LeetCode数据库50题总结（更新中……）

前言，主要是为了在实战中记语法，各位如果想刷这些题还是刷过再来看。

题目185.每个部门工资最高的三个人

表结构：

Id	Name	Salary	DepartmentId
1	Joe	70000	1
2	Henry	80000	2
3	Sam	60000	2
4	Max	90000	1
5	Janet	69000	1
6	Randy	85000	1

每个员工有id和部门id

from Department dep, Employee emp 
where emp.DepartmentId=dep.Id and 
    3 > (Select count(distinct Salary)
            From Employee 
            where DepartmentId = dep.Id and Salary>emp.Salary
        )

题目175. 将员工信息表和员工工资表并起来，其中有公共行PersonId，工资表没有记录的用null代替。

表结构：
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.

解答：

SELECT Person.FirstName, Person.LastName, Address.City, Address.State FROM Person left join Address on Person.PersonId=Address.PersonId

left join 类似于right join
join 等效于 inner join 只查两边都有的
full join 则是任一方有则加入结果

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(摘自评论)The SQL syntax for how to use Where vs. On

1. SELECT
   FROM
   WHERE
   SELECT
   FROM
   JOIN
   ON

   题目176: 查找薪水第二高的员工：

   表结构：

   Id	Salary
   1	100
   2	200
   3	300

   select Salary as SecondHighestSalary from Employee order by Salary DESC limit 1 offset 1
   

   先排序，再用select的关键字对结果进行处理，或者 嵌套select

   select MAX(Salary) as SecondHighestSalary from Employee 
   where Salary <(select MAX(Salary) from Employee)
   

   题目181. 工资比经理还高的员工每个员工都有一个经理或为空，找出所有工资比其经理高的员工

   表结构：

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

   解法：

   select emp.Name as Employee from Employee as emp, Employee as man where emp.ManagerId = man.Id and emp.Salary > man.Salary
   

   分别查两个Employee表从中去所属经理id与本体id相同的进行比较。
   题目182. 查找所有重复的邮件
   表格如下：

   Id	Email
   1	[email protected]
   2	[email protected]
   3	[email protected]

   count(column_name)返回指定列名中不同种值的种数数目
   count(*)返回列表的记录数
   解法1：distinct限制结果无重复，嵌套查找一个email在本表中匹配得到第二个，就是具有重复记录的email
   /*runtime 50.52%/

   select distinct Email from Person as p1
   where 
   ( 
   select count(*) from Person p2
   where 
   p1.Email = p2.Email
   )>1;
   

   解法2(优): having直接使用聚合函数，group by也是针对聚合函数就某一列或多列进行分组

   Select Email
   from person
   Group by Email
   Having count(*) > 1;