mysql的sql语句练习的2个网址

sql语句练习：

https://blog.csdn.net/mrbcy/article/details/68965271 完成。

https://blog.csdn.net/flycat296/article/details/63681089

hive语句综合练习

以下网址中的表为背景表，完成以下练习题：

https://blog.csdn.net/mrbcy/article/details/68965271

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

LEFT 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'  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'

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 ALL

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 techer 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 LEFT JOIN score b ON a.s_id = b.s_id WHERE b.c_id  IN  (

SELECT c.c_id

FROM course c LEFT JOIN techer t ON c.t_id = t.t_id WHERE t.t_name = '张三'

) ;

-- 8、查询没学过"张三"老师授课的同学的信息

SELECT s.*

FROM student s LEFT JOIN (

SELECT a.s_id

FROM student a LEFT JOIN score b ON a.s_id = b.s_id WHERE b.c_id  IN  (

SELECT c.c_id

FROM course c LEFT JOIN techer t ON c.t_id = t.t_id WHERE t.t_name = '张三'

)

) ss ON s.s_id = ss.s_id WHERE ss.s_id IS NULL;

-- 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 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 qq.*

FROM (

SELECT s.*

FROM student s LEFT JOIN score sco ON s.s_id = sco.s_id LEFT JOIN course c ON sco.c_id = c.c_id WHERE c.c_id='01'

) qq

LEFT JOIN (

SELECT stu.*

FROM student stu LEFT JOIN score mysco ON stu.s_id = mysco.s_id LEFT JOIN course cou ON mysco.c_id = cou.c_id WHERE cou.c_id='02'

) pp

ON qq.s_id = pp.s_id WHERE pp.s_id IS NULL;

-- 11、查询没有学全所有课程的同学的信息

SELECT ss.s_id

FROM (

SELECT stu.s_id,COUNT(stu.s_id) AS num

FROM student stu LEFT JOIN score sco ON stu.s_id = sco.s_id LEFT JOIN course cou ON sco.c_id = cou.c_id

GROUP BY stu.s_id

) ss WHERE ss.num < 3

-- 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 stu.*

FROM student stu LEFT JOIN

(

SELECT s.s_id

FROM score s WHERE s.c_id IN(

SELECT  c_id FROM score WHERE s_id = '01'

)GROUP BY s_id

) pp ON stu.s_id = pp.s_id WHERE pp.s_id IS NOT  NULL;