leetcode Database2 (四)

一、Duplicate Emails

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

+----+---------+

| Id | Email   |

+----+---------+

| 1  | a@b.com |

| 2  | c@d.com |

| 3  | a@b.com |

+----+---------+

For example, your query should return the following for the above table:

+---------+

| Email   |

+---------+

| a@b.com |

+---------+

Note: All emails are in lowercase.

分析：编写一个SQL查询从Person表中找出所有重复的邮箱地址。

解法一：（self join）

# Write your MySQL query statement below

select distinct a.Email from Person a, Person b where a.Email=b.Email and a.Id<>b.Id

一开始，写的时候没注意把distinct给漏了，导致出错：

Submission Result: Wrong AnswerMore Details

Input:{"headers": {"Person": ["Id", "Email"]}, "rows": {"Person": [[1, "paris@hilton.com"], [2, "paris@hilton.com"]]}}

Output:{"headers": ["Email"], "values": [["paris@hilton.com"], ["paris@hilton.com"]]}

Expected:{"headers": ["Email"], "values": [["paris@hilton.com"]]}　

解法二：

# Write your MySQL query statement below

Select Email

From Person

GROUP BY Email

Having count(Email)>1

二、Employees Earning More Than Their Managers

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      |

+----------+
分析：题意为雇员表记录了所有雇员的信息，包括他们的经理在内。每一个雇员都有一个Id，和他的经理的Id。
给定雇员表，编写一个SQL查询找出薪水大于经理的员工姓名。对于上表来说，Joe是唯一收入大于经理的员工。

使用自连接：

# Write your MySQL query statement below

select m.Name from Employee m,Employee n where m.ManagerId=n.Id and m.Salary>n.Salary;

　或：

Select emp.Name from

Employee emp inner join Employee manager

on emp.ManagerId = manager.Id

where emp.Salary > manager.Salary

　或：　

select a.name from Employee a left join Employee b on a.managerid=b.id where a.salary>b.salary

三、Second Highest Salary

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.

分析：题意为从员工表中找到工资第二高的数据（即比最高工资少的里面工资最高的）

代码如下：

# Write your MySQL query statement below

select max(Salary) from Employee

where Salary < (select max(Salary) from Employee);

其他解法：

# Write your MySQL query statement below

SELECT Salary FROM Employee GROUP BY Salary

UNION ALL (SELECT null AS Salary)

ORDER BY Salary DESC LIMIT 1 OFFSET 1

　或：

select (select distinct Salary from Employee order by salary desc limit 1,1) as Salary;

注：LIMIT 接受一个或两个数字参数。参数必须是一个整数常量。如果给定两个参数，第一个参数指定第一个返回记录行的偏移量，第二个参数指定返回记录行的最大数目。初始记录行的偏移量是 0(而不是 1)；offset偏移量　

四、Combine Two Tables

able: Person

+-------------+---------+

| Column Name | Type    |

+-------------+---------+

| PersonId    | int     |

| FirstName   | varchar |

| LastName    | varchar |

+-------------+---------+

PersonId is the primary key column for this table.

Table: Address

+-------------+---------+

| Column Name | Type    |

+-------------+---------+

| AddressId   | int     |

| PersonId    | int     |

| City        | varchar |

| State       | varchar |

+-------------+---------+

AddressId is the primary key column for this table.

Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:

FirstName, LastName, City, State

分析：题意为
有两个数据表：Person表和Address表。Person（人员）表主键为PersonId，Address（地址）表主键是AddressId，通过PersonId与Person表关联。编写一个SQL查询，对于Person表中的每一个人，取出FirstName, LastName, City, State属性，无论其地址信息是否存在。
思路：Person表是主表，Address表是从表，通过Left Outer Join左外连接即可。

# Write your MySQL query statement below

select p.FirstName,p.LastName,a.City,a.State

from Person p left outer join Address a using (PersonId);  

（using()用于两张表的join查询，要求using()指定的列在两个表中均存在，并使用之用于join的条件。）

　　等价于

# Write your MySQL query statement below

select p.FirstName,p.LastName,a.City,a.State

from Person p left outer join Address a on p.PersonId=a.PersonId

　　其他解法：

SELECT FirstName, LastName, City, State FROM Person NATURAL LEFT JOIN Address;

巴特西

leetcode Database2 (四)

一、Duplicate Emails

最新文章

热门文章