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中等
数据库

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题目描述

顾客表:Customers

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| customer_name | varchar |
| email         | varchar |
+---------------+---------+
customer_id 是这张表具有唯一值的列。
此表的每一行包含了某在线商店顾客的姓名和电子邮件。

 

联系方式表:Contacts

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | id      |
| contact_name  | varchar |
| contact_email | varchar |
+---------------+---------+
(user_id, contact_email) 是这张表的主键(具有唯一值的列的组合)。
此表的每一行表示编号为 user_id 的顾客的某位联系人的姓名和电子邮件。
此表包含每位顾客的联系人信息,但顾客的联系人不一定存在于顾客表中。

 

发票表:Invoices

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| invoice_id   | int     |
| price        | int     |
| user_id      | int     |
+--------------+---------+
invoice_id 是这张表具有唯一值的列。
此表的每一行分别表示编号为 user_id 的顾客拥有有一张编号为 invoice_id、价格为 price 的发票。

 

为每张发票 invoice_id 编写一个查询方案以查找以下内容:

  • customer_name:与发票相关的顾客名称。
  • price:发票的价格。
  • contacts_cnt:该顾客的联系人数量
  • trusted_contacts_cnt:可信联系人的数量:既是该顾客的联系人又是商店顾客的联系人数量(即:可信联系人的电子邮件存在于  Customers 表中)。

返回结果按照 invoice_id 排序

结果的格式如下例所示。

 

示例 1:

输入:
Customers table:
+-------------+---------------+--------------------+
| customer_id | customer_name | email              |
+-------------+---------------+--------------------+
| 1           | Alice         | [email protected] |
| 2           | Bob           | [email protected]   |
| 13          | John          | [email protected]  |
| 6           | Alex          | [email protected]  |
+-------------+---------------+--------------------+
Contacts table:
+-------------+--------------+--------------------+
| user_id     | contact_name | contact_email      |
+-------------+--------------+--------------------+
| 1           | Bob          | [email protected]   |
| 1           | John         | [email protected]  |
| 1           | Jal          | [email protected]   |
| 2           | Omar         | [email protected]  |
| 2           | Meir         | [email protected]  |
| 6           | Alice        | [email protected] |
+-------------+--------------+--------------------+
Invoices table:
+------------+-------+---------+
| invoice_id | price | user_id |
+------------+-------+---------+
| 77         | 100   | 1       |
| 88         | 200   | 1       |
| 99         | 300   | 2       |
| 66         | 400   | 2       |
| 55         | 500   | 13      |
| 44         | 60    | 6       |
+------------+-------+---------+
输出:
+------------+---------------+-------+--------------+----------------------+
| invoice_id | customer_name | price | contacts_cnt | trusted_contacts_cnt |
+------------+---------------+-------+--------------+----------------------+
| 44         | Alex          | 60    | 1            | 1                    |
| 55         | John          | 500   | 0            | 0                    |
| 66         | Bob           | 400   | 2            | 0                    |
| 77         | Alice         | 100   | 3            | 2                    |
| 88         | Alice         | 200   | 3            | 2                    |
| 99         | Bob           | 300   | 2            | 0                    |
+------------+---------------+-------+--------------+----------------------+
解释:
Alice 有三位联系人,其中两位(Bob 和 John)是可信联系人。
Bob 有两位联系人, 他们中的任何一位都不是可信联系人。
Alex 只有一位联系人(Alice),并是一位可信联系人。
John 没有任何联系人。

解法

方法一

MySQL

# Write your MySQL query statement below
SELECT
    invoice_id,
    t2.customer_name,
    price,
    COUNT(t3.user_id) AS contacts_cnt,
    COUNT(t4.email) AS trusted_contacts_cnt
FROM
    Invoices AS t1
    LEFT JOIN Customers AS t2 ON t1.user_id = t2.customer_id
    LEFT JOIN Contacts AS t3 ON t1.user_id = t3.user_id
    LEFT JOIN Customers AS t4 ON t3.contact_email = t4.email
GROUP BY invoice_id
ORDER BY invoice_id;