MySQL练习题

-- select s_id from student;
-- select c_id from course where c_name = 'python'
 
-- SELECT s_id,num from score where c_id in(select c_id from course where c_name = 'python') and s_id in(select s_id from student);
-- SELECT s_id,num from score where c_id in(select c_id from course where c_name = 'java') and s_id in(select s_id from student)
#查询学习课程"python"比课程 "java" 成绩高的学生的学号;
-- SELECT id from (SELECT s_id as id,num from score where c_id in(select c_id from course where c_name = 'python') <br>and s_id in(select s_id from student)) as p1
-- INNER JOIN <br>(SELECT s_id,num from score where c_id in(select c_id from course where c_name = 'java') <br>and s_id in(select s_id from student)) as p2 on p1.id = p2.s_id
-- where p1.num > p2.num;
#查询平均成绩大于65分的同学的姓名和平均成绩(保留两位小数); #ROUND(n,小数的位数)
-- SELECT avg(num) from score GROUP BY s_id ;
-- select s_name,ROUND(mid,2) from student s LEFT JOIN (SELECT avg(num)as mid,s_id from score GROUP BY s_id
-- ) as p on s.s_id = p.s_id where mid >65;
#查询所有同学的姓名、选课数、总成绩；
-- select s_name,c_num,s_sum from 
-- (select s_id,sum(num)as s_sum,count(c_id) as c_num from score GROUP BY s_id) as t
-- LEFT JOIN student on t.s_id = student.s_id
#查询所有的课程的名称以及对应的任课老师姓名;
-- select c_name,t_name from
-- (select c_name,t_id from course) as t LEFT JOIN teacher on t.t_id = teacher.t_id
#查询没学过“alex”老师课的同学的姓名;
#先查学过的，最后在否定一下。
-- select s_name from student where s_id not in
-- (SELECT s_id from score where c_id in (SELECT c_id from course where t_id  in(select t_id from teacher where t_name = 'alex')))
#in --or
#查询挂科超过两门(包括两门)的学生姓名
 
-- select s_name from student as s LEFT JOIN
-- (select s.s_id,sum(case WHEN s.num<= 60 THEN 1 ELSE 0 end) as number
-- from score as s GROUP BY s.s_id) as n on s.s_id = n.s_id WHERE number >=2
#查询学过'python'并且也学过编号'java'课程的同学的姓名;
-- select score.s_id,count(*)as number from course,score where course.c_id = score.c_id and course.c_name in ('python','java') <br>GROUP BY score.s_id HAVING number >=2
#在没有分组的时候的行数就是count(*)，就是count(*)相当于行数，但是不能都表示出来，如果要表示就要GROUP_CONCAT
-- SELECT student.s_name from student,
-- (select score.s_id,count(*)as number from course,score where course.c_id = score.c_id and course.c_name in ('python','java') <br>GROUP BY score.s_id HAVING number >=2) as s where student.s_id = s.s_id
#查询学过'貂蝉'同学全部课程 的其他同学姓名；
-- SELECT s_id,count(c_id) FROM score WHERE c_id in 
-- (SELECT n.c_id from student as s,score as n WHERE s.s_id =n.s_id and s.s_name = '貂蝉') <br>GROUP BY s_id HAVING count(c_id)>=2 and s_id!= (SELECT s_id from student where s_name = '貂蝉')
 
#用count(*)
-- select student.s_name from student,score where student.s_id = score.s_id and score.c_id in
-- (select c_id from student,score where student.s_id = score.s_id and student.s_name = '貂蝉') and student.s_name !='貂蝉'<br>GROUP BY student.s_id 
-- HAVING count(*) >= (select count(*) from student,score where student.s_id = score.s_id and student.s_name = '貂蝉')
#查询和'貂蝉'同学学习的课程完全相同的,其他同学姓名；
-- SELECT student.s_name from student,
-- (SELECT score.s_id,
-- sum(case when c_id in <br>(SELECT n.c_id from student as s,score as n WHERE s.s_id =n.s_id and s.s_name = '貂蝉')THEN 1 ELSE -1 END ) AS N
-- from score where score.s_id !=(SELECT student.s_id from student where s_name = '貂蝉')   GROUP BY score.s_id HAVING N=2) as m
-- where student.s_id = m.s_id
#先通过貂蝉的科目数 筛选 与貂蝉选相同科目的人 ，然后通过具体的科目来筛选；
 
##count(*)相当于没有分组前的行数
#SELECT student.s_id  from student,score where student.s_id = score.s_id GROUP BY score.s_id HAVING count(*)=
#(SELECT count(*) FROM student,score where student.s_id = score.s_id and student.s_name = '貂蝉')  ---(2,5)
 
-- SELECT DISTINCT student.s_name  from student,score where student.s_id = score.s_id and score.s_id in (2,5) AND score.c_id in
-- (SELECT score.c_id FROM student,score where student.s_id = score.s_id and student.s_name = '貂蝉') and student.s_name != '貂蝉'
 
#按平均成绩倒序显示所有学生的“python”、“java”、“linux”三门的课程成绩，按如下形式显示： 学生ID,python,java,linux,课程数,平均分
-- SELECT num from score,course where score.c_id = course.c_id and course.c_name ='python' and score.s_id = 1
 
#一个数据就可以作为一个字段，即把一个结果集作为一个字段；
-- SELECT s.s_id as '学生ID',
-- (SELECT num from score,course where score.c_id = course.c_id and course.c_name ='python' and score.s_id = s.s_id) as 'python',
-- (SELECT num from score,course where score.c_id = course.c_id and course.c_name ='java' and score.s_id = s.s_id) as 'java',
-- (SELECT num from score,course where score.c_id = course.c_id and course.c_name ='linux' and score.s_id = s.s_id) as 'linux',
-- count(*) as '课程数',
-- ROUND(avg(s.num),2) as '平均分'
-- from score as s GROUP BY s.s_id ORDER BY '平均分' DESC
-- 
#查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列
-- SELECT course.c_name,avg(num)  FROM score,course where score.c_id = course.c_id <br>GROUP BY score.c_id ORDER BY avg(num) ASC , score.c_id DESC
#按条件排序，可以按照多个条件，如果第一个条件不满足，则按照第二个条件；即 平均成绩相同时，按课程号降序排列
posted on 2019-03-24 20:05 小辉python 阅读(184) 评论(0) 编辑收藏举报