Hive全实现经典五十题（持续优化）

其他 2020-01-26 10:39:27 阅读次数: 0

学生课程成绩老师，四张表

  • 学生表

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  • 成绩表

在这里插入图片描述

  • 课程表

在这里插入图片描述

  • 老师表

在这里插入图片描述

  1. 查询"01"课程比"02"课程成绩高的学生的信息及课程分数

    //连接表时对连接的表加条件
select s.s_name,s.s_id,s.s_birth,s.s_sex,sc1.s_score
from student s
left join score sc1
on s.s_id=sc1.s_id and sc1.c_id=01
left join score sc2
on s.s_id=sc2.s_id and sc2.c_id=02
where sc1.s_score>sc2.s_score;
//
select sc1.s_id,sc1.c_id,sc1.s_score
from
(
select *
from score where c_id=01
)sc1
left join 
(
select *
from score where c_id=02
)sc2
on sc1.s_id=sc2.s_id
where sc1.s_score>sc2.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 s.s_name,s.s_id,s.s_birth,s.s_sex,sc1.s_score
from student s
left join score sc1
on s.s_id=sc1.s_id and sc1.c_id=01
left join score sc2
on s.s_id=sc2.s_id and sc2.c_id=02
where sc1.s_score < sc2.s_score;

select sc1.s_id,sc1.c_id,sc1.s_score
from
(
select *
from score where c_id=01
)sc1
left join 
(
select *
from score where c_id=02
)sc2
on sc1.s_id=sc2.s_id
where sc1.s_score<sc2.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 s.s_id,s.s_name,avg(sc.s_score) avg_score
from student s
left join score sc
on s.s_id=sc.s_id
group by s.s_id,s.s_name
having avg(sc.s_score)>60;
-一步应该最高效了普通办法

  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 s.s_id,s.s_name,avg(sc.s_score) avg_score
from student s
left join score sc
on s.s_id=sc.s_id
group by s.s_id,s.s_name
having avg(sc.s_score)<60;

  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 s.s_id,s.s_name,count(*) countSelect,sum(sc.s_score) sumScore
from student s
left join score sc
on s.s_id=sc.s_id
group by s.s_id,s.s_name;

  6. 查询"李"姓老师的数量

    select count(*) from teacher where t_name like "李%";

  7. 查询学过"张三"老师授课的同学的信息

    //打出后三张大表，student直接join
select *
from student s
where s.s_id in(
SELECT sc.s_id
from score sc
LEFT JOIN course co 
ON sc.c_id=co.c_id 
LEFT JOIN teacher te 
ON co.t_id=te.t_id
where t_name='张三'
);


Launching Job 3 out of 3
Time taken: 3.627 seconds, Fetched: 6 row(s)

//
select *
from student s
where s.s_id in
(
select sc.s_id
from score sc
join
(
select c.c_id
from course c
where c.t_id in
(
select t.t_id
from teacher t
where t.t_name='张三'
)
)t2
on sc.c_id=t2.c_id
)


//
select *
from student s
where exists
(
select sc.s_id
from score sc
join
(
select c.c_id
from course c
where c.t_id in
(
select t.t_id
from teacher t
where t.t_name='张三'
)
)t2
on sc.c_id=t2.c_id
where s.s_id=sc.s_id
)

//
SELECT stu.* 
FROM student stu 
JOIN score sc 
ON stu.s_id=sc.s_id 
JOIN course co 
ON sc.c_id=co.c_id 
JOIN teacher te 
ON co.t_id=te.t_id 
WHERE t_name='张三'




//全连接法
SELECT stu.s_id,stu.s_name,stu.s_birth,stu.s_sex
FROM student stu 
LEFT JOIN score sc 
ON stu.s_id=sc.s_id 
LEFT JOIN course co 
ON sc.c_id=co.c_id 
LEFT JOIN teacher te 
ON co.t_id=te.t_id
where t_name='张三';

Launching Job 3 out of 3
Time taken: 3.693 seconds, Fetched: 6 row(s)

//
select s.s_id,s.s_name,s.s_birth,s.s_sex
from student s,
(
select
s.s_id
from score s
left join
(
select c.c_id
from course c,(select t_id tid from teacher where t_name='张三')t
where c.t_id=t.tid
)t2
on s.c_id=t2.c_id
where s.c_id=t2.c_id
)t3
where s.s_id=t3.s_id

Launching Job 3 out of 3
Time taken: 3.689 seconds, Fetched: 6 row(s)

  8. 查询没学过"张三"老师授课的同学的信息

    select *
from student s
where s.s_id not in(
SELECT stu.s_id
FROM student stu 
LEFT JOIN score sc 
ON stu.s_id=sc.s_id 
LEFT JOIN course co 
ON sc.c_id=co.c_id 
LEFT JOIN teacher te 
ON co.t_id=te.t_id
where t_name='张三'
);

Launching Job 9 out of 9
Time taken: 10.695 seconds, Fetched: 2 row(s)

select *
from student s
where s.s_id not in(
SELECT sc.s_id
from score sc
LEFT JOIN course co 
ON sc.c_id=co.c_id 
LEFT JOIN teacher te 
ON co.t_id=te.t_id
where t_name='张三'
);

Launching Job 7 out of 7
Time taken: 8.351 seconds, Fetched: 2 row(s)

  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

    select *
from student s
left join score sc1
on s.s_id=sc1.s_id and sc1.c_id=01
left join score sc2
on s.s_id=sc2.s_id and sc2.c_id=02
where sc1.s_score is not null and sc2.s_score is not null;

  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 s.s_id,s.s_name,s.s_birth,s.s_sex
from student s
left join score sc1
on s.s_id=sc1.s_id and sc1.c_id=01
left join score sc2
on s.s_id=sc2.s_id and sc2.c_id=02
where sc1.s_score is not null and sc2.s_score is null;

  11. 查询没有学全所有课程的同学的信息

    select sc.s_id
from score sc,(select count(*) cou from course) cou_course
group by sc.s_id,cou_course.cou
having count(sc.s_score)<cou_course.cou

  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 s.*
from student s
left join score sc1
on s.s_id=sc1.s_id and sc1.c_id='01'
left join score sc2
on s.s_id=sc2.s_id and sc2.c_id='02'
left join score sc3
on s.s_id=sc3.s_id and sc3.c_id='03'
where sc1.c_id!=null or sc2.c_id!=null or sc3.c_id!=null and s.s_id!='01'
;

select distinct(s.s_name),s.s_sex,s.s_birth,s.s_id
from student s
join score sc
on s.s_id=sc.s_id
where s.s_id!='01' and sc.c_id in
(
select
c_id
from score 
where 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

    //将01同学的数课程转化为数组和其他所有同学按照课程转化为数组的表join
select t1.s_name,t1.s_id
from
(
select s.s_name,s.s_id,collect_list(sc.c_id) arr_cid
from student s
join score sc
on s.s_id=sc.s_id
group by s.s_name,s.s_id
)t1
join
(
select
collect_list(c_id) arr_cid
from score 
where s_id='01'
)t2
on t1.arr_cid=t2.arr_cid
where t1.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 *
from student s
where s.s_id not in(
SELECT sc.s_id
from score sc
LEFT JOIN course co 
ON sc.c_id=co.c_id 
LEFT JOIN teacher te 
ON co.t_id=te.t_id
where t_name='张三'
);

  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

    //先得到不及格的所有同学然后分组统计
select t.s_name,t.s_id,round(avg(t.s_score),2)
from
(
select s.s_name,s.s_id,sc.s_score
from
student s
join score sc
on s.s_id=sc.s_id
where sc.s_score < 60
)t
group by t.s_id,t.s_name
having count(*)>=2

  16. 检索"01"课程分数小于60，按分数降序排列的学生信息

    SELECT *
FROM student s
LEFT JOIN score sc
ON s.s_id=sc.s_id
WHERE c_id='01'
AND s_score<60
ORDER BY  s_score

  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

    //直接窗口函数取最前面到最后面的平均值
select s_id,s_score,
round(avg(nvl(s_score,0)) over(partition by s_id order by s_score desc rows between unbounded preceding and unbounded following ),1) avg_sc
from score

  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

    //按照成绩列将满足条件置为1不满足为0，然后进行求和就得出了每一个条件下的总人数，然后除以总人数

select c.c_id,c.c_name,
round(sum(case when sc.s_score >= 90 then 1 else 0 end)/count(*),1) exce,
round(sum(case when sc.s_score>=80 and sc.s_score<90 then 1 else 0 end)/count(*),1) good,
round(sum(case when sc.s_score >=70 and sc.s_score <80 then 1 else 0 end)/count(*),1) mid,
round(sum(case when sc.s_score >= 60 and sc.s_score <70 then 1 else 0 end)/count(*),1) seve,
max(sc.s_score) max_sc,
min(sc.s_score) min_sc,
round(avg(sc.s_score),1) avg_sc
from course c
left join score sc
on c.c_id=sc.c_id
group by c.c_id,c.c_name;


//

{"good":"0.33","notseve":"0.33","mid":"0.33"}
{"good":"0.5","seve":"0.17","notseve":"0.17","exce":"0.17"} 
{"good":"0.33","notseve":"0.33","exce":"0.33"}

//
一个新列是成绩的分类，然后是max等其他，将分类和人数除以总人数合并（cancat），然后将每一科的所有拼接的字符串转为list然后拼为总字符串然后转为map，

select t2.c_id,
t2.c_name,
t2.max_sc,
t2.min_sc,
t2.avg_sc,
nvl(t2.type['exce'],0) exec,
nvl(t2.type['good'],0) good,
nvl(t2.type['mid'],0) mid,
nvl(t2.type['seve'],0) seve
from
(
select t1.c_id,
t1.c_name,
str_to_map(concat_ws(',',collect_list(t1.type)),',','_') type,
t1.max_sc,
t1.min_sc,
t1.avg_sc
from
(
select t.c_id,
t.c_name,
concat(t.type,'_',round(count(*)/t.cou_,2)) type,
t.max_sc,
t.min_sc,
t.avg_sc
from
(
select c.c_id,c.c_name,
case when sc.s_score >= 90 then 'exce' 
when sc.s_score>=80 and sc.s_score<90 then 'good'
when sc.s_score >=70 and sc.s_score <80 then 'mid'
when sc.s_score >= 60 and sc.s_score <70 then 'seve'
else 'notseve' end type,
count(c.c_name) over(partition by c.c_name) cou_,
max(sc.s_score) over(partition by c.c_id) max_sc,
min(sc.s_score) over(partition by c.c_id) min_sc,
round(avg(sc.s_score) over(partition by c.c_id),1) avg_sc
from course c
left join score sc
on c.c_id=sc.c_id
)t
group by t.c_id,t.c_name,t.type,t.cou_,t.max_sc,t.min_sc,t.avg_sc
)t1
group by t1.c_id,t1.c_name,t1.max_sc,t1.min_sc,t1.avg_sc
)t2
group by t2.c_id,t2.c_name,t2.max_sc,t2.min_sc,t2.avg_sc,t2.type

  19. 按各科成绩进行排序，并显示排名(实现不完全)

    select s.*,rank() over(partition by s.c_id order by s.s_score) rank
from score s

  20. 查询学生的总成绩并进行排名

    select s_name,sum(nvl(s_score,0)) su_sc
from student s 
left join score sc
on s.s_id=sc.s_id
group by s_name
order by su_sc desc

  21. 查询不同老师所教不同课程平均分从高到低显示

    select c.t_id,c.c_name,round(avg(nvl(sc.s_score,0)),2) avg_score
from course c
left join score sc
on c.c_id=sc.c_id
group by c.t_id,c.c_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
t1.*
from
(
select s.*,sc.c_id,sc.s_score,
row_number() over(partition by sc.c_id order by sc.s_score desc) rank
from score sc
left join student s
on sc.s_id=s.s_id
)t1
where t1.rank>1 and rank <=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 c.c_id,c.c_name,
round(sum(case when sc.s_score>85 and sc.s_score <=100 then 1 else 0 end)/count(*),1) `[100-85]`,
round(sum(case when sc.s_score>85 and sc.s_score <=85 then 1 else 0 end)/count(*),1) `[85-70]`,
round(sum(case when sc.s_score>60 and sc.s_score <=70 then 1 else 0 end)/count(*),1) `[70-60]`,
round(sum(case when sc.s_score>0 and sc.s_score <=60 then 1 else 0 end)/count(*),1) `[0-60]`
from score sc
join course c
on sc.c_id=c.c_id
group by c.c_id,c.c_name

  24. 查询学生平均成绩及其名次

    select sc.s_id,
avg(if null(sc.s_score,0)) avg_sc ,
row_number() over(sort by sc.avg)
from score sc
group by sc.s_id

  25. 查询各科成绩前三名的记录

    select t.*
from
(
select sc.*,
row_number() over(partition by sc.c_id order by sc.s_score desc) rank
from score sc
)t
where t.rank <=3;

  26. 查询每门课程被选修的学生数

    select c_id,count(*) cou_per
from score
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

    select s_name,s.s_id
from score sc
left join student s
on sc.s_id=s.s_id
group by s.s_id,s_name
having count(*)=2;

  28. 查询男生、女生人数

    SELECT s_sex,COUNT(*) cou_per
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 s.s_name,s.s_sex,count(*) cou_per
from student s
group by s.s_name,s.s_sex

  31. 查询1990年出生的学生名单

    //
SELECT s_name 
FROM student
WHERE s_birth LIKE '1990%'

//
select s_name
from student
where year(s_birth)='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 avg(s_score)
from score
group by c_id
order by avg(s_score) desc, (c_id) avg

  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 s.s_id,s.s_name,avg(sc.s_score) avg_sc
from student s
left join score sc
on s.s_id = sc.s_id
group by sc.s_score
having avg_sc>=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 s_name,s_score
FROM student stu LEFT JOIN score sc ON stu.s_id=sc.s_id
LEFT JOIN course co ON sc.c_id=co.c_id
WHERE s_score < 60 AND c_name ='数学'

  35. 查询所有学生的课程及分数情况；

    SELECT s_name,c_name,s_score
FROM student stu 
LEFT JOIN score sc 
ON stu.s_id=sc.s_id

  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 s.s_name,c.c_name,sc.s_score
FROM student s
LEFT JOIN score sc 
ON s.s_id=sc.s_id
LEFT JOIN course c 
ON sc.c_id=c.c_id
where sc.s_score>70

  37. 查询不及格的课程

    //
SELECT s.s_name,sc.c_name,sc.s_score
FROM student s
LEFT JOIN score sc 
ON s.s_id=sc.s_id
WHERE sc.s_score<60

//
select c_id,avg(score)
from score 
group by c_id
having avg(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 stu.s_id,s_name
FROM student stu 
LEFT JOIN score sc 
ON stu.s_id=sc.s_id
LEFT JOIN course co 
ON sc.c_id=co.c_id
WHERE co.c_id=01
AND s_score>80

  39. 求每门课程的学生人数

    select c_id,count(*) cou_per
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

    select s.*,sc.s_score
FROM student s 
LEFT JOIN score sc 
ON s.s_id=sc.s_id 
LEFT JOIN course c 
ON sc.c_id=c.c_id 
LEFT JOIN teacher te 
ON c.t_id=te.t_id
where te.t_name='张三'
order by sc.s_score
limit 1;

  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(sc1.s_id),sc1.c_id,sc1.s_score
from score sc1,score sc2
where sc1.c_id!=sc2.c_id 
and sc1.s_score=sc2.s_score

  42. 查询每门功成绩最好的前两名

    select
t.s_id,t.c_id,
t.rank
from
(
select
s_id,c_id,
row_number() over(partition by c_id order by s_score desc) rank
from score
)t
where t.rank <= 2

  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(*)
FROM score
GROUP BY c_id
HAVING COUNT(*)>5
ORDER BY COUNT(*) DESC,c_id ASC

  44. 检索至少选修两门课程的学生学号

    select sc.s_id
from score sc
group by s.s_id
having count(*)>=2

  45. 查询选修了全部课程的学生信息

    select
from score sc,count(select count(*) cou_per from course c) cou_per
where 
group by sc.s_id
having count(*) > cou_per

  46. 查询各学生的年龄

    methon1：
自定义UDF函数上传使用
public class AgeCalculate extends UDF {
    public int evaluate(String str) {
        int age = 0;
        if (str.matches("\\d{4}(\\-|\\/|.)\\d{1,2}\\1\\d{1,2}")) {
            String[] dateStr = str.split("-");
            int year = Integer.parseInt(dateStr[0]);
            int month = Integer.parseInt(dateStr[1]);
            int day = Integer.parseInt(dateStr[2]);
            SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-MM-dd");
            String nowDate = simpleDateFormat.format(new Date());
            String[] now = nowDate.split("-");
            int nowYear = Integer.parseInt(now[0]);
            int nowmonth = Integer.parseInt(now[1]);
            int nowDay = Integer.parseInt(now[2]);
            if (nowmonth==month && nowDay <= day ){
                age=nowYear-year-1;
            }else if (nowmonth< month){
                age=nowYear-year-1;
            }else {
                age=nowYear-year;
            }
        } else {
            System.out.println("you input don't matche str");
        }
        return age;
    }

    public static void main(String[] args) {
        new AgeCalculate().evaluate("2010-9-2");
    }

}

//
methon2：
select s_name,from_unixtime(curdate,'yyyy')-year(s_birth) 'age'
from student

  47. 查询本周过生日的学生

    select *
from student
where weekofyear(s_birth)=weekofyear(current_date());

  48. 查询下周过生日的学生

    select *
from student
where weekofyear(s_birth)=weekofyear(current_date())+1;

  49. 查询本月过生日的学生

    select * from student where month(s_birth)=month(current_date);

  50. 查询下月过生日的学生

    select * from student where month(s_birth)=month(current_date)+1;
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