如何将选择表达式的结果保存为变量？答案

【问题标题】：How to save the results of a select expression as a variable?如何将选择表达式的结果保存为变量？
【发布时间】：2020-09-04 18:46:37
【问题描述】：

在下面的Postgresql sql中，有没有办法将mo.delivered_at - mo.created_at保存为变量，这样我就不用重复了？

SELECT 
    to_char(mo.created_at,'MM-YYYY') AS month,
    mo.sku_key as sku,
    c.name,
    COUNT(*) as total,
    COUNT(*) FILTER (WHERE mo.delivered_at - mo.created_at < interval '3 days') as three_days,
    COUNT(*) FILTER (WHERE mo.delivered_at - mo.created_at > interval '3 days' and mo.delivered_at - mo.created_at <= interval '6 days') as six_days,
    COUNT(*) FILTER (WHERE mo.delivered_at - mo.created_at > interval '6 days' and mo.delivered_at - mo.created_at <= interval '9 days') as nine_days,
    COUNT(*) FILTER (WHERE mo.delivered_at - mo.created_at > interval '9 days') as ten_days,
    min(mo.delivered_at - mo.created_at),
    max(mo.delivered_at - mo.created_at),
    percentile_disc(0.5) within group (order by mo.delivered_at - mo.created_at) as median,
    avg(mo.delivered_at - mo.created_at) as average
FROM medication_order mo
LEFT JOIN subscription s ON s.id=mo.subscription_id
LEFT JOIN condition c on s.condition_id = c.id
WHERE 
    mo.status = 'DELIVERED' AND 
    mo.payment_preference = 'INSURANCE' AND
    mo.created_at > '2020-01-01' AND
    mo.delivered_at IS NOT null AND
    mo.sku_key != 'manual_order_sku'
GROUP BY month, mo.sku_key, c.name

【问题讨论】：

您可以重新组织以使用 CTE 或子查询。
或者，如果您经常使用它，请将其设为生成的列，然后可以对其进行索引。 postgresql.org/docs/12/ddl-generated-columns.html
你按c.name分组。大概是一个冗长的文本列？ condition.name 是否定义为 UNIQUE？（可能应该是。）然后你可以让查询便宜一点......
@ErwinBrandstetter 是的，它是一个文本字段。不是独一无二的，现在我看看它。
那么你可以有两个distinct条件同名吗？ GROUP BY 不会产生误导性结果吗？

标签： sql postgresql datetime count subquery

【解决方案1】：

您可以按照建议在子查询或 CTE 中计算派生值。

但还有更多。这应该更快（并且正确）。并且也可以正确排序：

SELECT
    to_char(mo.month,'MM-YYYY') AS month,  -- optionally prettify
    mo.sku,
    s.condition_id,  -- I added this to make the result unambiguous
    (SELECT name FROM condition WHERE id = s.condition_id) AS condition_name,
    COUNT(*) AS total,
    COUNT(*) FILTER (WHERE mo.my_interval < interval '3 days') AS three_days,
    COUNT(*) FILTER (WHERE mo.my_interval > interval '3 days' AND mo.my_interval <= interval '6 days') AS six_days,
    COUNT(*) FILTER (WHERE mo.my_interval > interval '6 days' AND mo.my_interval <= interval '9 days') AS nine_days,
    COUNT(*) FILTER (WHERE mo.my_interval > interval '9 days') AS ten_days,
    min(mo.my_interval),
    max(mo.my_interval),
    percentile_disc(0.5) WITHIN GROUP (ORDER BY mo.my_interval) AS median,
    avg(mo.my_interval) AS average
FROM  (
   SELECT
       date_trunc('month', mo.created_at) AS month,         -- faster, keeps ORDER
       delivered_at - created_at          AS my_interval,   -- your core request
       sku_key                            AS sku
   FROM   medication_order mo
   WHERE  status = 'DELIVERED'                              -- filter early
   AND    payment_preference = 'INSURANCE'
   AND    created_at > '2020-01-01'
   AND    delivered_at IS NOT NULL
   AND    sku_key <> 'manual_order_sku'
   ) mo
LEFT   JOIN subscription s ON s.id = mo.subscription_id
GROUP  BY mo.month, mo.sku, s.condition_id  -- GROUP BY unique ID! Correct - and cheaper, too
ORDER  BY mo.month, mo.sku, s.condition_id; -- my addition: sorting by date works across years, 'MM-YYYY' does not

旁白：condition.name 应该是UNIQUE。而“名字”几乎从来都不是一个好名字。

【讨论】：

感谢您的回答。我想知道我遇到的排序问题。

【解决方案2】：

从表中选择时，您可以只计算子查询中的信息：

SELECT 
    to_char(mo.created_at,'MM-YYYY') AS month,
    mo.sku_key as sku,
    c.name,
    COUNT(*) as total,
    COUNT(*) FILTER (WHERE mo.delivery_interval < interval '3 days') as three_days,
    COUNT(*) FILTER (WHERE mo.delivery_interval > interval '3 days' and mo.delivery_interval <= interval '6 days') as six_days,
    COUNT(*) FILTER (WHERE mo.delivery_interval > interval '6 days' and mo.delivery_interval <= interval '9 days') as nine_days,
    COUNT(*) FILTER (WHERE mo.delivery_interval > interval '9 days') as ten_days,
    min(mo.delivery_interval),
    max(mo.delivery_interval),
    percentile_disc(0.5) within group (order by mo.delivery_interval) as median,
    avg(mo.delivery_interval) as average
FROM (
    SELECT mo.*, mo.delivery_interval delivery_interval   --> here
    FROM medication_order
) mo
LEFT JOIN subscription s ON s.id=mo.subscription_id
LEFT JOIN condition c on s.condition_id = c.id
WHERE 
    mo.status = 'DELIVERED' AND 
    mo.payment_preference = 'INSURANCE' AND
    mo.created_at > '2020-01-01' AND
    mo.delivered_at IS NOT null AND
    mo.sku_key != 'manual_order_sku'
GROUP BY month, mo.sku_key, c.name

【讨论】：