MySQL条件、排序、分组查询练习

阳光穿透心脏的1/2处 2024-04-06 10:39 78阅读 0赞

#### 一、条件查询 ####

1.查询员工号为176的员工的姓名和部门号和年薪

SELECT last_name,department_id,salary*12*(1+IFNULL(commission_pct,0)) AS 年薪
    FROM employees;

2.选择工资不在5000到12000的员工的姓名和工资

SELECT last_name,salary
    FROM employees
    WHERE salary<5000||salary>12000;

3.选择在20或50号部门工作的员工姓名和部门号

SELECT last_name,department_id
    FROM employees
    WHERE department_id=20||department_id=50;

4.选择姓名中有字母a和e的员工姓名

SELECT last_name
    FROM employees
    WHERE last_name LIKE'%a%' OR '%e%';

5.显示出表employees部门编号在80-100之间的姓名、职位

SELECT last_name,job_title,department_id
    FROM employees,jobs
    WHERE department_id>=80||department_id<=100;

#### 二、排序查询 ####

1.将姓名中包含e字符的年薪和姓名显示出来，并且按年薪进行降序

SELECT last_name,salary*12*(1+IFNULL(commission_pct,0)) 年薪
    FROM employees
    WHERE last_name LIKE '%e%'
    ORDER BY 年薪 DESC;

2.显示所有有奖金的员工姓名、奖金、姓名的长度，按姓名的长度从低到高排序

SELECT last_name,salary*commission_pct 奖金,LENGTH(last_name)
    FROM employees
    WHERE commission_pct IS NOT NULL
    ORDER BY LENGTH(last_name);

3.部门编号>50的按工资从高到低排序，如果一样，再按frist\_name升序

SELECT *
    FROM employees
    WHERE department_id>50
    ORDER BY salary DESC,first_name;

#### 三、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;

3.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;

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