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MySQL Demo 02

1. 排序查询

1.查询员工的姓名和部门号和年薪，按年薪降序 按姓名升序

SELECT last_name,department_id,salary*12*(1+IFNULL(commission_pct,0)) 年薪
FROM employees
ORDER BY 年薪 DESC,last_name ASC;

2.选择工资不在8000到17000的员工的姓名和工资，按工资降序

SELECT last_name,salary
FROM employees

WHERE salary NOT BETWEEN 8000 AND 17000
ORDER BY salary DESC;

3.查询邮箱中包含e的员工信息，并先按邮箱的字节数降序，再按部门号升序

SELECT *,LENGTH(email)
FROM employees
WHERE email LIKE '%e%'
ORDER BY LENGTH(email) DESC,department_id ASC;

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2. 分组查询

1.查询各job_id的员工工资的最大值，最小值，平均值，总和，并按job_id升序

SELECT MAX(salary),MIN(salary),AVG(salary),SUM(salary),job_id
FROM employees
GROUP BY job_id
ORDER BY job_id;

2.查询员工最高工资和最低工资的差距（DIFFERENCE）

SELECT MAX(salary)-MIN(salary) DIFFRENCE
FROM employees;

3.查询各个管理者手下员工的最低工资，其中最低工资不能低于6000，没有管理者的员工不计算在内

SELECT MIN(salary),manager_id
FROM employees
WHERE manager_id IS NOT NULL
GROUP BY manager_id
HAVING MIN(salary)>=6000;

4.查询所有部门的编号，员工数量和工资平均值,并按平均工资降序

SELECT department_id,COUNT(*),AVG(salary) a
FROM employees
GROUP BY department_id
ORDER BY a DESC;

5.选择具有各个job_id的员工人数

SELECT COUNT(*) 个数,job_id
FROM employees
GROUP BY job_id;

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3. 单行函数

1. 显示系统时间(注：日期+时间)

SELECT NOW();

2. 查询员工号，姓名，工资，以及工资提高百分之20%后的结果（new salary）

SELECT employee_id,last_name,salary,salary*1.2 "new salary"
FROM employees;

3. 将员工的姓名按首字母排序，并写出姓名的长度（length）

SELECT LENGTH(last_name) 长度,SUBSTR(last_name,1,1) 首字符,last_name
FROM employees
ORDER BY 首字符;

4. 做一个查询，产生下面的结果

<last_name> earns <salary> monthly but wants <salary*3>
Dream Salary
King earns 24000 monthly but wants 72000


SELECT CONCAT(last_name,' earns ',salary,' monthly but wants ',salary*3) AS "Dream Salary"
FROM employees
WHERE salary=24000;

5. 使用case-when，按照下面的条件：

job grade

AD_PRES A

ST_MAN B

IT_PROG C

SA_REP D

ST_CLERK E

产生下面的结果

Last_name Job_id Grade

king AD_PRES A

SELECT last_name,job_id AS  job,
CASE job_id
WHEN 'AD_PRES' THEN 'A' 
WHEN 'ST_MAN' THEN 'B' 
WHEN 'IT_PROG' THEN 'C' 
WHEN 'SA_PRE' THEN 'D'
WHEN 'ST_CLERK' THEN 'E'
END AS Grade
FROM employees
WHERE job_id = 'AD_PRES';

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4.分组函数

1.查询公司员工工资的最大值，最小值，平均值，总和

SELECT MAX(salary) 最大值,MIN(salary) 最小值,AVG(salary) 平均值,SUM(salary) 和
FROM employees;

2.查询员工表中的最大入职时间和最小入职时间的相差天数 （DIFFRENCE）

SELECT MAX(hiredate) 最大,MIN(hiredate) 最小,(MAX(hiredate)-MIN(hiredate))/1000/3600/24 DIFFRENCE
FROM employees;

SELECT DATEDIFF(MAX(hiredate),MIN(hiredate)) DIFFRENCE
FROM employees;

SELECT DATEDIFF('1995-2-7','1995-2-6');

3.查询部门编号为90的员工个数

SELECT COUNT(*) FROM employees WHERE department_id = 90;