python - Filter on hybrid property evaluates to "where false"

Question

I set up a mapping...

from sqlalchemy import Column, Integer, String
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import Session
from sqlalchemy.ext.hybrid import hybrid_property

Base = declarative_base()

class Chap(Base):
    __tablename__ = 'chap'

    id = Column(Integer, primary_key=True)
    tophats = Column(Integer)
    exclamation = Column(String)

    @hybrid_property
    def hat_brag(self):
        return "I have "+str(self.tophats)+" tophats!"

    @hybrid_property
    def cold_complain(self):
        return self.exclamation+", it's cold!"

...and I created a chap.

>>>c1 = Chap(tophats=5, exclamation="Blimey")

Now I want to query this chap based on how he complains about the weather:

>>>print(Session().query(Chap).filter(Chap.cold_complain == "Blimey, it's cold!"))
SELECT chap.id AS chap_id, chap.tophats AS chap_tophats, chap.gloves AS chap_gloves, chap.exclamation AS chap_exclamation 
FROM chap 
WHERE chap.exclamation || :exclamation_1 = :param_1

That SQL code looks right, but now I want to query him on another hybrid_property, one that contains an int converted to a str...

>>>print(Session().query(Chap).filter(Chap.hat_brag == "I have 5 tophats!"))
SELECT chap.id AS chap_id, chap.tophats AS chap_tophats, chap.exclamation AS chap_exclamation 
FROM chap 
WHERE false

WHERE false. That doesn't look like the right query! Anyone know what's going on?

标签： pythonsqlalchemy

Answer 1

在查询中，SQLAlchemy 将使用混合属性的 getter 方法而无需专用@...expression()方法来生成用于查询的必要 SQL 对象。getter 方法绑定到类，而不是实例，因此self在这种情况下将引用您的Chap类。

对于您的cold_complain属性，这意味着Chap.exclamation + ", it's cold!"返回，产生一个chap.exclamation || :exclamation_1SQL 表达式；+转换为 SQL 连接运算符。

但是对于你的hat_brag属性返回一个字符串；执行的表达式是真的"I have "+str(Chap.tophats)+" tophats!"，它变成'I have Chap.tophats tophats!'：

>>> "I have "+str(Chap.tophats)+" tophats!"
'I have Chap.tophats tophats!'

那是一个固定的静态字符串。接下来，这将成为您过滤器的一部分，使用== "I have 5 tophats!". 这两个静态字符串值不相等：

>>> "I have "+str(Chap.tophats)+" tophats!" == "I have 5 tophats!"
False

这就是发送到数据库的 SQL 查询中使用的内容。

您在这里要做的是使用该expression()选项并定义查询的 SQL 友好版本：

from sqlalchemy.sql.expression import cast

class Chap(Base):
    # ...

    @hybrid_property
    def hat_brag(self):
        return "I have "+str(self.tophats)+" tophats!"

    @hat_brag.expression
    def hat_brag(cls):
        return "I have " + cast(cls.tophats, String) + " tophats!"

现在，该expression方法用于查询，以及 Python 中实例上的原始函数：

>>> c1.hat_brag
'I have 5 tophats!'
>>> print(Session().query(Chap).filter(Chap.hat_brag == "I have 5 tophats!"))
SELECT chap.id AS chap_id, chap.tophats AS chap_tophats, chap.exclamation AS chap_exclamation
FROM chap
WHERE :param_1 || CAST(chap.tophats AS VARCHAR) || :param_2 = :param_3