题目
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 use_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 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 | 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。
解答
连接Invoices和Customers 得到顾客名称
把user_id选出是为了便于后边连接
select I.invoice_id, I.user_id,C.customer_name, I.price
from Invoices as I
left join Customers as C
on C.customer_id = I.user_id
order by I.invoice_id asc;
下面需要合并 contacts_cnt：该顾客的联系人数量
select CC.user_id, count(*) as contacts_cnt
from Contacts as CC
group by CC.user_id;
连接
select tmp.invoice_id, tmp.customer_name, tmp.price,
ifnull(tmp1.contacts_cnt, 0) as contacts_cnt
from (select I.invoice_id, I.user_id,C.customer_name, I.price
from Invoices as I
left join Customers as C
on C.customer_id = I.user_id
order by I.invoice_id asc) as tmp
left join (select CC.user_id, count(*) as contacts_cnt
from Contacts as CC
group by CC.user_id) as tmp1
on tmp.user_id = tmp1.user_id;
最后合并sted_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）。
先选出sted_contacts_cnt
select CCC.user_id, count(*) as trusted_contacts_cnt
from Contacts as CCC
where CCC.contact_name in (select customer_name from customers)
group by CCC.user_id;
最后再一次连接即可
select tmp.invoice_id, tmp.customer_name, tmp.price,
ifnull(tmp1.contacts_cnt, 0) as contacts_cnt,
ifnull(tmp2.trusted_contacts_cnt, 0) as trusted_contacts_cnt
from (select I.invoice_id, I.user_id,C.customer_name, I.price
from Invoices as I
left join Customers as C
on C.customer_id = I.user_id
order by I.invoice_id asc) as tmp
left join (select CC.user_id, count(*) as contacts_cnt
from Contacts as CC
group by CC.user_id) as tmp1
on tmp.user_id = tmp1.user_id
left join (select CCC.user_id, count(*) as trusted_contacts_cnt
from Contacts as CCC
where CCC.contact_name in (select customer_name from customers)
group by CCC.user_id) tmp2
on tmp.user_id = tmp2.user_id;
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