SQLAlchemy 按函数结果排序 [英] SQLAlchemy order by function result

问题描述

这是我拥有的代码并且它正在运行(返回按难度排序的所有问题):

This is the code I have and it is working (returns all problems ordered by difficulty):

def get_noteworthy_problems(self):

    ACategory = aliased(Category)
    AProblem = aliased(Problem)

    all_prob = DBSession.query(AProblem).filter(
        AProblem.parent_id == ACategory.id,
        ACategory.parent_id == self.id)

    noteworthy_problems = \
        sorted(all_prob, key=lambda x: x.difficulty(), reverse=True)

    return noteworthy_problems

但我想我必须优化这段代码.是否有可能更改具有 order_by 和我的函数 difficulty() 的代码?我的函数返回一个数字.我试过类似的东西:

But I think I must optimize this code. Is there a possibility to change the code having order_by and my function difficulty()? My function returns a number. I tried something like:

    result = DBSession.query(AProblem).filter(
        AProblem.parent_id == ACategory.id,
        ACategory.parent_id == self.id).order_by(
        AProblem.difficulty().desc())

但我收到错误 TypeError: 'NoneType' object is not callable.

推荐答案

混合属性是作为 Python 属性和 SQL 表达式的特殊方法.只要你的难度函数可以用SQL表达，就可以像普通列一样进行过滤排序.

Hybrid attributes are special methods that act as both a Python property and a SQL expression. As long as your difficulty function can be expressed in SQL, it can be used to filter and order like a normal column.

例如，如果您将难度计算为问题的鹦鹉数量，如果问题超过 30 天，则乘以 10，您可以使用:

For example, if you calculate difficulty as the number of parrots a problem has, times ten if the problem is older than 30 days, you would use:

from datetime import datetime, timedelta
from sqlalchemy import Column, Integer, DateTime, case
from sqlalchemy.ext.hybrid import hybrid_property

class Problem(Base):
    parrots = Column(Integer, nullable=False, default=1)
    created = Column(DateTime, nullable=False, default=datetime.utcnow)

    @hybrid_property
    def difficulty(self):
        # this getter is used when accessing the property of an instance
        if self.created <= (datetime.utcnow() - timedelta(30)):
            return self.parrots * 10

        return self.parrots

    @difficulty.expression
    def difficulty(cls):
        # this expression is used when querying the model
        return case(
            [(cls.created <= (datetime.utcnow() - timedelta(30)), cls.parrots * 10)],
            else_=cls.parrots
        )

并使用以下命令查询:

session.query(Problem).order_by(Problem.difficulty.desc())

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