如何在SQLAlchemy中实现空合并运算符？

Question

或者我该如何使这个东西工作？

我有一个Interval对象:

class Interval(Base):
    __tablename__ = 'intervals'
    id = Column(Integer, primary_key=True)
    start = Column(DateTime)
    end = Column(DateTime, nullable=True)
    task_id = Column(Integer, ForeignKey('tasks.id'))

@hybrid_property #used to just be @property
def hours_spent(self):
    end = self.end or datetime.datetime.now()
    return (end-start).total_seconds()/60/60

一个任务:

class Task(Base):
    __tablename__ = 'tasks'
    id = Column(Integer, primary_key=True)
    title = Column(String)
    intervals = relationship("Interval", backref="task")

@hybrid_property  # Also used to be just @property
def hours_spent(self):
    return sum(i.hours_spent for i in self.intervals)

当然,添加所有典型的设置代码.

现在,当我尝试做 session.query(Task).filter(Task.hours_spent > 3).all()

我NotImplementedError: <built-in function getitem>离开了sum(i.hours_spent....

所以我正在查看文档的这一部分,并且理论上可能有某种方式可以编写一些可以实现我想要的东西.这部分看起来也可能有用,我会在等待答案时看着它;)

Answer 1

有关SQLAlchemy的coalesce函数的简单示例,这可能会有所帮助:在SQLAlchemy查询中处理空值 - 相当于isnull,nullif或coalesce.

以下是该帖子的几行代码:

from sqlalchemy.sql.functions import coalesce
my_config = session.query(Config).order_by(coalesce(Config.last_processed_at, datetime.date.min)).first()

Answer 2

SQLAlchemy不够智能从这些操作数构建SQL表达式树,您必须使用显式propname.expression装饰器来提供它.但接下来又出现了另一个问题:没有可移植的方法将间隔转换为数据库中的小时数.你可以TIMEDIFF在MySQL,EXTRACT(EPOCH FROM ... ) / 3600PostgreSQL等中使用.我建议改变属性以返回timedelta,并将苹果与苹果进行比较.

from sqlalchemy import select, func


class Interval(Base):
    ...

    @hybrid_property
    def time_spent(self):
        return (self.end or datetime.now()) - self.start

    @time_spent.expression
    def time_spent(cls):
        return func.coalesce(cls.end, func.current_timestamp()) - cls.start


class Task(Base):
    ...

    @hybrid_property
    def time_spent(self):
        return sum((i.time_spent for i in self.intervals), timedelta(0))

    @time_spent.expression
    def hours_spent(cls):
        return (select([func.sum(Interval.time_spent)])
            .where(cls.id==Interval.task_id)
            .label('time_spent'))

最后的查询是:

session.query(Task).filter(Task.time_spent > timedelta(hours=3)).all()

转换为(在PostgreSQL后端):

SELECT task.id AS task_id, task.title AS task_title 
FROM task 
WHERE (SELECT sum(coalesce(interval."end", CURRENT_TIMESTAMP) - interval.start) AS sum_1 
FROM interval 
WHERE task.id = interval.task_id) > %(param_1)s