1795. 每个产品在不同商店的价格

表:Products

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| product_id  | int     |
| store1      | int     |
| store2      | int     |
| store3      | int     |
+-------------+---------+
在 SQL 中,这张表的主键是 product_id(产品Id)。
每行存储了这一产品在不同商店 store1, store2, store3 的价格。
如果这一产品在商店里没有出售,则值将为 null。

请你重构 Products 表,查询每个产品在不同商店的价格,使得输出的格式变为(product_id, store, price) 。如果这一产品在商店里没有出售,则不输出这一行。

输出结果表中的 顺序不作要求

查询输出格式请参考下面示例。

示例 1:

输入:
Products table:
+------------+--------+--------+--------+
| product_id | store1 | store2 | store3 |
+------------+--------+--------+--------+
| 0          | 95     | 100    | 105    |
| 1          | 70     | null   | 80     |
+------------+--------+--------+--------+
输出:
+------------+--------+-------+
| product_id | store  | price |
+------------+--------+-------+
| 0          | store1 | 95    |
| 0          | store2 | 100   |
| 0          | store3 | 105   |
| 1          | store1 | 70    |
| 1          | store3 | 80    |
+------------+--------+-------+
解释:
产品 0 在 store1、store2、store3 的价格分别为 95、100、105。
产品 1 在 store1、store3 的价格分别为 70、80。在 store2 无法买到。

题解

列转行

# Write your MySQL query statement below
select product_id,'store1' as store,store1 as price from products where store1 is not null
union
select product_id,'store2' as store,store2 as price from products where store2 is not null
union
select product_id,'store3' as store,store3 as price from products where store3 is not null;

行转列

# Write your MySQL query statement below
SELECT 
  product_id,
  SUM(IF(store = 'store1', price, NULL)) as 'store1',
  SUM(IF(store = 'store2', price, NULL)) as 'store2',
  SUM(IF(store = 'store3', price, NULL)) as 'store3' 
FROM
  products2 
GROUP BY product_id;