SQLAlchemy:在查询中覆盖关系定义的"order_by"

Question

所以,我有一个类似的模型:

class Foo(model):
    __tablename__ = "foo"
    id = Column(Integer, primary_key=True)
    data = relationship(
        "FooData",
        cascade="all, delete-orphan",
        backref="foo",
        lazy="dynamic",
        order_by="desc(FooData.timestamp)"
    )

    @property
    def first_item(self):
        # the problem is here:
        return self.data.order_by(asc("timestamp")).first()

    @property
    def latest_item(self):
        return self.data.first()


class FooData(Model):
    __tablename__ = "foo_data"
    foo_id = Column(Integer, ForeignKey("foo.id"), primary_key=True)
    timestamp = Column(DateTime, primary_key=True)
    actual_data = Column(Float, nullable=False)

所以,问题在于first_item那里的方法:当它如上定义时,SQL看起来像这样:

SELECT foo_data.timestamp AS foo_data_timestamp, foo_data.actual_data AS foo_data_actual_data, foo_data.foo_id AS foo_data_foo_id 
FROM foo_data 
WHERE :param_1 = foo_data.foo_id ORDER BY foo_data.timestamp DESC, foo_data.timestamp ASC
--                                                                 ^^^^^^^^^^^^^^^^^^^^^^

显然,order_by查询中指定的内容将附加到关系定义中指定的那个,而不是替换它; 有没有办法让查询覆盖原始的order_by？我知道我可以直接在FooData类上指定一个单独的查询,但我想尽可能避免这种情况.

Answer 1

根据文件:

可以通过传递来抑制所有现有的ORDER BY设置None- 这将抑制在映射器上配置的任何ORDER BY.

所以简单的解决方案是重置ORDER BY子句然后应用你需要的子句.喜欢:

self.data.order_by(None).order_by(asc("timestamp")).first()

如果您不想重置整个ORDER BY子句,但只想覆盖一个列顺序,AFAIK没有内置的方式.