牛客网数据开发题库_数据库刷题

21.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

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

思路:先分别用两次LEFT JOIN左连接employees与salaries，建立两张表，
分别存放员工当前工资（sCurrent）与员工入职时的工资（sStart），
再用INNER JOIN连接sCurrent与sStart，最后限定在同一员工下用当前工资减去入职工资。
法一：内层用LEFT JOIN，外层用INNER JOIN（内层也可以改用 INNER JOIN）
SELECT sCurrent.emp_no, (sCurrent.salary-sStart.salary) AS growth
FROM (SELECT s.emp_no, s.salary FROM employees e, salaries s WHERE e.emp_no = s.emp_no AND s.to_date = '9999-01-01') 
AS sCurrent INNER JOIN(SELECT s.emp_no, s.salary FROM employees e, salaries s WHERE e.emp_no = s.emp_no AND s.from_date = e.hire_date) 
AS sStart
ON sCurrent.emp_no = sStart.emp_no
ORDER BY growth;
法二：内外都层用FROM并列查询
SELECT sCurrent.emp_no, (sCurrent.salary-sStart.salary) AS growth
FROM (SELECT s.emp_no, s.salary FROM employees e, salaries s WHERE e.emp_no = s.emp_no AND s.to_date = '9999-01-01') 
AS sCurrent,(SELECT s.emp_no, s.salary FROM employees e, salaries s WHERE e.emp_no = s.emp_no AND s.from_date = e.hire_date) 
AS sStart
WHERE sCurrent.emp_no = sStart.emp_no
ORDER BY growth;

22.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

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 `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

SELECT dt.dept_no,dt.dept_name,COUNT(salary) AS SUM
FROM dept_emp de LEFT JOIN departments dt ON de.dept_no=dt.dept_noLEFT JOIN salaries s ON de.emp_no=s.emp_no
GROUP BY dt.dept_no,dt.dept_name;

23.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

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

法一-SQL Server数据库中开窗函数（未通过）：
1.
SELECT emp_no,salary,rank() over(order by salary ASC) as rank
FROM salary
WHERE to_date='9999-01-01'
ORDER BY emp_no;
2.
SELECT emp_no, salaries, DENSE_RANK() OVER(ORDER BY salary DESC) AS rank
WHERE to_date = '9999-01-01' 
ORDER BY salary DESC, emp_no ASC;
法二：复用salaries表进行比较排名
1、从两张相同的salaries表（分别为s1与s2）进行对比分析，先将两表限定条件设为to_date = '9999-01-01'，挑选出当前所有员工的薪水情况。
2、本题的精髓在于 s1.salary <= s2.salary，意思是在输出s1.salary的情况下，
有多少个s2.salary大于等于s1.salary，
比如当s1.salary=94409时，有3个s2.salary（分别为94692,94409,94409）大于等于它，
但由于94409重复，利用COUNT(DISTINCT s2.salary)去重可得工资为94409的rank等于2。其余排名以此类推。
3、千万不要忘了GROUP BY s1.emp_no，否则输出的记录只有一条（可能是第一条或者最后一条，根据不同的数据库而定），
因为用了合计函数COUNT()
4、最后先以 s1.salary 逆序排列，再以 s1.emp_no 顺序排列输出结果
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;

24.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'

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`));
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

重点：表连接,NOT IN
SELECT de.dept_no, s.emp_no, s.salary 
FROM (employees AS e INNER JOIN salaries AS s ON s.emp_no = e.emp_no AND s.to_date = '9999-01-01')
INNER JOIN dept_emp AS de 
ON e.emp_no = de.emp_no
WHERE de.emp_no NOT IN 
(SELECT emp_no 
FROM dept_manager 
WHERE to_date = '9999-01-01');

25.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',
结果第一列给出员工的emp_no，
第二列给出其manager的manager_no，
第三列给出该员工当前的薪水emp_salary,
第四列给该员工对应的manager当前的薪水manager_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 `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`));
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

思路：创建两张表（一张记录当前所有员工的工资，另一张只记录部门经理的工资）进行比较
1、先用INNER JOIN连接salaries和demp_emp，建立当前所有员工的工资记录sem
2、再用INNER JOIN连接salaries和demp_manager，建立当前所有员工的工资记录sdm
3、最后用限制条件sem.dept_no = sdm.dept_no AND sem.salary > sdm.salary找出同一部门中工资比经理高的员工，
并根据题意依次输出emp_no、manager_no、emp_salary、manager_salary
SELECT sem.emp_no AS emp_no, sdm.emp_no AS manager_no, 
sem.salary AS emp_salary, sdm.salary AS manager_salary
FROM (SELECT s.salary, s.emp_no, de.dept_no FROM salaries s INNER JOIN dept_emp de
ON s.emp_no = de.emp_no AND s.to_date = '9999-01-01' ) AS sem, 
(SELECT s.salary, s.emp_no, dm.dept_no FROM salaries s INNER JOIN dept_manager dm
ON s.emp_no = dm.emp_no AND s.to_date = '9999-01-01' ) AS sdm
WHERE sem.dept_no = sdm.dept_no AND sem.salary > sdm.salary；

26.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

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 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

思路：用GROUP BY同时对 de.dept_no 和 t.title 进行分组，
1、先用 INNER JOIN 连接 dept_emp 与 salaries，根据测试数据添加限定条件 
de.to_date = '9999-01-01' AND t.to_date = '9999-01-01'，即当前员工的当前头衔
2、再用 INNER JOIN 连接departments，限定条件为 de.dept_no = dp.dept_no，即部门编号相同
3、最后用 GROUP BY 同时对 de.dept_no 和 t.title 进行分组，
用 COUNT(t.title) 统计相同部门下相同头衔的员工个数
SELECT de.dept_no, dp.dept_name, t.title, COUNT(t.title) AS count
FROM titles AS t INNER JOIN dept_emp AS de ON t.emp_no = de.emp_no AND de.to_date = '9999-01-01' AND t.to_date = '9999-01-01'
INNER JOIN departments AS dp ON de.dept_no = dp.dept_no
GROUP BY de.dept_no, t.title;

27.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)

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

每年？？
正确答案：
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；

28.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

CREATE TABLE IF NOT EXISTS film (
film_id smallint(5)  NOT NULL DEFAULT '0',
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));

CREATE TABLE category  (
category_id  tinyint(3)  NOT NULL ,
name  varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));

CREATE TABLE film_category  (
film_id  smallint(5)  NOT NULL,
category_id  tinyint(3)  NOT NULL, `last_update` timestamp);

思路：1、找到对应电影数量>=5的所有分类，建立成虚表cc：
(select category_id, count(film_id) as category_num 
from  film_category  group by category_id having count(film_id)>=5) as cc
2、设定限制条件 f.description like '%robot%'
3、在表cc、f、fc、c中查找包括robot的电影对应的分类名称和对应的电影数目。
SELECT c.name, COUNT(fc.film_id) 
FROM (SELECT category_id, COUNT(film_id) AS category_num FROM film_category  GROUP BY category_id HAVING COUNT(film_id)>=5) AS cc,
film AS f, film_category AS fc, category AS c
WHERE  f.description LIKE '%robot%'
AND f.film_id = fc.film_id
AND c.category_id = fc.category_id
AND c.category_id=cc.category_id;

29.对表film、film_category，使用join查询方式找出没有分类的电影id以及名称

思路：运用 LEFT JOIN连接两表，用 IS NULL语句限定条件
SELECT f.film_id,f.title
FROM film f LEFT JOIN film_category fm
ON f.film_id=fm.film_id
WHERE fm.category_id IS NULL;

30.对表film、film_category、category，

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

SELECT title,description
FROM film
WHERE film_id IN (SELECT fm.film_id FROM film_category fm JOIN  category c ON fm.category_id=c.category_id WHERE c.name='Action');
多个子查询：
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 c.name = 'Action'));