SQL题练习

SQL语法总结：http://www.w3school.com.cn/sql/sql_func_min.asp
SQL练习：https://www.nowcoder.com/ta/sql

1.

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));

SELECT * FROM employees WHERE hire_date = (SELECT MAX(hire_date) FROM employees)

或

SELECT * FROM employees ORDER BY hire_date DESC LIMIT 1

2.

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));

LIMIT m,n : 表示从第m+1条开始，取n条数据；
LIMIT n ： 表示从第0条开始，取n条数据，是limit(0,n)的缩写。

SELECT * FROM employees ORDER BY hire_date DESC LIMIT 2,1

3.

查找各个部门当前(to_date='9999-01-01')领导当前薪水详情以及其对应部门编号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));

SELECT s.*, dept_no
FROM salaries s 
INNER JOIN dept_manager 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
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));

SELECT e.last_name, e.first_name, d.dept_no 
FROM dept_emp d, employees e
WHERE d.emp_no = e.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
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));

SELECT e.last_name, e.first_name, d.dept_no
FROM employees e
LEFT JOIN dept_emp d
ON e.emp_no = d.emp_no

6.查找所有员工入职时候的薪水情况

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
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));

SELECT e.emp_no, s.salary
FROM employees e, salaries s
WHERE e.emp_no = s.emp_no
AND hire_date = from_date
ORDER BY e.emp_no DESC

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));

HAVING用于对聚合结果做限制。这里有一个错误，涨薪大于15次，那么t应该>16。

SELECT emp_no, COUNT(*) AS t
FROM salaries s
GROUP BY emp_no
HAVING t > 15

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));

SELECT DISTINCT salary
FROM salaries
WHERE to_date = '9999-01-01'
ORDER BY salary DESC

.查找employees表

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
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));

SELECT * 
FROM employees
WHERE emp_no % 2 != 0
AND last_name != 'Mary'
ORDER BY hire_date DESC