PostgreSQL：查找租用电影总数的百分比

下面的代码给出了以下结果

Early: 7738
Late: 6586
On Time: 1720

我如何进一步添加第三列来查找百分比

以下是ERD和数据库设置的链接：

具有 t1 像 选择*，日期部分为“天”，返回日期-租赁日期为租赁天数 租金 , t2 像 选择租赁期限、租赁天数、， 租赁持续时间>租赁天数然后“提前”的情况 当租赁时间=租赁天数时，则“准时” 否则“迟到” 结束为租赁\退货\状态 来自胶片f，库存i，t1 其中f.film_id=i.film_id，t1.inventory_id=i.inventory_id 选择租赁\归还\状态，将*计为总租赁\电影 从t2开始 按1分组 按2描述订购； 使用a获取计数列sumcount*over的和，然后将计数除以该计数*/sumcount*over。乘以100使之成为一个百分比

psql (12.1 (Debian 12.1-1))
Type "help" for help.

testdb=# CREATE TABLE faket2 AS (
SELECT 'early' AS rental_return_status UNION ALL
SELECT 'early' UNION ALL
SELECT 'ontime' UNION ALL
SELECT 'late');
SELECT 4
testdb=# SELECT
  rental_return_status,
  COUNT(*) as total_films_rented,
  (100*count(*))/sum(count(*)) over () AS percentage
FROM faket2
GROUP BY 1
ORDER BY 2 DESC;
 rental_return_status | total_films_rented |     percentage      
----------------------+--------------------+---------------------
 early                |                  2 | 50.0000000000000000
 late                 |                  1 | 25.0000000000000000
 ontime               |                  1 | 25.0000000000000000
(3 rows)

---

您可以将窗口函数与一个CTE表（而不是2个）一起使用：

WITH raw_status AS (
  SELECT rental_duration - DATE_PART('day', return_date - rental_date) AS days_remaining
    FROM rental r
    JOIN inventory i ON r.inventory_id=i.inventory_id
    JOIN film f on f.film_id=i.film_id
  )
SELECT CASE WHEN days_remaining > 0 THEN 'Early'
            WHEN days_remaining = 0 THEN 'On Time'
            ELSE 'Late' END AS rental_status,
       count(*),
       (100*count(*))/sum(count(*)) OVER () AS percentage
FROM raw_status
GROUP BY 1;
 rental_status | count |     percentage      
---------------+-------+---------------------
 Early         |  7738 | 48.2298678633757168
 On Time       |  1720 | 10.7205185739217153
 Late          |  6586 | 41.0496135627025679
(3 rows)

披露：我为