网上流传的50道sql练习，发现其答案部分有问题，故在此记录下个人用mysql 的实现。
数据表介绍
--1.学生表
Student(SId,Sname,Sage,Ssex)
--SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
--2.课程表
Course(CId,Cname,TId)
--CId 课程编号,Cname 课程名称,TId 教师编号
--3.教师表
Teacher(TId,Tname)
--TId 教师编号,Tname 教师姓名
--4.成绩表
SC(SId,CId,score)
--SId 学生编号,CId 课程编号,score 分数
学生表 Student
-- ---------------------------- -- Table structure for `student` -- ---------------------------- DROP TABLE IF EXISTS `student`; CREATE TABLE `student` ( `SId` varchar(10) DEFAULT NULL, `Sname` varchar(10) DEFAULT NULL, `Sage` datetime DEFAULT NULL, `Ssex` varchar(10) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of student -- ---------------------------- INSERT INTO `student` VALUES ('01', '赵雷', '1990-05-18 00:00:00', '男'); INSERT INTO `student` VALUES ('02', '钱电', '1990-05-24 00:00:00', '男'); INSERT INTO `student` VALUES ('03', '孙风', '1990-05-20 00:00:00', '男'); INSERT INTO `student` VALUES ('04', '李云', '1990-05-25 00:00:00', '男'); INSERT INTO `student` VALUES ('05', '周梅', '1991-12-01 00:00:00', '女'); INSERT INTO `student` VALUES ('06', '吴兰', '1992-01-01 00:00:00', '女'); INSERT INTO `student` VALUES ('07', '郑竹', '1989-01-01 00:00:00', '女'); INSERT INTO `student` VALUES ('09', '张三', '2017-12-20 00:00:00', '女'); INSERT INTO `student` VALUES ('10', '李四', '2017-12-25 00:00:00', '女'); INSERT INTO `student` VALUES ('11', '李四', '2012-06-06 00:00:00', '女'); INSERT INTO `student` VALUES ('12', '赵六', '2013-06-13 00:00:00', '女'); INSERT INTO `student` VALUES ('13', '孙七', '2014-06-01 00:00:00', '女');
科目表 Course
-- ---------------------------- -- Table structure for `course` -- ---------------------------- DROP TABLE IF EXISTS `course`; CREATE TABLE `course` ( `CId` varchar(10) DEFAULT NULL, `Cname` varchar(10) DEFAULT NULL, `TId` varchar(10) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of course -- ---------------------------- INSERT INTO `course` VALUES ('01', '语文', '02'); INSERT INTO `course` VALUES ('02', '数学', '01'); INSERT INTO `course` VALUES ('03', '英语', '03');
教师表 Teacher
-- ---------------------------- -- Table structure for `teacher` -- ---------------------------- DROP TABLE IF EXISTS `teacher`; CREATE TABLE `teacher` ( `TId` varchar(10) DEFAULT NULL, `Tname` varchar(10) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of teacher -- ---------------------------- INSERT INTO `teacher` VALUES ('01', '张三'); INSERT INTO `teacher` VALUES ('02', '李四'); INSERT INTO `teacher` VALUES ('03', '王五');
成绩表 SC
-- ---------------------------- -- Table structure for `sc` -- ---------------------------- DROP TABLE IF EXISTS `sc`; CREATE TABLE `sc` ( `SId` varchar(10) DEFAULT NULL, `CId` varchar(10) DEFAULT NULL, `score` decimal(18,1) DEFAULT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of sc -- ---------------------------- INSERT INTO `sc` VALUES ('01', '02', '90.0'); INSERT INTO `sc` VALUES ('01', '01', '80.0'); INSERT INTO `sc` VALUES ('01', '03', '99.0'); INSERT INTO `sc` VALUES ('02', '02', '60.0'); INSERT INTO `sc` VALUES ('02', '01', '70.0'); INSERT INTO `sc` VALUES ('02', '03', '80.0'); INSERT INTO `sc` VALUES ('03', '01', '80.0'); INSERT INTO `sc` VALUES ('03', '02', '80.0'); INSERT INTO `sc` VALUES ('03', '03', '80.0'); INSERT INTO `sc` VALUES ('04', '01', '50.0'); INSERT INTO `sc` VALUES ('04', '02', '30.0'); INSERT INTO `sc` VALUES ('04', '03', '20.0'); INSERT INTO `sc` VALUES ('05', '01', '76.0'); INSERT INTO `sc` VALUES ('05', '02', '87.0'); INSERT INTO `sc` VALUES ('06', '01', '31.0'); INSERT INTO `sc` VALUES ('06', '03', '34.0'); INSERT INTO `sc` VALUES ('07', '02', '90.0'); INSERT INTO `sc` VALUES ('07', '03', '98.0');
下面是题目和mysql实现
-- 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 st.*, class1, class2 FROM student st JOIN (SELECT * FROM (SELECT SId s1, score class1 FROM sc WHERE CId = '01') c1, (SELECT SId s2, score class2 FROM sc WHERE CId = '02') c2 WHERE s1 = s2 AND class1 > class2) c ON st.SId = c.s1; -- 1.1 查询同时存在" 01 "课程和" 02 "课程的情况 SELECT * FROM ((SELECT * FROM sc WHERE CId = '01') t1 JOIN (SELECT * FROM sc WHERE CId = '02') t2 ON t1.Sid = t2.Sid); -- 1.2 查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null ) SELECT * FROM ((SELECT * FROM sc WHERE CId = '01') t1 left JOIN (SELECT * FROM sc WHERE CId = '02') t2 ON t1.Sid = t2.Sid); -- 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况 select * from sc where sc.CId = '02' AND sc.SId NOT IN (SELECT SId FROM sc WHERE sc.CId = '01'); -- 2.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 st.SId, Sname, AVG(score) 'avg' FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 60; -- 3.查询在 SC 表存在成绩的学生信息 SELECT * FROM student WHERE SId IN (SELECT DISTINCT SId FROM sc); -- 4.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和 SELECT st.sid, st.Sname, COUNT(sc.CId) 'nums', SUM(sc.score) 'sum' FROM student st JOIN sc on st.SId = sc.SId GROUP BY st.SId; -- 4.1显示没选课的学生(显示为NULL) SELECT st.sid, st.Sname, CASE WHEN COUNT(sc.SId) > 0 THEN COUNT(sc.SId) ELSE NULL END 'nums', SUM(sc.score) 'sum' FROM student st left JOIN sc on st.SId = sc.SId GROUP BY st.SId; -- 4.2查有成绩的学生信息 SELECT * FROM student WHERE SId IN (SELECT DISTINCT SId FROM sc); -- 适用于右表小 SELECT * FROM student st WHERE EXISTS (SELECT * FROM sc WHERE sc.SId = st.SId); -- 适用于右表大 -- 5.查询「李」姓老师的数量 SELECT COUNT(*) FROM teacher t WHERE t.Tname LIKE '李%'; -- 6.查询学过「张三」老师授课的同学的信息 SELECT * FROM student st WHERE st.SId IN ( SELECT SId from sc WHERE sc.CId = ( SELECT CId FROM course WHERE course.TId = ( SELECT TId FROM teacher WHERE teacher.Tname = '张三'))); -- 7.查询没有学全所有课程的同学的信息 SELECT * FROM student WHERE SId NOT IN( SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM course)); -- 8.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 * FROM student WHERE SId IN (SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01')); -- 9.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 -- 解法一 SELECT * FROM student WHERE SId IN (SELECT SId FROM sc GROUP BY SId HAVING SId <> '01' AND GROUP_CONCAT(CId ORDER BY CId) = (SELECT GROUP_CONCAT(CId ORDER BY CId) FROM sc WHERE SId = '01')); -- 解法二 SELECT * FROM student WHERE SId IN (SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01') AND SId <> '01' GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM sc WHERE SId = '01')); -- 10.查询没学过"张三"老师讲授的任一门课程的学生姓名 SELECT student.Sname FROM student WHERE SId NOT IN (SELECT SId FROM sc WHERE CId in (SELECT course.CId FROM course WHERE TId = (SELECT teacher.TId FROM teacher WHERE tname = '张三'))); -- 11.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 SELECT st.SId, sname, avg(score) FROM student st JOIN sc ON st.sid = sc.SId GROUP BY st.SId HAVING COUNT(score <= 60 OR NULL) >= 2; -- 错误解法： select student.sid, student.sname, AVG(sc.score) from student,sc where student.sid = sc.sid and sc.score< 60 group by sc.sid having count(*)>1; -- 12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息 SELECT st.*, score FROM student st JOIN sc ON st.SId = sc.SId AND sc.CId = '01' AND sc.score < 60 ORDER BY score DESC; -- 13.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 SELECT * from sc JOIN (SELECT SId, avg(score) avg FROM sc GROUP BY SId) avgs ON sc.SId = avgs.SId ORDER BY avgs.avg desc; -- 14.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 -- 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 sc.CId '课程 ID', count(*) '选修人数', course.Cname '课程 name', MAX(score) '最高分', MIN(score) '最低分', AVG(score) '平均分', COUNT(score >= 60 OR NULL) / COUNT(*) '及格率', COUNT(score >= 70 AND score < 80 OR NULL) / COUNT(*) '中等率', COUNT(score >= 80 AND score < 90 OR NULL) / COUNT(*) '优良率', COUNT(score >= 90 OR NULL) / COUNT(*) '优秀率' FROM sc join course ON sc.CId = course.CId GROUP BY sc.CId ORDER BY COUNT(*) desc, sc.CId; -- 15.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺 SELECT CId, SId, score , (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1 rank FROM sc o ORDER BY CId, score DESC; -- 16.查询学生的总成绩，并进行排名，总分重复时不保留名次空缺 SET @rank = 0; SELECT q.sid, q.sum, (@rank := @rank + 1) rank FROM (SELECT sid, sum(score) sum FROM sc GROUP BY SId ORDER BY sum) q; -- 17.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]，[60-0] 及所占百分比 SELECT s.CId, c.Cname, CONCAT(COUNT(score >= 85 OR NULL), ', ',COUNT(score >= 85 OR NULL) / COUNT(*)) '[100-85]', COUNT(score < 85 AND score >= 70 OR NULL) '[85-70]', COUNT(score < 70 AND score >= 60 OR NULL) '[70-60]', COUNT(score < 60 OR NULL) '[60-0]' FROM sc s JOIN course c ON s.CId = c.CId GROUP BY s.CId; -- 18. 查询各科成绩前三名的记录 SELECT o.* FROM sc o HAVING (SELECT COUNT(*) FROM sc WHERE o.CId = CId AND o.score < score) + 1 <= 3 ORDER BY CId, score DESC, SId; -- 19.查询每门课程被选修的学生数 SELECT CId, COUNT(*) FROM sc GROUP BY CId; -- 20.查询出只选修两门课程的学生学号和姓名 SELECT st.SId, st.Sname FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING COUNT(*) = 2; -- 21.查询男生、女生人数 select ssex, count(*) from student group by ssex; -- 22.查询名字中含有「风」字的学生信息 SELECT * FROM student WHERE Sname LIKE '%风%'; -- 23.查询同名学生名单，并统计同名人数 SELECT st.*, (SELECT COUNT(*) FROM student WHERE Sname = st.Sname) 'same' FROM student st WHERE st.SId IN (SELECT s1.SId FROM student s1 JOIN student s2 ON s1.SId != s2.SId AND s1.Sname = s2.Sname); -- 24.查询 1990 年出生的学生名单 SELECT * FROM student WHERE YEAR(Sage) = 1990; -- 25.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 CId, avg(score) avg FROM sc GROUP BY CId ORDER BY avg DESC, CId; -- 26.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 st.SId, Sname, AVG(score) avg FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 85; -- 27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数 SELECT st.Sname, s.score FROM student st, sc s, course c WHERE st.SId = s.SId AND s.CId = c.CId AND s.score < 60 AND c.Cname = '数学'; -- 28.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况） SELECT Sname, st.SId, score FROM student st LEFT JOIN sc s ON st.SId = s.SId; -- 29.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 -- 这个我没太理解，理解一是任意一门成绩均在70分以上 SELECT st.Sname, c.Cname, s.score FROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN (SELECT sid FROM sc GROUP BY SId HAVING MIN(score) > 70); -- 理解二是存在一门成绩在70分以上即可满足条件 SELECT st.Sname, c.Cname, s.score FROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN (SELECT sid FROM sc GROUP BY SId HAVING MAX(score) > 70); -- 理解三就是找出所有大于70分的得分。 select student.sname, course.cname,sc.score from student,course,sc where sc.score>70 and student.sid = sc.sid and sc.cid = course.cid; SELECT * FROM sc WHERE score > 70 -- 30.查询存在不及格的课程 SELECT cid FROM sc GROUP BY CId HAVING MIN(score) < 60; select cid from sc where score< 60 group by cid; SELECT DISTINCT cid FROM sc WHERE score < 60; -- 31.查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名 SELECT sid, sname FROM student WHERE sid IN (SELECT SId FROM sc WHERE CId = '01' AND score >= 80); -- 32.求每门课程的学生人数 SELECT cid, count(*) FROM sc GROUP BY cid; -- 33.成绩不重复，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 st.*, score FROM student st JOIN sc s ON st.SId = s.SId AND s.CId = (SELECT CId FROM course WHERE TId = (SELECT TId FROM teacher WHERE Tname = '张三')) ORDER BY score DESC LIMIT 1; -- 34.成绩有重复的情况下，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 st.*, score, CId FROM student st JOIN sc s ON st.SId = s.SId AND s.CId = (SELECT CId FROM course WHERE TId = (SELECT TId FROM teacher WHERE Tname = '张三')) WHERE score = (SELECT max(score) FROM sc WHERE CId = s.CId); -- 35.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 -- 这个问题其实一开始没太明白啥意思，后来理解为某个人的几科分数是一样的，需要把这个人找出来 select a.cid, a.sid, a.score from sc as a inner join sc as b on a.sid = b.sid and a.cid != b.cid and a.score = b.score group by cid, sid; -- 36.查询每门功成绩最好的前两名 SELECT o.* FROM sc o HAVING (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1 <= 2 ORDER BY o.cid, sid; -- 37.统计每门课程的学生选修人数（超过 5 人的课程才统计） SELECT CId, COUNT(*) sum FROM sc GROUP BY CId HAVING sum > 5; -- 38.检索至少选修两门课程的学生学号 SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) >= 2; -- 39.查询选修了全部课程的学生信息 SELECT * FROM student WHERE sid IN (SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROM course)); -- 40.查询各学生的年龄，只按年份来算 SELECT SId, Sname, Sage, YEAR(NOW()) - YEAR(Sage) FROM student; -- 41.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一 SELECT SId '学生编号', Sname '学生姓名', TIMESTAMPDIFF(YEAR, Sage, NOW()) '学生年龄' FROM student; -- 42.查询本周过生日的学生 -- 有点复杂，需要拼接出本周的起止日期 SELECT * FROM student WHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage))) BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) DAY)) AND DATE(DATE_ADD(NOW(),INTERVAL 6 - WEEKDAY(NOW()) DAY)); SELECT * FROM student -- 43. 查询下周过生日的学生 -- 同42 SELECT * FROM student WHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage))) BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) - 7 DAY)) AND DATE(DATE_ADD(NOW(),INTERVAL 13 - WEEKDAY(NOW()) DAY)); -- 44.查询本月过生日的学生 SELECT * FROM student WHERE month(sage) = month(NOW()) -- 45.查询下月过生日的学生 -- 注意本月是12月的话，下一个月份是1即可 SELECT * FROM student WHERE month(Sage) = (CASE WHEN month(NOW()) = 12 THEN 1 ELSE MONTH(NOW()) + 1 END);