【MySql】牛客SQL刷題（上）

本文轉載自查看原文 2019-10-06 12:43 598 MySql

牛客SQL題目

題目鏈接：https://www.nowcoder.com/ta/sql

查找最晚入職員工的所有信息

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

查找入職員工時間排名倒數第3的員工所有信息

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

查找各個部門當前(to_date='9999-01-01')的領導的當前薪水詳情以及其對應部門編號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';

查找所有已經分配部門的員工的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;

查找所有員工的last_name和first_name以及對應部門編號dept_no，也包括展示沒有分配具體部門的員工

select e.last_name,e.first_name,d.dept_no
from employees as e
left join dept_emp as d--坐表為主，右表補充，為空補NULL
on e.emp_no = d.emp_no;

查找所有員工入職時候的薪水情況，給出emp_no以及salary，並按照emp_no進行逆序

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;

查找薪水漲幅超過15次的員工號emp_no以及其對應的漲幅次數t

count( )：統計記錄的條數

還需要group by emp_no將每個員工的記錄顯示在一條記錄中
```
select emp_no,count(emp_no) as t
from salaries
group by
emp_no
having t > 15;
```
找出所有員工當前(to_date='9999-01-01')具體的薪水salary情況，對於相同的薪水只顯示一次,並按照逆序顯示

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

獲取所有部門當前manager的當前薪水情況，給出dept_no, emp_no以及salary，當前表示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';

獲取所有非manager的員工emp_no

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

獲取所有員工當前的manager，如果當前的manager是自己的話結果不顯示，當前表示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 -- 當前manager是自己不顯示，<>不等於
and e.to_date='9999-01-01'
and m.to_date='9999-01-01';

獲取所有部門中當前員工薪水最高的相關信息，給出dept_no, emp_no以及其對應的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;

從titles表獲取按照title進行分組，每組個數大於等於2，給出title以及對應的數目t。
```
select title,count(title) as t
from titles
group by
title
having t >= 2;
```
查找employees表所有emp_no為奇數，且last_name不為Mary的員工信息，並按照hire_date逆序排列
```
select * from employees
where emp_no%2 =1
and last_name <> 'Mary'
order by
hire_date
desc;
```

統計出當前各個title類型對應的員工當前（to_date='9999-01-01'）薪水對應的平均工資。結果給出title以及平均工資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;

獲取當前（to_date='9999-01-01'）薪水第二多的員工的emp_no以及其對應的薪水salary

select emp_no,salary from salaries
where to_date='9999-01-01'
order by
salary
desc
limit 1,1;  -- 降序，取第二個

查找當前薪水(to_date='9999-01-01')排名第二多的員工編號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;

不適用order by：需要用嵌套select和max結合

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

查找所有員工的last_name和first_name以及對應的dept_name，也包括暫時沒有分配部門的員工

雙左連接

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;

查找員工編號emp_no為10001其自入職以來的薪水salary漲幅值growth

通過入職時間來排序查詢：

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

查找所有員工自入職以來的薪水漲幅情況，給出員工編號emp_no以及其對應的薪水漲幅growth，並按照growth進行升序

--將兩個查詢的結果，當成兩個表，進行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  --所有員工的入職工資
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   --所有員工的當前工資
on S1.emp_no = S2.emp_no
order by growth;

統計各個部門對應員工漲幅的次數總和，給出部門編碼dept_no、部門名稱dept_name以及次數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;

對所有員工的當前(to_date='9999-01-01')薪水按照salary進行按照1-N的排名，相同salary並列且按照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

獲取所有非manager員工當前的薪水情況，給出dept_no、emp_no以及salary ，當前表示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';

獲取員工其當前的薪水比其manager當前薪水還高的相關信息，當前表示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

匯總各個部門當前員工的title類型的分配數目，結果給出部門編號dept_no、dept_name、其當前員工所有的title以及該類型title對應的數目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

給出每個員工每年薪水漲幅超過5000的員工編號emp_no、薪水變更開始日期from_date以及薪水漲幅值salary_growth，並按照salary_growth逆序排列。

提示：在sqlite中獲取datetime時間對應的年份函數為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

查找描述信息中包括robot的電影對應的分類名稱以及電影數目，而且還需要該分類對應電影數量>=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

使用join查詢方式找出沒有分類的電影id以及名稱

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

使用子查詢的方式找出屬於Action分類的所有電影對應的title,description

-- 非子查詢
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';

--子查詢
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'))

explain ，此題毫無意義 = =

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