1364.(Medium)顾客的可信联系人数量

顾客表:Customers

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| customer_name | varchar |
| email         | varchar |
+---------------+---------+
customer_id 是这张表的主键。
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。
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 是这张表的主键。
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SQL查询以查找以下内容:

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客户表中)。 将查询的结果按照 invoice_id 排序。

查询结果的格式如下例所示:

Customers table:
+-------------+---------------+--------------------+
| customer_id | customer_name | email              |
+-------------+---------------+--------------------+
| 1           | Alice         | alice@leetcode.com |
| 2           | Bob           | bob@leetcode.com   |
| 13          | John          | john@leetcode.com  |
| 6           | Alex          | alex@leetcode.com  |
+-------------+---------------+--------------------+
Contacts table:
+-------------+--------------+--------------------+
| user_id     | contact_name | contact_email      |
+-------------+--------------+--------------------+
| 1           | Bob          | bob@leetcode.com   |
| 1           | John         | john@leetcode.com  |
| 1           | Jal          | jal@leetcode.com   |
| 2           | Omar         | omar@leetcode.com  |
| 2           | Meir         | meir@leetcode.com  |
| 6           | Alice        | alice@leetcode.com |
+-------------+--------------+--------------------+
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       |
+------------+-------+---------+
Result table:
+------------+---------------+-------+--------------+----------------------+
| 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                    |
+------------+---------------+-------+--------------+----------------------+
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。

来源:力扣(LeetCode)

链接:https://leetcode-cn.com/problems/number-of-trusted-contacts-of-a-customer

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Solution

Invoices left join Customers left join Contacts

SELECT invoice_id,  
        max(cus.customer_name) customer_name, 
        max(price) price,
        count(DISTINCT con.contact_email) contacts_cnt,
        IFNULL(sum(con.contact_email in (SELECT email FROM Customers)),0) trusted_contacts_cnt
from Invoices i left JOIN 
    Customers cus on i.user_id = cus.customer_id LEFT JOIN
    Contacts con on i.user_id = con.user_id
GROUP BY invoice_id;

Table Schema

Create table If Not Exists Customers (customer_id int, customer_name varchar(20), email varchar(30));
Create table If Not Exists Contacts (user_id int, contact_name varchar(20), contact_email varchar(30));
Create table If Not Exists Invoices (invoice_id int, price int, user_id int);
Truncate table Customers;
insert into Customers (customer_id, customer_name, email) values ('1', 'Alice', 'alice@leetcode.com');
insert into Customers (customer_id, customer_name, email) values ('2', 'Bob', 'bob@leetcode.com');
insert into Customers (customer_id, customer_name, email) values ('13', 'John', 'john@leetcode.com');
insert into Customers (customer_id, customer_name, email) values ('6', 'Alex', 'alex@leetcode.com');
Truncate table Contacts;
insert into Contacts (user_id, contact_name, contact_email) values ('1', 'Bob', 'bob@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('1', 'John', 'john@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('1', 'Jal', 'jal@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('2', 'Omar', 'omar@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('2', 'Meir', 'meir@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('6', 'Alice', 'alice@leetcode.com');
Truncate table Invoices;
insert into Invoices (invoice_id, price, user_id) values ('77', '100', '1');
insert into Invoices (invoice_id, price, user_id) values ('88', '200', '1');
insert into Invoices (invoice_id, price, user_id) values ('99', '300', '2');
insert into Invoices (invoice_id, price, user_id) values ('66', '400', '2');
insert into Invoices (invoice_id, price, user_id) values ('55', '500', '13');
insert into Invoices (invoice_id, price, user_id) values ('44', '60', '6');

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