辣鸡刘的Leetcode之旅3(SQL Part)【两表合并查询，第二高薪，挣钱比经理多, 重复值查询】

1. 两表合并查询（Combine Two Tables）

题目描述：

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

这个还是挺简单的，直接上：

select Person.FirstName,Person.LastName,Address.City,Address.State from Person left join Address on Person.PersonId= Address.PersonId;

描述：
SQL中常见的JOIN

left join(左联接) 返回包括左表中的所有记录和右表中联结字段相等的记录 
right join(右联接) 返回包括右表中的所有记录和左表中联结字段相等的记录
inner join(等值连接) 只返回两个表中联结字段相等的行

上图：

2. 第二高薪（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 query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

找出第二高的，也就是在除去最高的剩余里面找出最高的；

select max(Salary) as SecondHighestSalary from Employee where  Salary not in (select max(Salary) from Employee);

3. 挣钱比经理多（Employees Earning More Than Their Managers）

题目描述：

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      |
+----------+

对于本题，Joe的经理是编号为3的Sam，Henry的经理为编号为4的Max，按照题目要求，Joe的工资是高于他的经理Sam的（7K>6K）,所以本例的目的是返回Joe：
类似于两表查询：

select name as Employee from Employee e1 where Salary > (select Salary from Employee e2 where e1.ManagerId=e2.Id) 

或者：

select name as Employee from Employee e1 
where exists (select * from Employee e2 where e2.Id=e1.ManagerId and e1.Salary>e2.Salary )

建议写exists语句时，子查询中直接用*，而不用对列进行任何函数操作，避免碰到官方bug，

4. 重复值查询

题目描述：

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

+----+---------+
| Id | Email   |
+----+---------+
| 1  | [e-mail] |
| 2  | [e-mail] |
| 3  | [e-mail] |
+----+---------+
For example, your query should return the following for the above table:

+---------+
| Email   |
+---------+
| [e-mail] |
+---------+
Note: All emails are in lowercase.

这个比较简单：

Select Email from Person group by Email having count(Email)>1;

这里注意三点：

1. group by:常规用法是配合聚合函数，利用分组信息进行统计，常见的是配合max等聚合函数筛选数据后分析，以及配合having进行筛选后过滤。
2. having:having字句可以让我们筛选成组后的各种数据，where字句在聚合前先筛选记录，也就是说作用在group by和having字句前。
3. count():它返回检索行的数目， 不论其是否包含 NULL值。