SQLAlchemy 不能在查询中使用 func.bigger 作为 func [英] SQLAlchemy can't use func.bigger as func in query

问题描述

描述我的问题.可以看到这个原始的sql:

To desc my problem. Can see this raw sql:

select datediff(now(), create_time) > 7 as is_new from test order by is_new desc limit 19; 

我尝试通过 SQLAlchemy 一步一步实现:

I try to implement by SQLAlchemy step by step:

diff_days = func.datediff(today, test.create_time).label("diff_days")
session.query(diff_days).filter(test.id.in_((1,2,3,33344))).order_by(diff_days.asc()).all()   

这个工作正常.但是当我想在 mysql 中描述 > 时.失败了:

This work fine. But when I want to desc > in mysql. It failed:

is_new = func.greater(func.datediff(today, test.create_time), 7).label("is_new")
session.query(is_new).filter(test.id.in_((1,2,3,33344))).order_by(is_new.asc()).all()

我知道 SQLAlchemy 将我的 sql 解释为 greater 而 mysql 不支持.那么我怎样才能得到我的答案 a >b 类似 greater(a, b)

I know SQLAlchemy explain my sql to greater while mysql don't support. So How can I to get my answer a > b with something like greater(a, b)

可能是简单的sql select a >b from test 也可以说明问题.虽然以上是我的起源需要.所以问题可以改变:

May be the simple sql select a > b from test can desc the problem too. While above is my origin need. So the problem can change :

如何使用SQLAIchemy orm来实现select a >b 来自测试.

How to using SQLAIchemy orm to implement select a > b from test.

推荐答案

SQLAlchemy 为您提供 丰富的运算符重载，所以就做

SQLAlchemy offers you rich operator overloading, so just do

is_new = (func.datediff(today, test.create_time) > 7).label("is_new")
session.query(is_new).\
    filter(test.id.in_([1, 2, 3, 33344])).\
    order_by(is_new.asc()).\
    all()

自从创建Function 也是一个 ColumnElement 等具有 ColumnOperators.

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