[MYSQL] JOIN 예제문제 풀어보기

예제문제 1 (LEFT JOIN 리트코드 문제)

Table: Customers

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id is the primary key column for this table.
Each row of this table indicates the ID and name of a customer.

Table: Orders

+-------------+------+
| Column Name | Type |
+-------------+------+
| id          | int  |
| customerId  | int  |
+-------------+------+
id is the primary key column for this table.
customerId is a foreign key of the ID from the Customers table.
Each row of this table indicates the ID of an order and the ID of the customer who ordered it.

Write an SQL query to report all customers who never order anything.

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
Customers table:
+----+-------+
| id | name  |
+----+-------+
| 1  | Joe   |
| 2  | Henry |
| 3  | Sam   |
| 4  | Max   |
+----+-------+
Orders table:
+----+------------+
| id | customerId |
+----+------------+
| 1  | 3          |
| 2  | 1          |
+----+------------+
Output: 
+-----------+
| Customers |
+-----------+
| Henry     |
| Max       |
+-----------+

예제문제 1 문제 풀이

# Write your MySQL query statement below

SELECT customers.name AS Customers
FROM customers
    LEFT JOIN orders ON customers.id = orders.customerid
WHERE orders.id IS NULL 

예제문제 2 SELF JOIN 리트코드 문제

Table: Employee

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| salary      | int     |
| managerId   | int     |
+-------------+---------+
id is the primary key column for this table.
Each row of this table indicates the ID of an employee, their name, salary, and the ID of their manager.

Write an SQL query to find the employees who earn more than their managers.

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
Employee table:
+----+-------+--------+-----------+
| id | name  | salary | managerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | Null      |
| 4  | Max   | 90000  | Null      |
+----+-------+--------+-----------+
Output: 
+----------+
| Employee |
+----------+
| Joe      |
+----------+

Explanation: Joe is the only employee who earns more than his manager.

예제문제 2 문제 풀이

# Write your MySQL query statement below

SELECT Employee.Name AS Employee

FROM Employee 
    INNER JOIN employee as Manager ON Employee.managerid = Manager.id
WHERE Employee.Salary > Manager.Salary

예제문제 3 SELF JOIN + MySQL 날짜 데이터 더하고 빼기

Table: Weather

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| recordDate    | date    |
| temperature   | int     |
+---------------+---------+
id is the primary key for this table.
This table contains information about the temperature on a certain day.

Write an SQL query to find all dates' Id with higher temperatures compared to its previous dates (yesterday).

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
Weather table:
+----+------------+-------------+
| id | recordDate | temperature |
+----+------------+-------------+
| 1  | 2015-01-01 | 10          |
| 2  | 2015-01-02 | 25          |
| 3  | 2015-01-03 | 20          |
| 4  | 2015-01-04 | 30          |
+----+------------+-------------+
Output: 
+----+
| id |
+----+
| 2  |
| 4  |
+----+

Explanation:
In 2015-01-02, the temperature was higher than the previous day (10 -> 25).
In 2015-01-04, the temperature was higher than the previous day (20 -> 30).

예제문제 3 문제 풀이

# Write your MySQL query statement below

SELECT today.id AS id
FROM weather AS today
    INNER JOIN Weather As yesterday ON today.recordDate = DATE_ADD(yesterday.recordDate, INTERVAL 1 DAY)
WHERE today.temperature > yesterday.temperature

사용한 개념