photo'
    id = Column(Integer, primary_key=True)
    report_id = Column(Integer, ForeignKey(Report.id), nullable=False)

    report = relationship(Report, uselist=False, backref=backref('report_photo', uselist=True))

我想在Report模型中添加一列，表明ReportPhoto中是否有任何记录。我尝试以这种方式使用column_property：

class Report(Base):
    __tablename__ = 'report'
    id = Column(Integer, primary_key=True)

    has_photo = column_property(
        select(ReportPhoto.any())
    )

但收到错误NameError: name 'ReportPhoto' is not defined。我如何解决这个问题？

Answer 1

这样的事情应该有效：

    class ReportPhoto(Base):
        __tablename__ = 'report_photo'
        id = Column(Integer, primary_key=True)
        report_id = Column(Integer, ForeignKey('report.id'), nullable=False)

    class Report(Base):
        __tablename__ = 'report'
        id = Column(Integer, primary_key=True)
        report_photos = relationship(ReportPhoto, backref='report')
        has_photo = column_property(
            exists().where(ReportPhoto.report_id==id)
        )

Answer 2

我将添加@Vladimir lliev的回复，并对可能看不到如何执行此操作的其他人做出一些澄清。

在引用的表之后放置将具有“外部表引用”column_property 的表。在这种情况下，它意味着在 ReportPhoto之后放置Report 。这将解决您的NameError，但是，您将在ReportPhoto外键引用上留下新错误。要解决此问题，请将外键表引用放在引号中。您可以通过引用声明性文档（例如，declarative.py）并查看“配置关系”来阅读更多内容 - 具体来说，请阅读引用外部引用的部分。

使用您的代码，这看起来像：

class ReportPhoto(Base): # This now goes first __tablename__ = 'report_photo' id = Column(Integer, primary_key=True) # Notice the quotations around Report references here report_id = Column(Integer, ForeignKey("Report.id"), nullable=False) # Notice the quotations around Report references here report = relationship("Report", uselist=False, backref=backref("report_photo", uselist=True)) class Report(Base): # This is now _after_ ReportPhoto __tablename__ = 'report' id = Column(Integer, primary_key=True) # ReportPhoto now exists and we will not trip a NameError exception has_photo = column_property( select(ReportPhoto.any()) )

SQLAlchemy column_property基础知识

2 个答案: