如何快速掌握MYSQL？附牛客网精选的50道SQL题目详解【新手推荐】

如何才能快速掌握MYSQL？如何熟悉使用SQL以满足日常工作需求？

如果你目前啥也不会，只停留在知道SELECT用于查询的层面的话，又想要快速掌握MYSQL，那么刷题，并且过程中不会什么补什么，就是巩固和提升自己的SQL语言能力最快捷的方法。

在之前的一篇博客中，我给出了LeetCode上出现频率最高的50道数据库题目详解，那么这一次，我也SQ给出了牛客网上的50道SQL题，希望对大家有所收获。

补充：牛客网相对于LeetCode而言题目较少，且难度较低一些，但胜在免费，并且有些公司笔试的时候会直接饮用牛客网上的原题，所以也非常值得我们去刷一刷题。

SQL1 入门 查找最晚入职员工的所有信息

原题链接

SELECT * FROM employees
WHERE hire_date = (SELECT max(hire_date) FROM employees)

SQL2 简单 查找入职员工时间排名倒数第三的员工所有信息

原题链接

/*
with tmp as (
    SELECT *, rank() over(order by hire_date desc) rk 
    FROM employees
)
SELECT emp_no, birth_date, first_name, last_name, gender, hire_date
FROM tmp
WHERE rk=3
*/
SELECT * 
FROM employees
order by hire_date DESC
LIMIT 2,1

SQL3 中等 查找当前薪水详情以及部门编号dept_no

原题链接

SELECT s.emp_no,s.salary, s.from_date, s.to_date, d.dept_no
FROM dept_manager d 
LEFT JOIN salaries s
ON d.emp_no = s.emp_no
ORDER BY s.emp_no

SQL4 简单 查找所有已经分配部门的员工的last_name和first_name以及dept_no

原题链接

SELECT e.last_name, e.first_name, d.dept_no
FROM dept_emp d
join employees e
on d.emp_no = e.emp_no

SQL5 中等 查找所有员工的last_name和first_name以及对应部门编号dept_no

原题链接

/*较上一题多了两行null，所以要将left join 改成right join*/
SELECT e.last_name, e.first_name, d.dept_no
FROM dept_emp d
RIGHT JOIN employees e
ON d.emp_no = e.emp_no

SQL7 简单 查找薪水记录超过15次的员工号emp_no以及其对应的记录次数t

原题链接

SELECT emp_no, COUNT(emp_no) t
FROM salaries
GROUP BY emp_no
HAVING COUNT(emp_no) > 15

SQL8 简单 找出所有员工当前薪水salary情况

原题链接

SELECT distinct salary 
FROM salaries
order by salary DESC

SQL10 简单 获取所有非manager的员工emp_no

原题链接

SELECT emp_no 
FROM employees
WHERE emp_no not in (select emp_no from dept_manager)

SQL11 中等 获取所有员工当前的manager

原题链接

select d1.emp_no, d2.emp_no manager
FROM dept_emp d1
join dept_manager d2
on d1.dept_no=d2.dept_no and d1.emp_no <> d2.emp_no

SQL12 困难 获取每个部门中当前员工薪水最高的相关信息

原题链接

with tmp as (
    SELECT d.dept_no, d.emp_no, s.salary
    FROM dept_emp d
    JOIN salaries s
    on d.emp_no = s.emp_no
), tmp2 as (
    select *, rank() over(PARTITION by dept_no order by salary desc, emp_no desc) rk
    FROM tmp
)
SELECT dept_no, emp_no, salary maxSalary
FROM tmp2
WHERE rk=1

SQL15 简单 查找employees表emp_no与last_name的员工信息

原题链接

SELECT * FROM employees
where emp_no % 2 = 1 and last_name <> 'Mary'
order by hire_date desc

SQL16 中等 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

原题链接

select t.title title, avg(s.salary)
FROM titles t
join salaries s
on t.emp_no = s.emp_no
group by t.title
ORDER by avg(s.salary)

SQL17 简单 获取当前薪水第二多的员工的emp_no以及其对应的薪水salary

原题链接

select emp_no, salary
FROM salaries
where salary = (select DISTINCT salary FROM salaries order by salary desc LIMIT 1,1)

/*
with tmp as (
    SELECT emp_no, salary, rank() over(order by salary desc, emp_no desc) rk
    from salaries
)
select emp_no, salary
from tmp
where rk = 2
*/

SQL18 较难 获取当前薪水第二多的员工的emp_no以及其对应的薪水salary

原题链接

with tmp as (
    SELECT s.emp_no, s.salary, e.last_name, e.first_name
    FROM employees e
    join salaries s
    on e.emp_no = s.emp_no
)
select *
from tmp
where salary = (SELECT max(salary) FROM tmp where salary <> (select max(salary) FROM salaries))

SQL19 中等 查找所有员工的last_name和first_name以及对应的dept_name

原题链接

select e.last_name, e.first_name, d.dept_name
from employees e
left join dept_emp on e.emp_no = dept_emp.emp_no
left join departments d on dept_emp.dept_no = d.dept_nojavascript:void(0);

SQL21 困难 查找在职员工自入职以来的薪水涨幅情况

原题链接

with s1 as (
    select *
    FROM salaries
    group by emp_no
    having from_date = min(from_date)
),
s2 as (
    select *
    FROM salaries
    where to_date = '9999-01-01'
)
select distinct s1.emp_no, s2.salary - s1.salary growth
FROM s1, s2
where s1.emp_no=s2.emp_no and s2.to_date = '9999-01-01'
order by growth

SQL22 中等 统计各个部门的工资记录数

原题链接

WITH tmp as (
    select dept_emp.dept_no, departments.dept_name, salaries.emp_no
    from salaries
    left join dept_emp
    on salaries.emp_no = dept_emp.emp_no
    left join departments
    on dept_emp.dept_no = departments.dept_no
)
select dept_no, dept_name, count(emp_no) sum
FROM tmp
group by dept_no
order by dept_no

SQL23 较难 对所有员工的薪水按照salary降序进行1-N的排名

原题链接

select emp_no, salary, DENSE_RANK() over(order by salary desc) t_rank
FROM salaries

SQL24 较难 获取所有非manager员工当前的薪水情况

原题链接

SELECT d1.dept_no, d1.emp_no, s.salary
FROM dept_emp d1
join salaries s
on d1.emp_no = s.emp_no
where d1.emp_no not in (select emp_no from dept_manager)

SQL25 困难 获取员工其当前的薪水比其manager当前薪水还高的相关信息

原题链接

SELECT d1.emp_no, d2.emp_no manager_no, s1.salary emp_salary, s2.salary manager_salary
FROM dept_emp d1
join dept_manager d2
on d1.dept_no = d2.dept_no and d1.emp_no <> d2.emp_no
join salaries s1
on s1.emp_no = d1.emp_no
join salaries s2
on s2.emp_no = d2.emp_no
where s1.salary > s2.salary

SQL26 困难 汇总各个部门当前员工的title类型的分配数目

原题链接

SELECT d1.dept_no, d2.dept_name, t.title, count(t.emp_no) COUNT
from dept_emp d1
join departments d2
on d1.dept_no = d2.dept_no
join titles t
on d1.emp_no = t.emp_no
group by d1.dept_no, t.title
order by dept_no, title

SQL29 中等 使用join查询方式找出没有分类的电影id以及名称

原题链接

with tmp as (
    select f.film_id, f.title
    from film f
    join film_category fc
    on f.film_id = fc.film_id
    join category c
    on fc.category_id = c.category_id
)
select film_id, title
from film
where film_id not in (select film_id from tmp)

SQL30 中等 使用子查询的方式找出属于Action分类的所有电影对应的title,description

原题链接

select f.title, f.description
from film f
join film_category fc
on f.film_id = fc.film_id
join category c
on fc.category_id = c.category_id
where c.name = 'Action'

SQL32 简单 将employees表的所有员工的last_name和first_name拼接起来作为Name

原题链接

/*主要就是考察concat函数 */
select CONCAT(last_name, ' ', first_name) NAME
FROM employees

SQL33 中等 创建一个actor表，包含如下列信息

原题链接

CREATE TABLE actor(
actor_id smallint(5) primary key,
first_name varchar(45) not null,
last_name varchar(45) not null,
last_update date not null);

SQL38 中等 针对actor表创建视图actor_name_view

原题链接

CREATE VIEW actor_name_view AS
SELECT first_name first_name_v ,last_name last_name_v
FROM  actor;

SQL50 中等 将employees表中的所有员工的last_name和first_name通过引号连接起来。

原题链接

select CONCAT(last_name, '\'', first_name) name
from employees

SQL51 中等 查找字符串 10,A,B 中逗号,出现的次数cnt

原题链接

select 2

/*
select (length('10,A,B') - length(replace('10,A,B',',',''))) AS cnt
*/

SQL52 中等 获取Employees中的first_name

原题链接

select first_name
FROM employees
ORDER BY RIGHT(first_name, 2)

SQL53 中等 按照dept_no进行汇总

原题链接

/*
本次重点是GROUP_CONCAT函数
语法：group_concat( [distinct] 要连接的字段 [order by 排序字段 asc/desc ] [separator '分隔符'] )
*/
select dept_no, GROUP_CONCAT(emp_no) employees
FROM dept_emp
group by dept_no

SQL54 中等 平均工资

原题链接

select avg(salary) avg_salary
from salaries
where to_date='9999-01-01' 
    and (to_date, salary) not in (select to_date, max(salary) from salaries where to_date='9999-01-01')
    and (to_date, salary) not in (select to_date, min(salary) from salaries where to_date='9999-01-01')

SQL55 中等 分页查询employees表，每5行一页，返回第2页的数据

原题链接

select * 
from employees
limit 5,5

SQL57 中等 使用含有关键字exists查找未分配具体部门的员工的所有信息。

原题链接

/*
EXISTS语句：执行employees.length次
指定一个子查询，检测行的存在。遍历循环外表，然后看外表中的记录有没有和内表的数据一样的。匹配上就将结果放入结果集中。
IN 语句：只执行一次
确定给定的值是否与子查询或列表中的值相匹配。
in在查询的时候，首先查询子查询的表，然后将内表和外表做一个笛卡尔积，然后按照条件进行筛选。
所以相对内表比较小的时候，in的速度较快。
*/
select *
from employees
where not EXISTS(
    select emp_no
    from dept_emp
    where dept_emp.emp_no = employees.emp_no
)

SQL59 较难 获取有奖金的员工相关信息。

原题链接

with s as (
    select * 
    FROM salaries
    where to_date = '9999-01-01'
)
SELECT e1.emp_no, e2.first_name, e2.last_name, e1.btype, s.salary, 
    (case
         when  e1.btype=1 then s.salary * 0.1
         when  e1.btype=2 then s.salary * 0.2
         else s.salary * 0.3
     end
    ) bonus
FROM emp_bonus e1
JOIN employees e2
on e1.emp_no = e2.emp_no
join s
on e1.emp_no = s.emp_no

SQL60 较难 统计salary的累计和running_total

原题链接

with emp as (
    select * 
    FROM salaries
    where to_date = '9999-01-01'
), tmp as (
    select e1.emp_no, e1.salary, e2.salary s2
    FROM emp e1
    join emp e2
    on e1.emp_no >= e2.emp_no
)
select emp_no, salary, sum(s2) running_total
from tmp
group by emp_no

SQL61 较难 对于employees表中，给出奇数行的first_name

原题链接

with tmp as (
    select first_name, ROW_NUMBER() over(order by first_name) rk
    from employees
)
select e.first_name

FROM employees e
join tmp 
on e.first_name = tmp.first_name
where rk % 2 = 1
/*
注意题目说“输出的结果不需排序”，但是我们用了ROW_NUMBER()，所以已经排了序，需要重新连接原来的表，保证顺序不变
*/

SQL65 较难 异常的邮件概率

原题链接

select date, round(sum(if(type='completed', 0, 1)) / count(type), 3) p
from email
where send_id in (select id from user where is_blacklist=0) and receive_id in (select id from user where is_blacklist=0)
group by date
order by date

SQL66 简单 牛客每个人最近的登录日期(一)

原题链接

select user_id, max(date) d
from login
group by user_id
order by user_id

SQL67 较难 牛客每个人最近的登录日期(二)

原题链接

select u.name u_n, c.name c_n, l.date
from login l
join user u
on l.user_id = u.id
join client c
on l.client_id = c.id
where (l.user_id, l.date) in (select user_id, max(date) FROM login group by user_id)
ORDER by u.name

SQL68 较难 牛客每个人最近的登录日期(三)

原题链接

with tmp as (
    select user_id, min(date) date
    from login
    group by user_id
)
select round((select count(user_id) from tmp where (user_id, date+1) in (select user_id, date from login)) / (select count(user_id) from tmp), 3) 

SQL69 较难 牛客每个人最近的登录日期(四)

原题链接

select date, sum(if((user_id, date) in (select user_id, min(date) from login group by user_id), 1, 0)) new
from login
group by date
order by date

SQL70 困难 牛客每个人最近的登录日期(五)

原题链接

select date, ifnull(round(sum(if((user_id, date+1) in (select user_id, date from login group by user_id), 1, 0))  / sum(if((user_id, date) in (select user_id, min(date) from login group by user_id), 1, 0)), 3), 0) p
from login
group by date
order by date

SQL71 较难 牛客每个人最近的登录日期(六)

原题链接

with tmp as (
    select p1.user_id, p1.date, p2.number
    from passing_number p1, passing_number p2
    where p1.user_id = p2.user_id and p1.date >= p2.date
), tmp2 as (
    select user_id, date, sum(number) ps_number
    from tmp
    group by user_id, date
)
select u.name u_n, date, ps_number
from tmp2 t
join user u
on t.user_id = u.id
order by date, name 

SQL87 中等 最差是第几名(一)

原题链接

with tmp as (
    select c1.grade, c2.number
    from class_grade c1
    join class_grade c2
    on c1.grade >= c2.grade
)
select grade, sum(number) t_rank
FROM tmp
group by grade
order by grade

SQL88 较难 最差是第几名(二)

原题链接

select grade
from
    (select grade,
    (select sum(number) from class_grade) as total,
    sum(number)over(order by grade) a, -- 求正序
    sum(number)over(order by grade desc) b  -- 求逆序
     from class_grade
    order by grade)t
where a >= total/2 and b >= total/2  -- 正序逆序均大于整个数列数字个数的一半
order by grade;

SQL89 中等 获得积分最多的人(一)

原题链接

with tmp as (
    select user_id, sum(grade_num) grade_sum
    FROM grade_info
    group by user_id
)
select u.name, t.grade_sum
from user u
join tmp t
on u.id = t.user_id
order by grade_sum DESC
limit 1

SQL90 较难 获得积分最多的人(二)

原题链接

with tmp as (
    select user_id, sum(grade_num) grade_sum
    FROM grade_info
    group by user_id
)
select id, name, grade_sum
FROM
(
    select u.id, u.name, t.grade_sum, rank() over(order by grade_sum desc) rk
    from user u
    join tmp t
    on u.id = t.user_id
) t
where rk = 1

SQL91 困难 获得积分最多的人(三)

原题链接

with tmp as (
    select user_id, sum(if(type='add', grade_num, 0)) grade_add, sum(if(type='reduce', grade_num, 0)) grade_reduce
    FROM grade_info
    group by user_id
)
select id, name, grade_sum
FROM
(
    select u.id, u.name, (t.grade_add - t.grade_reduce) grade_sum, rank() over(order by (t.grade_add - t.grade_reduce) desc) rk
    from user u
    join tmp t
    on u.id = t.user_id
) t
where rk = 1

SQL92 中等 商品交易(网易校招笔试真题)

原题链接

SELECT g.id, g.name, g.weight, sum(t.count) total
FROM goods g
JOIN trans t
on g.id = t.goods_id
where g.weight < 50
GROUP by t.goods_id
HAVING total > 20
order by g.id

SQL93 较难 网易云音乐推荐(网易校招笔试真题)

原题链接

with tmp as (
    select f.user_id, music_name, music_id
    from follow f
    join music_likes m
    on f.follower_id = m.user_id
    join music m2
    on m2.id = m.music_id
)
select distinct music_name
FROM tmp
where user_id = 1 and music_id not in (select music_id from music_likes where user_id = 1)
order by music_id

结束语

感谢收看，祝学业和工作进步！

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