group by 和 having 用法

例如 1:

create table if not exists employee(eid integer auto_increment primary key,name varchar(5),age,integer,salary integer,depart varchar(5),workage integer);

insert into employee (name,age,salary,depart,workage)

values ('崔铭','25',1500,'研发部',3),
       ('佳伟','23',1000,'市场部',2),
       ('刘涵','30',10600,'人事部',6),
       ('孙铭泽','25',2000,'运营部',5),
       ('张吉龙','21',15000,'生产部',12),
       ('从好平','22',1500,'质量部',3),
       ('杨忠','22',5000,'财务部',4),
       ('芦淞','24',6000,'采购部',7),
       ('马玉','25',450000,'销售部',29),
       ('成林','21',12000,'安全部',10),
       ('张龙','32',17000,'研发部',21),
       ('王建业','25',11000,'研发部',7),
       ('王佳敏','22',10000,'市场部',9),
       ('姜佳伟','27',10000,'人事部',13),
       ('王国栋','20',10012,'研发部',2),
       ('周昌洋','38',10560,'研发部',1),
       ('刘鑫鑫','18',1900,'人事部',6),
       ('刘博','21',2000,'研发部',11),
       ('乔鑫','19',13000,'运营部',12),
       ('宇航','20',10500,'生产部',21),
       ('赵浩然','33',10400,'总经理',21),
       ('常盛','24',1000,'生产部',2),
       ('刘麟','25',3000,'武装部',8);

# 1. 查询每个部门的总薪资
select depart,SUM(salary) from employee
group by depart;
# 2. 员工数超过3人的部门的最高薪资和最低薪资
select depart, MAX(salary),MIN(salary) from employee
group by depart
having count(*) >3;

# 3. 工龄超过3年的员工中,薪资最低的所有员工信息
select * from employee
where salary in (
select MIN(salary)from emploree
where workage>3);

# 4. 工龄超过3年的员工数大于3的部门

select depart,count(*) from employee
where workage > 3
group by depart
having count(*)>3;

例如 2:

create table if not exists score(sname varchar(10), cname varchar(5),grade integer) charset=utf8;
insert into score(sname,cname,grade)
values ('张三','数学',80),
       ('张三','语文',90),
       ('张三','英语',70),
       ('张三','物理',60),
       ('李四','数学',66),
       ('李四','语文',60),
       ('李四','英语',80),
       ('李四','物理',90),
       ('刘志麟','语文',99),
       ('刘志麟','数学',50),
       ('刘志麟','英语',50),
       ('刘志麟','物理',89),
       ('罗宇航','语文',99),
       ('罗宇航','数学',80),
       ('罗宇航','物理',78),
       ('罗宇航','英语',96),
       ('许振东','数学',96),
       ('许振东','语文',96),
       ('许振东','英语',96),
       ('许振东','物理',96);

# 1. 查询90分以上的学生的课程名和成绩
select * from score
where grade > 90;

# 2. 查询每个学生的成绩在90分以上的各有多少门
select sname,count(cname) from score
where grade > 90
group by sname
having count(cname);
# 3. 至少有两门课程在90分以上的学员以及90分以上的课程数 
select sname,count(cname) from socre
where grade > 90
group by sname
having count(cname) >2;

# 4. 平均成绩比张三的平均成绩高的学员和其平均分

select sname, AVG(grade) from score 
group by sname
having AVG(grade) > (select AVG(grade) from score
where sname = '张三');

# 5. 查询平均成绩大于90分并且语文课95分以上的学生名和平均成绩
select sname, AVG(grade) from score
where sname in (
select sname from score
where cname='语文'and grade>95)
group by sname
having AVG(grade)>90;

# 6. 查询每个学员的平均分和学生名
select sname, AVG(grade) from score
group by sname;

# 7. 查询每门课的最好成绩和平均分
select cname MAX(grade),AVG(grade) from score
group by cname;

# 8. 查询数学课成绩最好的学员的所有成绩
select * from score
where sname in (
select sname from score
where grade =(
select MAX(grade)from score
where cname='数学')and cname='数学');

# 9. 查询学员及其总分,按总分降序排列
select sname ,SUM(grade) from score
group by sname
order by SUM(grade) desc;