如何将 SQL 标量子查询转换为 SQLAlchemy 表达式 [英] How to convert SQL scalar subquery to SQLAlchemy expression

问题描述

我需要一点帮助来用 SQLAlchemy 语言表达我的代码，如下所示:

I need a litle help with expressing in SQLAlchemy language my code like this:

SELECT
    s.agent_id,
    s.property_id,
    p.address_zip,
    (
        SELECT v.valuation
        FROM property_valuations v WHERE v.zip_code = p.address_zip
        ORDER BY ABS(DATEDIFF(v.as_of, s.date_sold))
        LIMIT 1
    ) AS back_valuation,
FROM sales s
JOIN properties p ON s.property_id = p.id

内部子查询旨在从表 propert_valuations 中获取属性值，其中列(zip_code INT、估价 DECIMAL、as_if DATE)最接近表 的销售日期销售.我知道如何重写它，但我完全坚持 order_by 表达式 - 我无法准备子查询以稍后传递排序成员.

Inner subquery aimed to get property value from table propert_valuations with columns (zip_code INT, valuation DECIMAL, as_if DATE) closest to the date of sale from table sales. I know how to rewrite it but I completely stuck on order_by expression - I cannot prepare subquery to pass ordering member later.

目前我有以下疑问:

subquery = (
    session.query(PropertyValuation)
    .filter(PropertyValuation.zip_code == Property.address_zip)
    .order_by(func.abs(func.datediff(PropertyValuation.as_of, Sale.date_sold)))
    .limit(1)
)

query = session.query(Sale).join(Sale.property_)

如何将这些查询组合在一起?

How to combine these queries together?

推荐答案

如何将这些查询组合在一起?

How to combine these queries together?

使用 as_scalar() 或 标签():

subquery = (
    session.query(PropertyValuation.valuation)
    .filter(PropertyValuation.zip_code == Property.address_zip)
    .order_by(func.abs(func.datediff(PropertyValuation.as_of, Sale.date_sold)))
    .limit(1)
)

query = session.query(Sale.agent_id,
                      Sale.property_id,
                      Property.address_zip,
                      # `subquery.as_scalar()` or
                      subquery.label('back_valuation'))\
        .join(Property)

使用 as_scalar() 将返回的列和行限制为 1，因此您无法使用它获取整个模型对象(因为 query(PropertyValuation) 是所有PropertyValuation 的属性)，但只获取 valuation 属性有效.

Using as_scalar() limits returned columns and rows to 1, so you cannot get the whole model object using it (as query(PropertyValuation) is a select of all the attributes of PropertyValuation), but getting just the valuation attribute works.

但我完全坚持使用 order_by 表达式 - 我无法准备子查询以稍后传递排序成员.

but I completely stuck on order_by expression - I cannot prepare subquery to pass ordering member later.

以后就不用传了.您当前声明子查询的方式很好，因为 SQLAlchemy 可以 自动将 FROM 对象与封闭查询的对象相关联.我尝试创建在某种程度上代表您所拥有内容的模型，以下是上述查询的工作方式(添加换行符和缩进以提高可读性):

There's no need to pass it later. Your current way of declaring the subquery is fine as it is, since SQLAlchemy can automatically correlate FROM objects to those of an enclosing query. I tried creating models that somewhat represent what you have, and here's how the query above works out (with added line-breaks and indentation for readability):

In [10]: print(query)
SELECT sale.agent_id AS sale_agent_id,
       sale.property_id AS sale_property_id,
       property.address_zip AS property_address_zip,
       (SELECT property_valuations.valuation
        FROM property_valuations 
        WHERE property_valuations.zip_code = property.address_zip
        ORDER BY abs(datediff(property_valuations.as_of, sale.date_sold))
        LIMIT ? OFFSET ?) AS back_valuation 
FROM sale
JOIN property ON property.id = sale.property_id

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