标签:max inner 排序 相同 where esc 去重 姓名 mysql
1、查询所有的课程的名称以及对应的任课老师姓名 2、查询学生表中男女生各有多少人 3、查询物理成绩等于100的学生的姓名 4、查询平均成绩大于八十分的同学的姓名和平均成绩 5、查询所有学生的学号,姓名,选课数,总成绩 6、 查询姓李老师的个数 7、 查询没有报李平老师课的学生姓名 8、 查询物理课程比生物课程高的学生的学号 9、 查询没有同时选修物理课程和体育课程的学生姓名 10、查询挂科超过两门(包括两门)的学生姓名和班级 、查询选修了所有课程的学生姓名 12、查询李平老师教的课程的所有成绩记录 13、查询全部学生都选修了的课程号和课程名 14、查询每门课程被选修的次数 15、查询之选修了一门课程的学生姓名和学号 16、查询所有学生考出的成绩并按从高到低排序(成绩去重) 17、查询平均成绩大于85的学生姓名和平均成绩 18、查询生物成绩不及格的学生姓名和对应生物分数 19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名 20、查询每门课程成绩最好的前两名学生姓名 21、查询不同课程但成绩相同的学号,课程号,成绩 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称; 23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名; 24、任课最多的老师中学生单科成绩最高的学生姓名
#1、查询所有的课程的名称以及对应的任课老师姓名 SELECT course.cname, teacher.tname FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid; #2、查询学生表中男女生各有多少人 SELECT gender 性别, count(1) 人数 FROM student GROUP BY gender; #3、查询物理成绩等于100的学生的姓名 SELECT student.sname FROM student WHERE sid IN ( SELECT student_id FROM score INNER JOIN course ON score.course_id = course.cid WHERE course.cname = ‘物理‘ AND score.num = 100 ); #4、查询平均成绩大于八十分的同学的姓名和平均成绩 SELECT student.sname, t1.avg_num FROM student INNER JOIN ( SELECT student_id, avg(num) AS avg_num FROM score GROUP BY student_id HAVING avg(num) > 80 ) AS t1 ON student.sid = t1.student_id; #5、查询所有学生的学号,姓名,选课数,总成绩(注意:对于那些没有选修任何课程的学生也算在内) SELECT student.sid, student.sname, t1.course_num, t1.total_num FROM student LEFT JOIN ( SELECT student_id, COUNT(course_id) course_num, sum(num) total_num FROM score GROUP BY student_id ) AS t1 ON student.sid = t1.student_id; #6、 查询姓李老师的个数 SELECT count(tid) FROM teacher WHERE tname LIKE ‘李%‘; #7、 查询没有报李平老师课的学生姓名(找出报名李平老师课程的学生,然后取反就可以) SELECT student.sname FROM student WHERE sid NOT IN ( SELECT DISTINCT student_id FROM score WHERE course_id IN ( SELECT course.cid FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid WHERE teacher.tname = ‘李平老师‘ ) ); #8、 查询物理课程比生物课程高的学生的学号(分别得到物理成绩表与生物成绩表,然后连表即可) SELECT t1.student_id FROM ( SELECT student_id, num FROM score WHERE course_id = ( SELECT cid FROM course WHERE cname = ‘物理‘ ) ) AS t1 INNER JOIN ( SELECT student_id, num FROM score WHERE course_id = ( SELECT cid FROM course WHERE cname = ‘生物‘ ) ) AS t2 ON t1.student_id = t2.student_id WHERE t1.num > t2.num; #9、 查询没有同时选修物理课程和体育课程的学生姓名(没有同时选修指的是选修了一门的,思路是得到物理+体育课程的学生信息表,然后基于学生分组,统计count(课程)=1) SELECT student.sname FROM student WHERE sid IN ( SELECT student_id FROM score WHERE course_id IN ( SELECT cid FROM course WHERE cname = ‘物理‘ OR cname = ‘体育‘ ) GROUP BY student_id HAVING COUNT(course_id) = 1 ); #10、查询挂科超过两门(包括两门)的学生姓名和班级(求出<60的表,然后对学生进行分组,统计课程数目>=2) SELECT student.sname, class.caption FROM student INNER JOIN ( SELECT student_id FROM score WHERE num < 60 GROUP BY student_id HAVING count(course_id) >= 2 ) AS t1 INNER JOIN class ON student.sid = t1.student_id AND student.class_id = class.cid; #11、查询选修了所有课程的学生姓名(先从course表统计课程的总数,然后基于score表按照student_id分组,统计课程数据等于课程总数即可) SELECT student.sname FROM student WHERE sid IN ( SELECT student_id FROM score GROUP BY student_id HAVING COUNT(course_id) = (SELECT count(cid) FROM course) ); #12、查询李平老师教的课程的所有成绩记录 SELECT * FROM score WHERE course_id IN ( SELECT cid FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid WHERE teacher.tname = ‘李平老师‘ ); #13、查询全部学生都选修了的课程号和课程名(取所有学生数,然后基于score表的课程分组,找出count(student_id)等于学生数即可) SELECT cid, cname FROM course WHERE cid IN ( SELECT course_id FROM score GROUP BY course_id HAVING COUNT(student_id) = ( SELECT COUNT(sid) FROM student ) ); #14、查询每门课程被选修的次数 SELECT course_id, COUNT(student_id) FROM score GROUP BY course_id; #15、查询之选修了一门课程的学生姓名和学号 SELECT sid, sname FROM student WHERE sid IN ( SELECT student_id FROM score GROUP BY student_id HAVING COUNT(course_id) = 1 ); #16、查询所有学生考出的成绩并按从高到低排序(成绩去重) SELECT DISTINCT num FROM score ORDER BY num DESC; #17、查询平均成绩大于85的学生姓名和平均成绩 SELECT sname, t1.avg_num FROM student INNER JOIN ( SELECT student_id, avg(num) avg_num FROM score GROUP BY student_id HAVING AVG(num) > 85 ) t1 ON student.sid = t1.student_id; #18、查询生物成绩不及格的学生姓名和对应生物分数 SELECT sname 姓名, num 生物成绩 FROM score LEFT JOIN course ON score.course_id = course.cid LEFT JOIN student ON score.student_id = student.sid WHERE course.cname = ‘生物‘ AND score.num < 60; #19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名 SELECT sname FROM student WHERE sid = ( SELECT student_id FROM score WHERE course_id IN ( SELECT course.cid FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid WHERE teacher.tname = ‘李平老师‘ ) GROUP BY student_id ORDER BY AVG(num) DESC LIMIT 1 ); #20、查询每门课程成绩最好的前两名学生姓名 #查看每门课程按照分数排序的信息,为下列查找正确与否提供依据 SELECT * FROM score ORDER BY course_id, num DESC; #表1:求出每门课程的课程course_id,与最高分数first_num SELECT course_id, max(num) first_num FROM score GROUP BY course_id; #表2:去掉最高分,再按照课程分组,取得的最高分,就是第二高的分数second_num SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id; #将表1和表2联合到一起,得到一张表t3,包含课程course_id与该们课程的first_num与second_num SELECT t1.course_id, t1.first_num, t2.second_num FROM ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t1 INNER JOIN ( SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id ) AS t2 ON t1.course_id = t2.course_id; #查询前两名的学生(有可能出现并列第一或者并列第二的情况) SELECT score.student_id, t3.course_id, t3.first_num, t3.second_num FROM score INNER JOIN ( SELECT t1.course_id, t1.first_num, t2.second_num FROM ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t1 INNER JOIN ( SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id ) AS t2 ON t1.course_id = t2.course_id ) AS t3 ON score.course_id = t3.course_id WHERE score.num >= t3.second_num AND score.num <= t3.first_num; #排序后可以看的明显点 SELECT score.student_id, t3.course_id, t3.first_num, t3.second_num FROM score INNER JOIN ( SELECT t1.course_id, t1.first_num, t2.second_num FROM ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t1 INNER JOIN ( SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id ) AS t2 ON t1.course_id = t2.course_id ) AS t3 ON score.course_id = t3.course_id WHERE score.num >= t3.second_num AND score.num <= t3.first_num ORDER BY course_id; #可以用以下命令验证上述查询的正确性 SELECT * FROM score ORDER BY course_id, num DESC; -- 21、查询不同课程但成绩相同的学号,课程号,成绩 -- 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称; -- 23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名; -- 24、任课最多的老师中学生单科成绩最高的学生姓名
标签:max inner 排序 相同 where esc 去重 姓名 mysql
原文地址:https://www.cnblogs.com/zhuangshenhao/p/11997059.html