SQL：如何获得给定队列的分组项目总和

Question

我有以下orders表：

(1, 2, '2021-03-05', 15, 'books'),
(1, 13, '2022-03-07', 3, 'music'),
(1, 14, '2022-06-15', 900, 'travel'),
(1, 11, '2021-11-17', 25, 'books'),
(1, 16, '2022-08-03', 32, 'books'),
(2, 4, '2021-04-12', 4, 'music'),
(2, 7, '2021-06-29', 9, 'music'),
(2, 20, '2022-11-03', 8, 'music'),
(2, 22, '2022-11-07', 575, 'food'),
(2, 24, '2022-11-20', 95, 'food'),
(3, 3, '2021-03-17', 25, 'books'),
(3, 5, '2021-06-01', 650, 'travel'),
(3, 17, '2022-08-17', 1200, 'travel'),
(3, 19, '2022-10-02', 6, 'music'),
(3, 23, '2022-11-08', 7, 'food'),
(4, 9, '2021-08-20', 3200, 'travel'),
(4, 10, '2021-10-29', 2750, 'travel'),
(4, 15, '2022-07-15', 1820, 'travel'),
(4, 21, '2022-11-05', 8000, 'travel'),
(4, 25, '2022-11-29', 2300, 'travel'),
(5, 1, '2021-01-04', 3, 'music'),
(5, 6, '2021-06-09', 820, 'travel'),
(5, 8, '2021-07-30', 19, 'books'),
(5, 12, '2021-12-10', 22, 'music'),
(5, 18, '2022-09-19', 20, 'books'),
(6, 26, '2023-01-09', 700, 'travel'),
(6, 27, '2023-01-23', 1900, 'travel')

这是一个小提琴： http://sqlfiddle.com/#!17/71698/3

我想在那些曾经购买过travel产品的客户中获得按product分类的revenue总和。

本例中，客户1、3、4、5、6购买了travel产品。 因此，所需的结果集将如下所示：

客户ID	收入账簿	收入_音乐	收入_食物
1个	72	3个	0
3个	25	6个	7
4个	0	0	0
5个	39	25	0
6个	0	0	0

我该怎么做？ 谢谢你！

Answer 1

SELECT
  user_id
  ,sum(case when product='books' then revenue else 0 end) as revenue_books    
  ,sum(case when product='music' then revenue else 0 end) as revenue_music    
  ,sum(case when product='food' then revenue else 0 end) as revenue_food    
  FROM
  orders
  where user_id in (select user_id from orders where product='travel')
  group by user_id

http://sqlfiddle.com/#!17/71698/5

编辑

正如建议的那样，这是另一种选择。 但它显示 null 而不是零。 如果需要零，你应该使用 coalesce()

SELECT
  user_id
  ,coalesce(sum(revenue) filter (where product = 'books'),0) as revenue_books    
  ,coalesce(sum(revenue) filter (where product = 'music'),0)  as revenue_music    
  ,coalesce(sum(revenue) filter (where product = 'food'),0)  as revenue_food    
  FROM  orders
  where user_id in (select user_id from orders where product='travel')
  group by user_id

http://sqlfiddle.com/#!17/71698/7

Answer 2

在我的回答中，我展示了如何思考问题以获得结果——将其分解为多个部分，然后将其组合起来。 一些答案给出了一个不太冗长的查询，但我认为它们不会更快。 对于 SQL 的新手来说，这应该更容易理解

首先购买旅游产品的人

SELECT DISTINCT user_id
FROM orders 
WHERE product = 'travel'

你关心书籍、音乐和食物，你可以得到这样的总数：

SELECT user_id, product, SUM(revenue) as TOT
FROM orders
GROUP BY user_id, product 
WHERE product in ('books', 'music', 'food'),

现在加入他们

WITH sums AS (
  SELECT user_id, product, SUM(revenue) as TOT
  FROM orders
  GROUP BY user_id, product 
  WHERE product in ('books', 'music', 'food'),
)
SELECT u.user_id, books.TOT as book_total, music.TOT as music_total, food.TOT as food_total
FROM (
  SELECT DISTINCT user_id
  FROM orders 
  WHERE product = 'travel'
) as U
LEFT JOIN sums as books ON u.userid = books.user_id and books.product = 'books'
LEFT JOIN sums as music ON u.userid = music.user_id and music.product = 'music'
LEFT JOIN sums as food ON u.userid = food.user_id and food.product = 'food'