1.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
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
));
解答:select distinct salary from salaries where to_date=‘9999-01-01’ order by salary desc;
2.查找最晚入职员工的所有信息
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
));
解答:select * from employees where hire_date=(select max(hire_date) from employees);
3.查找入职员工时间排名倒数第三的员工所有信息
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
));
解答:select * from employees order by hire_date desc limit 2,1;
4.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
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
));
解答;select emp_no,count(emp_no) as t from salaries group by emp_no having t>15;
5.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’
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
));
解答:
6.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。
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);
解答:select b.title,count (b.title) as t from
(select distinct (a.emp_no),a.title from a) as b group by b.title having t>=2;
或select title,count(distinct emp_no) from titles group by title having count(distinct emp_no)>=2;