Database Leetcode习题集解析

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本文提供多个SQL实战案例，包括查找重复邮箱、结合两张表获取完整信息、找出薪水高于上级的员工等，通过具体例子帮助读者掌握SQL技巧。

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<1>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 |
+---------+

Solution:

# 找到person表中重复的Email
# 方案一：
# SELECT Email FROM Person GROUP BY Email WHERE count(*)>1 #由于合计函数 GROUP BY不能和 WHERE 同时使用所以引入HAVING。
SELECT Email FROM Person GROUP BY Email HAVING count(*)>1

# 方案二
#使用JOIN ON
SELECT DISTINCT a.Email FROM Person aINNER JOIN Person b ON (a.Email = b.Email) WHERE a.Id != b.Id

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

<2>Combine Two Tables

Table: 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

Solution:

# Write your MySQL query statement below
SELECT Person.FirstName,Person.LastName,Address.City,Address.StateFROM PersonLEFT JOIN Address ON Person.personId = Address.PersonId;

<3>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      |
+----------+

Solution:

SELECT e1.NAME as Employee FROM Employee e1 ,Employee e2 where e1.ManagerId = e2.Idand e1.Salary>e2.Salary

<4>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.

Solution:

SELECT IFNULL( (SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1,1) ,NULL) AS SecondHighestSalary

<5>Nth Highest Salary

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

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

For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.

Solution:

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
DECLARE M INT;
SET M=N-1;
RETURN (
# Write your MySQL query statement below.
SELECT IFNULL((SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT M,1),NULL) AS NHighestSalary
);
END