Question

1364. Number of Trusted Contacts of a Customer

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Description

Table: Customers

+---------------+---------+
| Column Name   | Type  |
+---------------+---------+
| customer_id   | int   |
| customer_name | varchar |
| email     | varchar |
+---------------+---------+
customer_id is the column of unique values for this table.
Each row of this table contains the name and the email of a customer of an online shop.

 

Table: Contacts

+---------------+---------+
| Column Name   | Type  |
+---------------+---------+
| user_id     | id    |
| contact_name  | varchar |
| contact_email | varchar |
+---------------+---------+
(user_id, contact_email) is the primary key (combination of columns with unique values) for this table.
Each row of this table contains the name and email of one contact of customer with user_id.
This table contains information about people each customer trust. The contact may or may not exist in the Customers table.

 

Table: Invoices

+--------------+---------+
| Column Name  | Type  |
+--------------+---------+
| invoice_id   | int   |
| price    | int   |
| user_id    | int   |
+--------------+---------+
invoice_id is the column of unique values for this table.
Each row of this table indicates that user_id has an invoice with invoice_id and a price.

 

Write a solution to find the following for each invoice_id:

• customer_name: The name of the customer the invoice is related to.
• price: The price of the invoice.
• contacts_cnt: The number of contacts related to the customer.
• trusted_contacts_cnt: The number of contacts related to the customer and at the same time they are customers to the shop. (i.e their email exists in the Customers table.)

Return the result table ordered by invoice_id.

The result format is in the following example.

 

Example 1:

Input: 
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     |
+------------+-------+---------+
Output: 
+------------+---------------+-------+--------------+----------------------+
| 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          |
+------------+---------------+-------+--------------+----------------------+
Explanation: 
Alice has three contacts, two of them are trusted contacts (Bob and John).
Bob has two contacts, none of them is a trusted contact.
Alex has one contact and it is a trusted contact (Alice).
John doesn't have any contacts.

Solutions

Solution 1

# 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;