心得

題目要求刪除同樣Email的資料，在DELETE的條件WHERE裡面不能包含異動的表(Person)，所以這裡再包了一層SELECT來排除編譯錯誤的問題，而這層必須給予別名(強制)否則會編譯錯誤。

問題

Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | [email protected] |
| 2  | [email protected]  |
| 3  | [email protected] |
+----+------------------+
Id is the primary key column for this table.

For example, after running your query, the above Person table should have the following rows:

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | [email protected] |
| 2  | [email protected]  |
+----+------------------+

答案

DELETE 
FROM    Person
WHERE   Id 
NOT IN (SELECT y.*
        FROM (  SELECT MIN(z.Id)
                FROM Person AS z
                GROUP BY z.Email) AS y)

心得

題目要求找出溫度比前一天高的資料，這裡用了DATE_SUB與TO_DAYS兩種方式來解答。

題目

Given a Weather table, write a SQL query to find all dates’ Ids with higher temperature compared to its previous (yesterday’s) dates.

+---------+------------+------------------+
| Id(INT) | Date(DATE) | Temperature(INT) |
+---------+------------+------------------+
|       1 | 2015-01-01 |               10 |
|       2 | 2015-01-02 |               25 |
|       3 | 2015-01-03 |               20 |
|       4 | 2015-01-04 |               30 |
+---------+------------+------------------+

For example, return the following Ids for the above Weather table:

+----+
| Id |
+----+
|  2 |
|  4 |
+----+

答案

1. Sub Query

   # Write your MySQL query statement below
   SELECT  a.Id
   FROM    Weather AS a
   WHERE   (   SELECT  z.Temperature
               FROM    Weather AS z
               WHERE   DATE_SUB(a.Date, INTERVAL 1 DAY) = z.Date) < a.Temperature

2. Join

   # Write your MySQL query statement below
   SELECT  a.Id
   FROM    Weather AS a
   JOIN    Weather AS b
   ON      DATE_SUB(a.Date, INTERVAL 1 DAY) = b.Date
   WHERE   a.Temperature > b.Temperature

3. TO_DAYS

   # Write your MySQL query statement below
   SELECT  a.Id
   FROM    Weather AS a
   JOIN    Weather AS b
   ON      TO_DAYS(a.Date) - TO_DAYS(b.Date) = 1
   WHERE   a.Temperature > b.Temperature

    

參考：

1. DATE_ADD() 與 DATE_SUB() 日期的加法與減法
2. Mysql日期和時間函數不求人

心得

這題要show出沒有訂過東西的客戶，所以可以使用LEFT JOIN來關聯兩張表。

問題

Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.

Table: Customers.

+----+-------+
| Id | Name  |
+----+-------+
| 1  | Joe   |
| 2  | Henry |
| 3  | Sam   |
| 4  | Max   |
+----+-------+

Table: Orders.

+----+------------+
| Id | CustomerId |
+----+------------+
| 1  | 3          |
| 2  | 1          |
+----+------------+

Using the above tables as example, return the following:

+-----------+
| Customers |
+-----------+
| Henry     |
| Max       |
+-----------+

答案

1. LEFT JOIN

   # Write your MySQL query statement below
   SELECT      a.Name AS Customers
   FROM        Customers AS a
   LEFT JOIN   Orders AS b
   ON          a.Id = b.CustomerId
   WHERE       b.CustomerId IS NULL

2. Sub Query

   # Write your MySQL query statement below
   SELECT  a.Name AS Customers
   FROM    Customers AS a
   WHERE   (   SELECT  COUNT(1)
               FROM    Orders AS z
               WHERE   z.CustomerId = a.Id) = 0

心得：

找出該資料表內第二高薪水的筆數，若無則回傳null。

問題：

Write a SQL query to get the second highest salary from the Employee table.

+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

For example, given the above Employee table, the second highest salary is 200. If there is no second highest salary, then the query should return null.

答案：

1. Sub Query

   # Write your MySQL query statement below
   SELECT  MAX(a.Salary) AS SecondHighestSalary
   FROM    Employee AS a
   WHERE   a.Salary < (SELECT  MAX(z.Salary)
                       FROM    Employee AS z)

2. DISTINCT > 不重複

   # Write your MySQL query statement below
   SELECT
   (
       SELECT DISTINCT Salary
       FROM            Employee
       ORDER BY        Salary DESC
       LIMIT           1, 1
   ) AS SecondHighestSalary

3. GROUP BY > 不重複

   # Write your MySQL query statement below
   SELECT
   (
       SELECT          Salary
       FROM            Employee
       GROUP BY        Salary
       ORDER BY        Salary DESC
       LIMIT           1, 1
   ) AS SecondHighestSalary

    

心得：

題目要求找出有哪些人的薪水比主管高，找出名字並修改欄位名稱為Employee。

題目：

The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.

+----+-------+--------+-----------+
| Id | Name  | Salary | ManagerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | NULL      |
| 4  | Max   | 90000  | NULL      |
+----+-------+--------+-----------+

Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.

+----------+
| Employee |
+----------+
| Joe      |
+----------+

答案：

1. Sub Select

   # Write your MySQL query statement below
   SELECT  a.Name AS Employee
   FROM    Employee AS a
   WHERE   a.Salary > (SELECT  z.Salary 
                       FROM    Employee AS z
                       WHERE   z.Id = a.ManagerId);

2. Join

   # Write your MySQL query statement below
   SELECT  a.Name AS Employee
   FROM    Employee AS a
   JOIN    Employee AS b
   ON      a.ManagerId = b.Id
   WHERE   a.Salary > b.Salary

3. From

   # Write your MySQL query statement below
   SELECT  a.Name AS Employee
   FROM    Employee AS a, Employee AS b
   WHERE   a.ManagerId = b.Id
   AND     a.Salary > b.Salary

    

心得：

題目要求找出Email重複的資料並Show出來，這裡要注意的是沒辦法使用WHERE而必須使用HAVING，因為WHERE在GROUP BY前面，HAVING在GROUP BY後面。

問題：

Write a SQL query to find all duplicate emails in a table named Person.

答案：

SELECT Email
FROM Person
GROUP BY Email
HAVING COUNT(Email) > 1

參考：

1. SQL語法中WHERE與HAVING有何差異?