作者:嘉娜杰_877 | 来源:互联网 | 2023-10-12 15:31
1. 第二高的薪水
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SQL架构
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 |
+---------------------+
答案:select (select distinct salary from Employee order by salary desc limit 1,1) as SecondHighestSalary
2. 超过经理收入的员工
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SQL架构
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 |
+----------+
答案:select Name as 'Employee' from Employee e where Salary > ( select Salary from Employee where id = e.ManagerId )
3. 查找重复的电子邮箱
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SQL架构
编写一个 SQL 查询,查找 Person
表中所有重复的电子邮箱。
示例:
+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+
根据以上输入,你的查询应返回以下结果:
+---------+
| Email |
+---------+
| a@b.com |
+---------+
说明:所有电子邮箱都是小写字母。
答案:select email from person group by email having count(email)>1
4. 组合两个表
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SQL架构
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:
FirstName, LastName, City, State
答案:select FirstName, LastName, City, State from Person left join Address on Person.PersOnId= Address.PersonId
5. 两个表
班级表class (cid,sid)
分数表source(sid,source)
1,查询出每个班级的学生人数:
select count(sid) from class group by cid;
2,查询出1001班级下,成绩大于80的人数
select count(s.sid) from class c,source s where c.sid=s.sid and c.cid = 1001 and s.source>80;
6. 部门工资最高的员工
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SQL架构
Employee
表包含所有员工信息,每个员工有其对应的 Id, salary 和 department Id。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Jim | 90000 | 1 |
| 3 | Henry | 80000 | 2 |
| 4 | Sam | 60000 | 2 |
| 5 | Max | 90000 | 1 |
+----+-------+--------+--------------+
Department
表包含公司所有部门的信息。
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
编写一个 SQL 查询,找出每个部门工资最高的员工。对于上述表,您的 SQL 查询应返回以下行(行的顺序无关紧要)。
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Jim | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
解释:
Max 和 Jim 在 IT 部门的工资都是最高的,Henry 在销售部的工资最高。
答案:
方法:使用 JOIN 和 IN 语句
算法
因为 Employee 表包含 Salary 和 DepartmentId 字段,我们可以以此在部门内查询最高工资。
SELECT
DepartmentId, MAX(Salary)
FROM
Employee
GROUP BY DepartmentId;
注意:有可能有多个员工同时拥有最高工资,所以最好在这个查询中不包含雇员名字的信息。
| DepartmentId | MAX(Salary) |
|--------------|-------------|
| 1 | 90000 |
| 2 | 80000 |
然后,我们可以把表 Employee 和 Department 连接,再在这张临时表里用 IN 语句查询部门名字和工资的关系。
MySQL
SELECT
Department.name AS 'Department',
Employee.name AS 'Employee',
Salary
FROM
Employee
JOIN
Department ON Employee.DepartmentId = Department.Id
WHERE
(Employee.DepartmentId , Salary) IN
( SELECT
DepartmentId, MAX(Salary)
FROM
Employee
GROUP BY DepartmentId
)
;