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leetcode

1、组合两个表

表1：Person

+-------------+---------+
| 列名         | 类型     |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
PersonId 是上表主键

表2：Address

+-------------+---------+
| 列名         | 类型    |
+-------------+---------+
| AddressId   | int     |
| PersonId    | int     |
| City        | varchar |
| State       | varchar |
+-------------+---------+
AddressId 是上表主键

编写一个 SQL 查询，满足条件：无论 person 是否有地址信息，都需要基于上述两表提供 person 的以下信息：

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

2、第二个高薪水

编写一个 SQL 查询，获取 Employee 表中第二高的薪水（Salary） 。

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

例如上述 Employee 表，SQL查询应该返回 200 作为第二高的薪水。如果不存在第二高的薪水，那么查询应返回 null。

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+
select (select distinct(Salary) 
from Employee
order by Salary desc
limit 1,1) as SecondHighestSalary

3、超过经理收入的员工

Employee 表包含所有员工，他们的经理也属于员工。每个员工都有一个 Id，此外还有一列对应员工的经理的 Id。

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

给定 Employee 表，编写一个 SQL 查询，该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中，Joe 是唯一一个收入超过他的经理的员工。

+----------+
| Employee |
+----------+
| Joe      |
+----------+
select Name Employee
from Employee a
where Salary > (select Salary from Employee where Id = a.ManagerId);

4、查找重复的电子邮箱

编写一个 SQL 查询，查找 Person 表中所有重复的电子邮箱。

+----+---------+
| Id | Email   |
+----+---------+
| 1  | [email protected] |
| 2  | [email protected] |
| 3  | [email protected] |
+----+---------+

根据以上输入，你的查询应返回以下结果：

+---------+
| Email   |
+---------+
| [email protected] |
+---------+
select email from Person group by email having count(email)>1;
select distinct(p1.Email) from Person p1 
join Person p2 on p1.email = p2.email and p1.Id != p2.Id;