牛客网—数据库SQL实战（2）

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原创最新推荐文章于 2026-05-13 09:15:22 发布 289 阅读 AI 写同款· GEO 优化 ›

牛客网SQL题目解析（答案+解析+理解）本文记录了牛客网sql全部题目的答案与难题解析，部分题目包含多种解法，并且涵盖了开窗函数等各种语法点的理解标题中高亮的题目，是易错题 牛客网刷题链接：牛客网sql在线练习参考教程资源： mysql教程1 mysql教程2 1.where+子查询 select * from employees order by hire_date desc limit 0,1; select * from employees where hire_date = . 阅读详情

11. 获取所有员工当前的manager

获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date='9999-01-01'。
结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_no。
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));

输入描述:

无

输出描述:

emp_no	manager_no
10001	10002
10003	10004
10009	10010

select d1.emp_no, d2.emp_no as manager_no
from dept_emp as d1, dept_manager as d2
where d1.dept_no = d2.dept_no
    and d1.emp_no <> d2.emp_no
    and d1.to_date = '9999-01-01'
    and d2.to_date = '9999-01-01'

12. 获取所有部门中当前员工薪水最高的相关信息

题目描述

获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

输入描述:

无

输出描述:

dept_no	emp_no	salary
d001	10001	88958
d002	10006	43311
d003	10005	94692
d004	10004	74057
d005	10007	88070
d006	10009	95409

select d.dept_no, d.emp_no, max(s.salary) as salary
from dept_emp as d, salaries as s
where d.emp_no = s.emp_no
    and d.to_date = '9999-01-01'
    and s.to_date = '9999-01-01'
group by d.dept_no

13. 从titles表获取按照title进行分组

题目描述

从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);

输入描述:

无

输出描述:

title	t
Assistant Engineer	2
Engineer	4
省略	省略
Staff	3

select title, count(title) as t
from titles
group by title
having t >= 2

14. 从titles表获取按照title进行分组，注意对于重复的emp_no进行忽略

题目描述

从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。
注意对于重复的emp_no进行忽略。
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);

输入描述:

无

输出描述:

title	t
Assistant Engineer	2
Engineer	3
省略	省略
Staff	3

select title, count(distinct emp_no) as t
from titles
group by title
having t >= 2

15. 查找employees表

题目描述

查找employees表所有emp_no为奇数，且last_name不为Mary的员工信息，并按照hire_date逆序排列
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

输入描述:

无

输出描述:

emp_no	birth_date	first_name	last_name	gender	hire_date
10011	1953-11-07	Mary	Sluis	F	1990-01-22
10005	1955-01-21	Kyoichi	Maliniak	M	1989-09-12
10007	1957-05-23	Tzvetan	Zielinski	F	1989-02-10
10003	1959-12-03	Parto	Bamford	M	1986-08-28
10001	1953-09-02	Georgi	Facello	M	1986-06-26
10009	1952-04-19	Sumant	Peac	F	1985-02-18

select emp_no, birth_date, first_name, last_name, gender, hire_date
from employees
where emp_no % 2 = 1
    and last_name <> 'Mary'
order by hire_date desc

16. 统计出当前各个title类型对应的员工当前薪水对应的平均工资

题目描述

统计出当前各个title类型对应的员工当前薪水对应的平均工资。结果给出title以及平均工资avg。
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);

输入描述:

无

输出描述:

title	avg
Engineer	94409.0
Senior Engineer	69009.2
Senior Staff	91381.0
Staff	72527.0

select t.title, avg(s.salary) as avg
from titles as t, salaries as s
where t.emp_no = s.emp_no
    and t.to_date = '9999-01-01'
    and s.to_date = '9999-01-01'
group by t.title

17. 获取当前薪水第二多的员工的emp_no以及其对应的薪水salary

题目描述

获取当前（to_date='9999-01-01'）薪水第二多的员工的emp_no以及其对应的薪水salary
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

输入描述:

无

输出描述:

emp_no	salary
10009	94409

select emp_no, salary
from salaries
order by salary desc limit 1,1

18. 查找当前薪水排名第二多的员工编号emp_no以及其对应的薪水salary，不准使用order_by

题目描述

查找当前薪水(to_date='9999-01-01')排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

输入描述:

无

输出描述:

emp_no	salary	last_name	first_name
10009	94409	Peac	Sumant

select e.emp_no, s.salary, e.last_name, e.first_name
from employees as e, salaries as s
where e.emp_no = s.emp_no
    and s.to_date = '9999-01-01'
    and s.salary = (select max(salary) 
              from salaries
              where to_date = "9999-01-01"
                  and salary != (select max(salary) 
                             from salaries
                             where to_date = "9999-01-01"))

19. 查找所有员工的last_name和first_name以及对应的dept_name

题目描述

查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

输入描述:

无

输出描述:

last_name	first_name	dept_name
Facello	Georgi	Marketing
省略	省略	省略
Sluis	Mary	NULL

select e.last_name, e.first_name, d2.dept_name
from employees as e left join dept_emp as d1
    on e.emp_no = d1.emp_no left join departments as d2
        on d1.dept_no = d2.dept_no

20. 查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth

题目描述

查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

输入描述:

无

输出描述:

growth
28841

select ( 
(select salary from salaries where emp_no = 10001 order by to_date desc limit 1) -
(select salary from salaries where emp_no = 10001 order by to_date asc limit 1)
) as growth

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