表名和字段

准备工作：

表设计：

    –1.学生表 
    Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别 
    –2.课程表  
    Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号 
    –3.教师表  
    Teacher(t_id,t_name) –教师编号,教师姓名 
    –4.成绩表  
    Score(s_id,c_id,s_score) –学生编号,课程编号,分数 

 建表语句：

[code]    --建库
create database querySql;
use querySql;
--建表
--学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`s_name` VARCHAR(20) NOT NULL DEFAULT '',
`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
`c_id`  VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id`  VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);

数据插入：

[code]--插入学生表测试数据
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

结果图如下：

数据查询：

[code]-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

select a.* ,b.s_score as 01_score,c.s_score as 02_score from
student a
join score b on a.s_id=b.s_id and b.c_id='01'
left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score

-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数

select a.* ,b.s_score as 01_score,c.s_score as 02_score from
student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL
join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score

-- 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
student b
join score a on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)>=60;

-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
-- (包括有成绩的和无成绩的)

select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
student b
left join score a on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)<60
union
select a.s_id,a.s_name,0 as avg_score from
student a
where a.s_id not in (
select distinct s_id from score);

-- 5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from
student a
left join score b on a.s_id=b.s_id
GROUP BY a.s_id,a.s_name;

-- 6、查询"李"姓老师的数量
select count(t_id) from teacher where t_name like '李%';

-- 7、查询学过"张三"老师授课的同学的信息
select a.* from
student a
join score b on a.s_id=b.s_id where b.c_id in(
select c_id from course where t_id =(
select t_id from teacher where t_name = '张三'));

-- 8、查询没学过"张三"老师授课的同学的信息
select * from
student c
where c.s_id not in(
select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(
select c_id from course where t_id =(
select t_id from teacher where t_name = '张三')));
-- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息（好题）

select a.* from
student a,score b,score c
where a.s_id = b.s_id  and a.s_id = c.s_id and b.c_id='01' and c.c_id='02';

-- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息

select a.* from
student a
where a.s_id in (select s_id from score where c_id='01' ) and a.s_id not in(select s_id from score where c_id='02')

-- 11、查询没有学全所有课程的同学的信息

select s.* from
student s where s.s_id in(
select s_id from score where s_id not in(
select a.s_id from score a
join score b on a.s_id = b.s_id and b.c_id='02'
join score c on a.s_id = c.s_id and c.c_id='03'
where a.c_id='01'))

-- 12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息

select * from student where s_id in(
select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')
);

-- 13、查询和"01"号的同学学习的课程完全相同的其他同学的信息

select a.* from student a where a.s_id in(
select distinct s_id from score where s_id!='01' and c_id in(select c_id from score where s_id='01')
group by s_id
having count(1)=(select count(1) from score where s_id='01'));
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select a.s_name from student a where a.s_id not in (
select s_id from score where c_id =
(select c_id from course where t_id =(
select t_id from teacher where t_name = '张三'))
group by s_id);

-- 15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from
student a
left join score b on a.s_id = b.s_id
where a.s_id in(
select s_id from score where s_score<60 GROUP BY  s_id having count(1)>=2)
GROUP BY a.s_id,a.s_name

-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
select a.*,b.c_id,b.s_score from
student a,score b
where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;

-- 17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,
(select s_score from score where s_id=a.s_id and c_id='02') as 数学,
(select s_score from score where s_id=a.s_id and c_id='03') as 英语,
round(avg(s_score),2) as 平均分 from score a  GROUP BY a.s_id ORDER BY 平均分 DESC;

-- 18.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率
--及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name

分析：ROUND() 函数         ROUND 函数用于把数值字段舍入为指定的小数位数。
SQL ROUND() 语法    SELECT ROUND(column_name,decimals) FROM table_name

-- 19、按各科成绩进行排序，并显示排名(实现不完全)
-- mysql没有rank函数
select a.s_id,a.c_id,
@i:=@i +1 as i保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id='02' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
union
select a.s_id,a.c_id,
@i:=@i +1 as i,
@k:=(case when @score=a.s_score then @k else @i end) as rank,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score WHERE c_id='03' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s

-- 20、查询学生的总成绩并进行排名
select a.s_id,
@i:=@i+1 as i,
@k:=(case when @score=a.sum_score then @k else @i end) as rank,
@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
(select @k:=0,@i:=0,@score:=0)s

-- 21、查询不同老师所教不同课程平均分从高到低显示 (为啥avg的参数b.s_score？)

select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
left join score b on a.c_id=b.c_id
left join teacher c on a.t_id=c.t_id
GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
-- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03'
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3;

-- 23、统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比

select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)b on a.c_id=b.c_id
left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)c on a.c_id=c.c_id
left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)d on a.c_id=d.c_id
left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)e on a.c_id=e.c_id
left join course f on a.c_id = f.c_id

-- 24、查询学生平均成绩及其名次

select a.s_id,
@i:=@i+1 as '不保留空缺排名',
@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
@avg_score:=avg_s as '平均分'
from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;

分析：

《select sid,ROUND(AVG(score),2) as avg_s from SC GROUP BY sid)a:根据sid算出score列的平均数，保留两位小数，结果起名a；

(select @avg_score:=0,@i:=0,@k:=0)b：定义@avg_score，@i，@k三个标量，初始值=0,起名b,@avg_score是保存avg_s的值，@i，@k是排序的序列值（第一名，第二名这类）

@i:=@i+1 as '不保留空缺排名'：执行时@i的值在上一次基础上加1,初始值0，第一行@i=1,第二行@i=2,以此类推，这一列的名字起名'不保留空缺排名'，也就是'按照成绩排序，成绩相同的排在后面一位’

@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名'：判断@avg_score和a.avg_s相不相等，@avg_score初始值0，第一行是0，第二行的值是第一行avg_s的值，也就是说判断下一个sid的成绩和上一个相不相等，要是相等返回@k,不相等返回@i,@k的值初始值是0，要是 @avg_score和a.avg_s不相等，@k的值就是@i的值，要是 @avg_score和a.avg_s相等，@k的值就是上一行的@i,结果起名'保留空缺排名'，也就是‘成绩相同排名并列’  总结来说就是根据平均分产生不并列排名的成绩和并列排名的成绩，》

-- 25、查询各科成绩前三名的记录
-- 1.选出b表比a表成绩大的所有组
-- 2.选出比当前id成绩大的 小于三个的
select a.s_id,a.c_id,a.s_score from score a
left join score b on a.c_id = b.c_id and a.s_score<b.s_score
group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
ORDER BY a.c_id,a.s_score DESC

-- 26、查询每门课程被选修的学生数

select c_id,count(s_id) from score a GROUP BY c_id

-- 27、查询出只有两门课程的全部学生的学号和姓名 （多练）
select s_id,s_name from student where s_id in(
select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);

-- 28、查询男生、女生人数
select s_sex,COUNT(s_sex) as 人数  from student GROUP BY s_sex

-- 29、查询名字中含有"风"字的学生信息

select * from student where s_name like '%风%';

-- 30、查询同名同性学生名单，并统计同名人数

select a.s_name,a.s_sex,count(*) from student a  JOIN
student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex

-- 31、查询1990年出生的学生名单

select s_name from student where s_birth like '1990%'

-- 32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

select c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC

-- 33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩

select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85

-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数

select a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=(
select c_id from course where c_name ='数学') and b.s_score<60

思考：什么时候用GROUP BY ？

-- 35、查询所有学生的课程及分数情况；

select a.s_id,a.s_name,
SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
SUM(b.s_score) as  '总分'
from student a left join score b on a.s_id = b.s_id
left join course c on b.c_id = c.c_id
GROUP BY a.s_id,a.s_name
《《《 分析：Case具有两种格式。简单Case函数和Case搜索函数。
--简单Case函数
CASE sex
WHEN '1' THEN '男'
WHEN '2' THEN '女'
ELSE '其他' END
--Case搜索函数
CASE WHEN sex = '1' THEN '男'
WHEN sex = '2' THEN '女'
ELSE '其他' END》》》

-- 36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数；
select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id
left join student a on a.s_id=c.s_id where c.s_score>=70

-- 37、查询不及格的课程
select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id
where a.s_score<60

--38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名；
select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
where a.c_id = '01' and a.s_score>80

-- 39、求每门课程的学生人数
select count(*) from score GROUP BY c_id;

-- 40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩
-- 查询老师id
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
-- 查询最高分（可能有相同分数）
select MAX(s_score) from score where c_id='02'
-- 查询信息（综合以上两条）
select a.*,b.s_score,b.c_id,c.c_name from student a
LEFT JOIN score b on a.s_id = b.s_id
LEFT JOIN course c on b.c_id=c.c_id
where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')
and b.s_score in (select MAX(s_score) from score where c_id='02')

-- 41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score

-- 42、查询每门功课成绩最好的前两名（不懂）
-- 牛逼的写法
select a.s_id,a.c_id,a.s_score from score a
where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id

-- 43、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列
select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC

-- 44、检索至少选修两门课程的学生学号
select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2

-- 45、查询选修了全部课程的学生信息
select * from student where s_id in(
select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))

--46、查询各学生的年龄
-- 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') -
(case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age
from student;

-- 47、查询本周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

-- 48、查询下周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)

-- 49、查询本月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)

-- 50、查询下月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

-- 51.扩展这里着重介绍Not Exists（Exists用法类似），Not Exists 不存在

用法：Select  * from TableA  a  where Not Exists （Select * from TableB  b where a.id=b.id and a.name=b.name）;

1、Not Exists 用在where之后，且后面紧跟子查询语句（带括号）；

2、Not Exists(Exists) 并不关心子查询的结果具体是什么，只关心子查询有没有结果；

3、这条语句的意思，把TableA的记录逐条代入到子查询，如果子查询结果集为空，说明不存在，那么这条TableA的记录出现在最终结果集，否则被排除；

用法：Select * from TableA a where Not Exists (Select 1 from TableB);
--  用left join查询不在黑名单里边用户信息

SELECT a.uid,a.name FROM USER a LEFT JOIN banuser b ON a.uid=b.uid WHERE a.uid NOT IN(SELECT b.uid FROM banuser WHERE a.uid=b.uid)
--  用NOT EXISTS  方式查询不在黑名单里边用户信息
SELECT a.uid,a.name FROM USER a WHERE uid NOT EXISTS(SELECT uid FROM banuser ) 有问题--

-- 面试题
CREATE TABLE Course_score(
id VARCHAR(15),
name VARCHAR(20),
course VARCHAR(20),
c_score DOUBLE(5,2),
t_term VARCHAR(20)
);

INSERT INTO Course_score VALUES('1','张三','数学',80,'2015');
INSERT INTO Course_score VALUES('2','李四','语文',90,'2016');
INSERT INTO Course_score VALUES('3','王五','化学',70,'2017');
INSERT INTO Course_score VALUES('4','张三','语文',80,'2015');
INSERT INTO Course_score VALUES('5','张三','化学',90,'2015');

-- 51 查出每个学期每门课程分最高分记录（结果集包含全部5个字段）
SELECT c.* FROM (SELECT t_term term,course course,MAX(c_score) mscore
FROM course_score
GROUP BY t_term,course) maxsc LEFT JOIN course_score c ON maxsc.course = c.`course` AND maxsc.term = c.`t_term`
AND maxsc.mscore = c.`c_score` ORDER BY c.t_term asc

-- 第二种方法
SELECT a.*,MAX(a.c_score)AS max_score FROM course_score a LEFT JOIN course_score b ON a.course=b.course AND a.t_term = b.t_term
GROUP BY  a.t_term,a.course  ORDER BY b.t_term ASC

-- 52 HAVING和group搭配使用（新的表见网址https://www.jianshu.com/p/8b135d373df1）
当然提到GROUP BY 我们就不得不提到HAVING，HAVING相当于条件筛选，但它与WHERE筛选不同，HAVING是对于GROUP BY对象进行筛选。
我们举个例子：
每个部门人数都有了，那如果我们想要进一步知道员工人数大于30000的部门是哪些，这个时候就得用到HAVING了。
语句如下：
SELECT
( SELECT d.dept_name FROM departments d WHERE de.dept_no = d.dept_no ) AS 部门,
count( de.emp_no ) AS 人数
FROM
dept_emp de
WHERE
de.to_date = '9999-01-01'
GROUP BY
de.dept_no
HAVING
count( de.emp_no ) > 30000

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