数据库mysql no.4

1.流程控制函数

①if(条件表达式,表达式1,表达式2):

如果条件表达式成立,返回表达式1,否则返回表达式2

case情况1

case 变量或表达式或字段

when 常量1 then 值1

when 常量2 then 值2

...

else 值n

end

case情况2

case

when 条件1 then 值1

when 条件2 then 值2

...

else 值n

end

2.分组函数

++sum 求和 avg 平均值++ ++max 最大值 min 最小值 count 计算个数++

只能处理数值型 任何类型都可以处理

他们都忽略null值

3.分组查询 group by

筛选条件

|-------|---------|-------------|--------|
| | 数据源 | 位置 | 关键字 |
| 分组前筛选 | 原始表 | group by子句前 | where |
| 分组后筛选 | 分组后的结果集 | group by子句后 | having |

注意:分组函数作为筛选条件的时候一定是放在having子句中

分组查询也可以用排序order by,一般置于最后

优先使用分组前筛选,即where筛选

4.连接查询

内连接:等值连接,非等值连接(betweenand..)自连接(两个表为一个)

外连接:左外连接(left outer join),右外连接,全外连接(full outer join)

交叉连接(cross outer join)

注意:如果给表起了别名,则查询的字段就不能使用原来的表名去限定

5.SQL练习

sql 复制代码
-- 分组查询GROUP BY
-- 查询每个部门的平均工资
SELECT AVG(salary),job_id FROM employees GROUP BY job_id;
-- 查询每个位置上的部门个数
SELECT COUNT(*),location_id FROM departments GROUP BY location_id;

-- 添加筛选条件  查询邮箱中包含a字符的,每个部门的平均工资
SELECT AVG(salary),department_id FROM employees WHERE email LIKE '%a%' GROUP BY department_id;

-- 查询有奖金的每个领导手下员工的最高工资
SELECT MAX(salary),manager_id FROM employees WHERE commission_pct IS NOT NULL GROUP BY manager_id;

SELECT  COUNT(*),department_id FROM employees GROUP BY department_id HAVING count(*)>2;

-- 查询每个工种有奖金的员工的最高工资》12000的工种编号和最高工资
SELECT MAX(salary),job_id FROM employees WHERE commission_pct IS NOT NULL GROUP BY job_id HAVING max(salary)>12000;


-- 查询领导编号>102的每个领导手下的最低工资》5000的领导编号是哪个以及最低工资

SELECT MIN(salary),manager_id FROM employees WHERE manager_id>102 GROUP BY manager_id HAVING MIN(salary)>5000; 

-- 按员工姓名的长度分组,查询每一组的员工个数筛选员工个数>5的有哪些
SELECT COUNT(*) c,LENGTH(last_name) a from employees GROUP BY a HAVING c>5;

-- 按多个字段分组
-- 查询每个部门每个工种的员工的平均工资
SELECT AVG(salary),department_id,job_id FROM employees WHERE department_id IS NOT NULL GROUP BY department_id,job_id HAVING AVG(salary)>12000 ORDER BY AVG(salary) DESC;

-- 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;
-- 查询员工最高工资和最低工资的差距
SELECT max(salary)-MIN(salary) difference FROM employees;
-- 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;
-- 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) FROM employees GROUP BY department_id ORDER BY AVG(salary) DESC
-- 选择具有各个job_id的员工人数
SELECT COUNT(*),job_id FROM employees GROUP BY job_id;


SELECT last_name,department_name FROM employees,departments where employees.department_id=departments.department_id;

-- 查询员工名 工种号,工种名
SELECT last_name,jobs.job_id,job_title FROM employees,jobs WHERE employees.job_id=jobs.job_id;
-- 别名
SELECT last_name,a.job_id,job_title FROM employees a,jobs b WHERE a.job_id=b.job_id;

-- 查询有奖金的员工名、部门名
SELECT last_name,department_name, commission_pct FROM employees e,departments d WHERE e.department_id=d.department_id AND e.commission_pct is not null;

-- 查询城市名中第二个字符为o的部门名和城市名
SELECT department_name,city FROM departments d,locations l WHERE d.location_id=l.location_id and city LIKE '_o%'

-- 查询每个城市的部门个数
SELECT COUNT(*) 个数,city FROM departments,locations
WHERE departments.location_id=locations.location_id GROUP BY city;

-- 查询每个工种的工种名和员工的个数,并且按员工个数降序
SELECT job_title,COUNT(*) FROM jobs,employees WHERE
jobs.job_id=employees.job_id GROUP BY job_title ORDER BY COUNT(*) DESC;

-- 查询员工名,部门名和所在的城市
SELECT last_name,department_name,city FROM employees e,departments d,locations l WHERE
e.department_id=d.department_id and d.location_id=l.location_id GROUP BY last_name,department_name,city ORDER BY department_name DESC;

-- 自连接
-- 查询员工名和上级的名称
SELECT e.employee_id,e.last_name,m.employee_id,m.last_name
FROM employees e,employees m WHERE e.manager_id=m.employee_id;

-- 显示员工表的最大工资,工资平均值
SELECT max(salary),ROUND(AVG(salary),2) FROM employees;

SELECT employee_id,job_id,last_name FROM employees ORDER BY department_id DESC,salary ASC;

 SELECT job_id FROM employees WHERE job_id LIKE '%a%e%';

SELECT NOW();
SELECT TRIM(' s s s   ');
SELECT TRIM('a' FROM 'aaa hejun aaa');
SELECT SUBSTR('hejun',3,3);

-- 显示所有员工的姓名,部门号和部门名称
SELECT last_name,e.department_id,d.department_name FROM employees e,departments d WHERE e.department_id=d.department_id; 

-- 查询90号部门员工的job_id和90号部门的location_id
SELECT job_id,location_id FROM employees e,departments d WHERE
e.department_id=d.department_id AND e.department_id=90;


-- 查询每个国家下的部门个数大于2的国家编号
SELECT country_id,COUNT(*) 部门个数 FROM departments d,locations l WHERE d.location_id=l.location_id GROUP BY country_id HAVING 部门个数>2;

SELECT e.last_name employees,e.employee_id "Emp#",m.last_name manager,m.manager_id "Mgr#" FROM employees e,employees m WHERE
e.manager_id=m.employee_id AND e.last_name='kochhar';

-- 用内连接实现等值连接
-- 查询员工名 部门名
SELECT last_name,department_name FROM employees e INNER JOIN
departments d on e.department_id=d.department_id;

-- 查询名字中包含e的员工名和工种名
SELECT last_name,job_title FROM employees e INNER JOIN
jobs j on e.job_id=j.job_id WHERE e.last_name LIKE '%e%';

-- 查询员工名部门名工种名并按部门名降序
SELECT last_name,department_name,job_title FROM employees e
INNER JOIN departments d on e.department_id=d.department_id
INNER JOIN jobs j on e.job_id=j.job_id ORDER BY department_name
DESC;

-- 查询姓名中包含字符k的员工的名字,上级的名字
SELECT e.last_name,m.last_name FROM employees e
INNER JOIN employees m on e.manager_id=m.employee_id
AND e.last_name like '%k%';
SELECT e.last_name,m.last_name FROM employees e
INNER JOIN employees m on e.manager_id=m.employee_id
WHERE e.last_name like '%k%';

SELECT * FROM beauty;
SELECT * FROM boys;

SELECT b.`name`,bo.* FROM beauty b left OUTER JOIN
boys bo on b.boyfriend_id=bo.id WHERE bo.id is not NULL;

SELECT b.`name`,bo.* FROM boys bo right OUTER JOIN
beauty b on b.boyfriend_id=bo.id WHERE bo.id is not NULL;

-- 查询哪个部门没有员工
SELECT d.*,e.employee_id 
FROM employees e 
RIGHT OUTER JOIN departments d 
on d.department_id=e.department_id 
WHERE e.department_id is null;

SELECT b.id,b.name,bo.*
FROM beauty b LEFT JOIN
boys bo on b.boyfriend_id=bo.id
WHERE b.id>3;

SELECT city,d.* FROM locations l 
left outer join departments d ON
l.location_id=d.location_id
WHERE d.location_id is NULL;
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