问题描述
在下面的Postgresql sql中，是否有一种方法可以将mo.delivered_at - mo.created_at保存为变量，这样我就不必重复自己了？
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
解决方法
从表中选择时，您可以只在子查询中计算信息：
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
您可以按照建议在子查询或CTE中计算得出的值。
但是还有更多。这应该更快（正确）。并且也可以正确排序：
SELECT
to_char(mo.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,s.condition_id; -- my addition: sorting by date works across years,'MM-YYYY' does not
此外：condition.name应该是UNIQUE。而且“名字”几乎永远不是好名字。