comments | difficulty | edit_url | tags | |
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true |
中等 |
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Customers
表:
+---------------------+---------+ | Column Name | Type | +---------------------+---------+ | customer_id | int | | customer_name | varchar | +---------------------+---------+ customer_id 是这张表中具有唯一值的列。 customer_name 是顾客的名称。
Orders
表:
+---------------+---------+ | Column Name | Type | +---------------+---------+ | order_id | int | | customer_id | int | | product_name | varchar | +---------------+---------+ order_id 是这张表中具有唯一值的列。 customer_id 是购买了名为 "product_name" 产品顾客的id。
请你编写解决方案,报告购买了产品 "A","B" 但没有购买产品 "C" 的客户的 customer_id 和 customer_name,因为我们想推荐他们购买这样的产品。
返回按 customer_id
排序 的结果表。
返回结果格式如下所示。
示例 1:
输入: Customers table: +-------------+---------------+ | customer_id | customer_name | +-------------+---------------+ | 1 | Daniel | | 2 | Diana | | 3 | Elizabeth | | 4 | Jhon | +-------------+---------------+ Orders table: +------------+--------------+---------------+ | order_id | customer_id | product_name | +------------+--------------+---------------+ | 10 | 1 | A | | 20 | 1 | B | | 30 | 1 | D | | 40 | 1 | C | | 50 | 2 | A | | 60 | 3 | A | | 70 | 3 | B | | 80 | 3 | D | | 90 | 4 | C | +------------+--------------+---------------+ 输出: +-------------+---------------+ | customer_id | customer_name | +-------------+---------------+ | 3 | Elizabeth | +-------------+---------------+ 解释: 只有 customer_id 为 3 的顾客购买了产品 A 和产品 B ,却没有购买产品 C 。
我们可以用 LEFT JOIN
将 Customers
表和 Orders
表连接起来,然后按照 customer_id
进行分组,最后筛选出购买了产品 A 和产品 B 却没有购买产品 C 的顾客 🔒。
# Write your MySQL query statement below
SELECT customer_id, customer_name
FROM
Customers
LEFT JOIN Orders USING (customer_id)
GROUP BY 1
HAVING SUM(product_name = 'A') > 0 AND SUM(product_name = 'B') > 0 AND SUM(product_name = 'C') = 0
ORDER BY 1;