【MySql】牛客SQL刷题（上）

牛客SQL题目

题目链接：https://www.nowcoder.com/ta/sql

  1. 查找最晚入职员工的所有信息

    select * 
from 
employees
where 
hire_date = (select max(hire_date) from employees);

  2. 查找入职员工时间排名倒数第3的员工所有信息

    select * 
from 
employees 
order by 
hire_date 
desc --降序
limit 2,1;--从3项，偏移1项，即第3项

  3. 查找各个部门当前(to_date='9999-01-01')的领导的当前薪水详情以及其对应部门编号dept_no

    select s.*,d.dept_no
from salaries as s
join dept_manager as d
on d.emp_no=s.emp_no
where s.to_date='9999-01-01'
and d.to_date='9999-01-01';

  4. 查找所有已经分配部门的员工的last_name和first_name

    select e.last_name,e.first_name,d.dept_no
from employees as e
join dept_emp as d
on e.emp_no = d.emp_no;

  5. 查找所有员工的last_name和first_name以及对应部门编号dept_no，也包括展示没有分配具体部门的员工

    select e.last_name,e.first_name,d.dept_no
from employees as e
left join dept_emp as d--坐表为主，右表补充，为空补NULL
on e.emp_no = d.emp_no;

  6. 查找所有员工入职时候的薪水情况，给出emp_no以及salary， 并按照emp_no进行逆序

    select e.emp_no,s.salary
from salaries as s
join employees as e
where e.emp_no = s.emp_no
and e.hire_date = s.from_date
order by
e.emp_no
desc;

  7. 查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t

    count( )：统计记录的条数

    还需要group by emp_no将每个员工的记录显示在一条记录中

    select emp_no,count(emp_no) as t
from salaries
group by
emp_no
having t > 15;

  8. 找出所有员工当前(to_date='9999-01-01')具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示

    distinct：去除重复

    select distinct salary-- 去除重复
from salaries
where to_date = '9999-01-01'
order by
salary
desc;

  9. 获取所有部门当前manager的当前薪水情况，给出dept_no, emp_no以及salary，当前表示to_date='9999-01-01'

    select d.dept_no,d.emp_no,s.salary
from dept_manager as d
join salaries as s
on d.emp_no=s.emp_no
where d.to_date='9999-01-01'
and s.to_date='9999-01-01';

  10. 获取所有非manager的员工emp_no

    select emp_no
from employees
where emp_no not in (select emp_no from dept_manager);

  11. 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'。

    select e.emp_no,m.emp_no as manager_no
from dept_emp as e
join dept_manager as m
on e.dept_no=m.dept_no
where e.emp_no <> m.emp_no -- 当前manager是自己不显示，<>不等于
and e.to_date='9999-01-01'
and m.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，给出dept_no, emp_no以及其对应的salary

    select d.dept_no,d.emp_no,max(s.salary)
from dept_emp as d
join salaries as s
on d.emp_no = s.emp_no
group by
d.dept_no;

  13. 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。

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

  14. 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

    select * from employees
where emp_no%2 =1
and last_name <> 'Mary'
order by
hire_date
desc;

  15. 统计出当前各个title类型对应的员工当前（to_date='9999-01-01'）薪水对应的平均工资。结果给出title以及平均工资avg。

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

  16. 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

    select emp_no,salary from salaries
where to_date='9999-01-01'
order by
salary
desc
limit 1,1;  -- 降序，取第二个

  17. 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

    select e.emp_no,s.salary,e.last_name,e.first_name
from employees as e
join salaries as s
on e.emp_no=s.emp_no
where s.to_date='9999-01-01'
order by
s.salary
desc
limit 1,1;

    不适用order by：需要用嵌套select和max结合

    select e.emp_no,max(salary) ,e.last_name,e.first_name --这里用max(salary)
from  salaries as s,employees as e
where s.emp_no = e.emp_no
and salary<  -- 小于最大薪水中的最大，就是第二大
(
select max(salary)
from salaries
where to_date = '9999-01-01'
);

  18. 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

    双左连接

    select e.last_name,e.first_name,de.dept_name
from employees as e
left join dept_emp as d
on e.emp_no=d.emp_no
left join departments as de
on d.dept_no=de.dept_no;

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

    通过入职时间来排序查询：

    如果直接用薪水最大值-最小值，有可能最后一次是降薪。

    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

  20. 查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序

    --将两个查询的结果，当成两个表，进行join
select S2.emp_no, (S2.salary-S1.salary) as growth
from 
(select e.emp_no,s.salary 
 from employees as e 
 join salaries as s 
 on e.emp_no=s.emp_no 
 and e.hire_date=s.from_date) as S1  --所有员工的入职工资
join 
(select e.emp_no,s.salary 
 from employees as e 
 join salaries as s 
 on e.emp_no=s.emp_no 
 and s.to_date='9999-01-01') as S2   --所有员工的当前工资
on S1.emp_no = S2.emp_no
order by growth;

  21. 统计各个部门对应员工涨幅的次数总和，给出部门编码dept_no、部门名称dept_name以及次数sum

    select dp.dept_no,dp.dept_name,count(s.emp_no) as sum
from departments as dp
join dept_emp as de on de.dept_no = dp.dept_no
join salaries as s
on de.emp_no = s.emp_no
group by
de.dept_no;

  22. 对所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名，相同salary并列且按照emp_no升序排列

    SELECT s1.emp_no, s1.salary, COUNT(DISTINCT s2.salary) AS rank
FROM salaries AS s1, salaries AS s2
WHERE s1.to_date = '9999-01-01'  AND s2.to_date = '9999-01-01' AND s1.salary <= s2.salary
GROUP BY s1.emp_no
ORDER BY s1.salary DESC, s1.emp_no ASC

  23. 获取所有非manager员工当前的薪水情况，给出dept_no、emp_no以及salary ，当前表示to_date='9999-01-01'

    select de.dept_no,de.emp_no,s.salary
from dept_emp as de
join salaries as s
on de.emp_no=s.emp_no
where de.emp_no not in (select emp_no from dept_manager where to_date='9999-01-01')
and de.to_date='9999-01-01'
and s.to_date='9999-01-01';

  24. 获取员工其当前的薪水比其manager当前薪水还高的相关信息，当前表示to_date='9999-01-01',

    select S1.emp_no as emp_no,S2.emp_no as manager_no,S1.salary as emp_salary,S2.salary as manager_salary
from
(select s.salary,e.emp_no,e.dept_no from salaries as s join dept_emp as e on e.emp_no = s.emp_no and s.to_date='9999-01-01') as S1,
(select s.salary,m.emp_no,m.dept_no from salaries as s join dept_manager as m on m.emp_no = s.emp_no and s.to_date='9999-01-01') as S2
where S1.dept_no = S2.dept_no and S1.salary > S2.salary

  25. 汇总各个部门当前员工的title类型的分配数目，结果给出部门编号dept_no、dept_name、其当前员工所有的title以及该类型title对应的数目count

    select d.dept_no,d.dept_name,t.title,count(t.title) as count
from titles as t
join dept_emp as e
on e.emp_no = t.emp_no and e.to_date='9999-01-01' and t.to_date='9999-01-01'
join
departments as d
on d.dept_no = e.dept_no
group by d.dept_no,T.title

  26. 给出每个员工每年薪水涨幅超过5000的员工编号emp_no、薪水变更开始日期from_date以及薪水涨幅值salary_growth，并按照salary_growth逆序排列。

    提示：在sqlite中获取datetime时间对应的年份函数为strftime('%Y', to_date)

    select s2.emp_no,s2.from_date,s2.salary-s1.salary as salary_growth
from salaries as s1,salaries as s2
where s1.emp_no = s2.emp_no
and salary_growth > 5000
and (strftime("%Y",s2.to_date) - strftime("%Y",s1.to_date) = 1
OR strftime("%Y",s2.from_date) - strftime("%Y",s1.from_date) = 1)
order by salary_growth desc

  27. 查找描述信息中包括robot的电影对应的分类名称以及电影数目，而且还需要该分类对应电影数量>=5部

    SELECT c.name AS name, COUNT(f.film_id) AS amount
FROM film AS f, film_category AS fc, category AS c,
(SELECT category_id FROM film_category GROUP BY category_id HAVING COUNT(category_id) >= 5) AS cc
WHERE f.description LIKE '%robot%'
AND f.film_id = fc.film_id
AND fc.category_id = c.category_id
AND c.category_id = cc.category_id

  28. 使用join查询方式找出没有分类的电影id以及名称

    select f.film_id,f.title
from film as f
left join 
film_category as fc
on fc.film_id = f.film_id
where fc.category_id is null

  29. 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

    -- 非子查询
select f.title,f.description
from film as f,film_category as fc,
(select category_id from category where name = 'Action') as ac
where f.film_id = fc.film_id and fc.category_id = ac.category_id
    --join
select f.title,f.description
from film as f 
inner join 
film_category as fc 
on f.film_id = fc.film_id
inner join 
category as c 
on c.category_id = fc.category_id
where c.name = 'Action';

     

    --子查询
select f.title,f.description
from film as f where f.film_id in 
    (select fc.film_id from film_category as fc where fc.category_id in 
    (select c.category_id from category as c where name ='Action'))

  30. explain ，此题毫无意义 = =

posted @ 2019-10-06 12:43  mussessein  阅读(2719)  评论(0)  编辑  收藏  举报