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一、 题目篇
1.1 第十六题
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，不可使用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`));
答案
select s.emp_no, s.salary, e.last_name, e.first_name
from salaries s join employees e
on s.emp_no = e.emp_no
where s.salary =
(select max(salary) from salaries where salary <
(select max(salary) from salaries where to_date = '9999-01-01' )
and to_date = '9999-01-01')
and s.to_date = '9999-01-01'
1.2 第十七题
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`));
答案
思路：列出employees表里所有员工last_name, first_name, 根据employees中emp_no对应dept_emp中的dept_no,没有分配的员工找不到对应–>采用LEFT JOIN 再根据dept_no对应departments表中的dept_name,没有分配的员工找不到对应–>采用LEFT JOIN
SELECT last_name, first_name, dept_name
FROM employees AS e LEFT JOIN (SELECT emp_no, dept_name
FROM dept_emp AS de LEFT JOIN departments AS d
ON de.dept_no = d.dept_no) k
ON e.emp_no = k.emp_no;
1.3 第十八题
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(总共涨了多少)growth(可能有多次涨薪，没有降薪)
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`));
答案
select (select max(s1.salary) from salaries s1 where s1.emp_no=10001) -
(select min(s2.salary) from salaries s2 where s2.emp_no=10001) as growth
1.4 第十九题
题目描述
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 （注:可能有employees表和salaries表里存在记录的员工，有对应的员工编号和涨薪记录，但是已经离职了，离职的员工salaries表的最新的to_date!=‘9999-01-01’，这样的数据不显示在查找结果里面）
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, -- '入职时间'
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL, -- '一条薪水记录开始时间'
`to_date` date NOT NULL, -- '一条薪水记录结束时间'
PRIMARY KEY (`emp_no`,`from_date`));
答案
select a.emp_no, (b.salary - c.salary) as growth
from employees as a
inner join salaries as b
on a.emp_no = b.emp_no and b.to_date = '9999-01-01'
inner join salaries as c
on a.emp_no = c.emp_no and a.hire_date = c.from_date
order by growth asc
1.5 第二十题
题目描述
统计各个部门的工资记录数，给出部门编码dept_no、部门名称dept_name以及部门在salaries表里面有多少条记录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`));
答案
SELECT dept_no, dept_name, (SELECT COUNT()
FROM dept_emp AS de
INNER JOIN salaries AS s
ON de.emp_no=s.emp_no
WHERE de.dept_no=d.dept_no) AS sum
FROM departments AS d;
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