create table student(
s_id string comment ‘学生编号‘,
s_name string comment ‘学生姓名‘,
s_birth string comment ‘学生生日‘,
s_sex string comment ‘学生性别‘) comment ‘学生表‘
row format delimited
fields terminated by ‘ ‘;

create table score(
s_id string comment ‘学生编号‘,
c_id string comment ‘课程编号‘,
s_score int comment ‘分数‘) comment ‘课程表‘
row format delimited
fields terminated by ‘ ‘;

 

create table course(
c_id string comment ‘学生编号‘,
c_name string comment ‘课程名称‘,
t_id string comment ‘教师编号‘) comment ‘课程表‘
row format delimited
fields terminated by ‘ ‘;

 

create table teacher(
t_id string comment ‘教师编号‘,
t_name string comment ‘教师姓名‘) comment ‘教师表‘
row format delimited
fields terminated by ‘ ‘;

 

[root@ke04 0516]# cat student 
01 赵雷 1990-01-01 男
02 张三 1990-12-21 男
03 迪丽热巴 2000-01-01 女
04 雪见 2005-03-04 女
05 梅超风 1988-04-06 女
06 李云龙 1999-12-12 男
07 蒸煮 1989-07-01 男
08 王菊 1990-01-20 女

[root@ke04 0516]# cat score
01 01 80
01 02 90
01 03 99
02 01 70
02 02 60
02 03 80
03 01 80
03 02 80
03 03 80
04 01 50
04 02 30
04 03 20
05 01 76
05 02 87
06 01 31
06 03 34
07 02 89
07 03 98

 

[root@ke04 0516]# cat course
01 语文 02
02 数学 01
03 英语 03

 

[root@ke04 0516]# cat teacher
01 张三
02 李四
03 王五

load data local inpath ‘/tmp/bigdata/0516/student‘ into table student;
load data local inpath ‘/tmp/bigdata/0516/score‘ into table score;
load data local inpath ‘/tmp/bigdata/0516/course‘ into table course;
load data local inpath ‘/tmp/bigdata/0516/teacher‘ into table teacher;

select stu.*, a.s_score as 01_score, b.s_score as 02_score
from student stu
join score a on stu.s_id = a.s_id and a.c_id = ‘01‘
join score b on stu.s_id = b.s_id and b.c_id = ‘02‘
where a.s_score > b.s_score;

02    张三    1990-12-21    男    70    60
04    雪见    2005-03-04    女    50    30


方案二: hive不支持子查询 所以查询报错

select stu.* from student stu 
where stu.s_id in 
(select a.s_id from score a 
where a.c_id = ‘01‘ and a.s_score > 
(select max(b.s_score) from score b where b.c_id = ‘02‘));

select stu.s_id, stu.s_name, round(avg(score.s_score), 1) as avg 
from student stu 
join score b 
on stu.s_id=b.s_id group by stu.s_id,stu.s_name having avg >=60;


select stu.s_id, stu.s_name, tmp.avgscore as score from student stu 
join (select score.s_id, round(avg(score.s_score), 1) as avgscore
from score group by s_id) as tmp 
on tmp.s_id = stu.s_id where tmp.avgscore >= 60;

01    赵雷    89.7
02    张三    70.0
03    迪丽热巴    80.0
05    梅超风    81.5
07    蒸煮    93.5

hive不支持not in, 可以使用1. left join isnull  2.not exist 解决
select stu2.s_id, stu2.s_name, 0 as avgScore from student stu2 
where stu2.s_id not in
(select distinct(sc2.s_id) from score sc2);
替换一:
select stu2.s_id, stu2.s_name, 0 as avgScore from student stu2 
left join score sc2 
on stu2.s_id = sc2.s_id 
where sc2.s_id is null;
替换二:
select stu2.s_id, stu2.s_name, 0 as avgScore from student stu2 
where not exists
(select distinct(sc2.s_id) from score sc2 where sc2.s_id = stu2.s_id);




方案一:
select stu.s_id, stu.s_name, tmp.avgScore from student as stu
join (select sc.s_id, round(avg(sc.s_score), 1) as avgScore from score sc 
group by sc.s_id having round(avg(sc.s_score), 1) < 60 ) as tmp
on stu.s_id = tmp.s_id 
union all
select stu2.s_id, stu2.s_name, 0 as avgScore from student stu2 
left join score sc2 
on stu2.s_id = sc2.s_id 
where sc2.s_id is null;


方案二: having不支持别名查询
select stu.s_id, stu.s_name, round(avg(sc.s_score), 1) avgScore from student stu
join score sc on stu.s_id = sc.s_id 
group by stu.s_id, stu.s_name having round(avg(sc.s_score), 1) < 60
union all
select stu2.s_id, stu2.s_name, 0 as avgScore from student stu2 
where not exists
(select distinct(sc2.s_id) from score sc2 where sc2.s_id = stu2.s_id);


08    王菊    0.0
04    雪见    33.3
06    李云龙    32.5

select t_name, count(1) from
teacher where t_name like ‘李%‘ group by t_name;

李四    1

select stu.s_id, stu.s_name, count(sc.c_id) as total_count, sum(sc.s_score) as total_score
from student stu left join score sc 
on stu.s_id = sc.s_id group by stu.s_id, stu.s_name;


01    赵雷    3    269
02    张三    3    210
03    迪丽热巴    3    240
04    雪见    3    100
05    梅超风    2    163
06    李云龙    2    65
07    蒸煮    2    187
08    王菊    0    NULL

select stu.s_id, stu.s_name, stu.s_birth, stu.s_sex
from student stu 
join score on stu.s_id = score.s_id
join course on score.c_id = course.c_id 
join teacher t on course.t_id = t.t_id and t.t_name =‘张三‘;


01    赵雷    1990-01-01    男
02    张三    1990-12-21    男
03    迪丽热巴    2000-01-01    女
04    雪见    2005-03-04    女
05    梅超风    1988-04-06    女
07    蒸煮    1989-07-01    男

7.查询没学过‘张三‘老师授课的同学的信息

select stu.s_id, stu.s_name, stu.s_birth, stu.s_sex
from student stu left join 
(select s_id from score 
join course on course.c_id = score.c_id
join teacher on course.t_id =teacher.t_id and teacher.t_name = ‘张三‘ )tmp
on stu.s_id = tmp.s_id where tmp.s_id is null;

select stu.s_id, stu.s_name, stu.s_birth, stu.s_sex
from student stu where not exists
(select s_id from score 
join course on course.c_id = score.c_id
join teacher on course.t_id =teacher.t_id and teacher.t_name = ‘张三‘ 
where score.s_id = stu.s_id);

06    李云龙    1999-12-12    男
08    王菊    1990-01-20    女

select stu.s_id, stu.s_name, stu.s_birth, stu.s_sex
from student stu join
(select s_id from score where c_id = ‘1‘) tmp1 
on stu.s_id = tmp1.s_id
join (select s_id from score where c_id = ‘2‘) tmp2
on stu.s_id = tmp2.s_id;

select stu.* from student stu join
(select s_id from score where c_id = ‘1‘)tmp1
on stu.s_id = tmp1.s_id 
left join (select s_id from score where c_id = ‘2‘)tmp2
on stu.s_id = tmp2.s_id where tmp2.s_id is null;

select stu.* from student stu join
(select count(c_id) num from course) tmp1 left join
(select s_id, count(c_id) num from score group by s_id) tmp2
on stu.s_id = tmp2.s_id and tmp1.num = tmp2.num
where tmp2.s_id is null;


select stu.* from student stu 
left join (select s_id, count(c_id) num from score group by s_id) tmp2 on stu.s_id = tmp2.s_id
join (select count(c_id) num from course) tmp1 
on tmp1.num = tmp2.num 
where tmp2.s_id is null;

select stu.* from student stu join
(select s_id, c_id from score) tmp1 join 
(select c_id from score where score.s_id=‘01‘) tmp2 
on tmp1.c_id = tmp2.c_id and stu.s_id = tmp1.s_id
group by stu.s_id, stu.s_name, stu.s_birth, stu.s_sex;

01    赵雷    1990-01-01    男
02    张三    1990-12-21    男
03    迪丽热巴    2000-01-01    女
04    雪见    2005-03-04    女
05    梅超风    1988-04-06    女
06    李云龙    1999-12-12    男
07    蒸煮    1989-07-01    男

select stu.*, tmp1.course_id from student stu join
(select s_id, concat_ws(‘|‘, collect_set(c_id)) course_id 
from score group by s_id having s_id not in (‘1‘)) tmp1
on stu.s_id = tmp1.s_id join
(select concat_ws(‘|‘, collect_set(c_id)) course_id 
from score where s_id = ‘1‘) tmp2
on tmp2.course_id = tmp1.course_id;

select stu.* from student stu join
(select s_id from score tmp1 join 
(select c_id from course join teacher on course.t_id = teacher.t_id
where teacher.t_name=‘张三‘) tmp2
on tmp1.c_id = tmp2.c_id) tmp3
on stu.s_id = tmp3.s_id where tmp3.s_id is null;