LeetCode-1364. 顾客的可信联系人数量（中等）

顾客表：Customers

联系方式表：Contacts

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 use_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 | [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 |
+------------+-------+---------+
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
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

审题：

联系表：编号为 use_id 的顾客的某位联系人的姓名和电子邮件。

查询：与发票相关的顾客名称。

思考：

针对于所要求的问题，逐步化解

第一步连接Invoices表和Customers表，按升序排列invoice_id。

select invoice_id, c.customer_name,i.price
from Invoices as i left join customers as c
on i.user_id=c.customer_id
order by invoice_id asc

第二步求解contacts_cnt（顾客的联系人数）：连接Contacts（联系方式表）表和Customers（顾客表）表，这里我按照customer_name分组，是为了最后与第一步的表连接。

select customer_name,count(user_id) as contacts_cnt
from customers as cu left join Contacts as co
on cu.customer_id=co.user_id
group by customer_name

第三步求解trusted_contacts_cnt（既是该顾客的联系人又是商店顾客的联系人数量（可信联系人的电子邮件存在于客户表中））：连接Contacts表和Customers表，这里同样按照customer_name分组，
求出trusted_contacts_cnt的关键在于where条件的选择。

select customer_name,count(*) as trusted_contacts_cnt
from customers as cu left join Contacts as co
on cu.customer_id=co.user_id 
where contact_name in (select customer_name from customers) 
group by customer_name

最后三表合并，以customer_name作为连接条件。

select invoice_id, c.customer_name,i.price ,a.contacts_cnt,ifnull(b.trusted_contacts_cnt,0) 
as trusted_contacts_cnt
from Invoices as i left join customers as c
on i.user_id=c.customer_id
left join (select customer_name,count(user_id) as contacts_cnt
from customers as cu left join Contacts as co
on cu.customer_id=co.user_id
group by customer_name) a
on c.customer_name =a.customer_name
left join 
(select customer_name,count(*) as trusted_contacts_cnt
from customers as cu left join Contacts as co
on cu.customer_id=co.user_id 
where contact_name in (select customer_name from customers) 
group by customer_name) b
on a.customer_name=b.customer_name
order by invoice_id asc

知识点：

红楼终究一场梦

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LeetCode-1364. 顾客的可信联系人数量（中等）

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