SET FOREIGN_KEY_CHECKS=0;
-- ----------------------------
-- Table structure for student
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
`Sid` varchar(6) NOT NULL DEFAULT '',
`Sname` varchar(10) DEFAULT NULL,
`Sage` datetime DEFAULT NULL,
`Ssex` varchar(10) DEFAULT NULL,
PRIMARY KEY (`Sid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
`Tid` varchar(10) NOT NULL DEFAULT '',
`Tname` varchar(10) DEFAULT NULL,
PRIMARY KEY (`Tid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
`Cid` varchar(10) NOT NULL DEFAULT '',
`Tid` varchar(10) NOT NULL DEFAULT '',
`Cname` varchar(10) DEFAULT NULL,
PRIMARY KEY (`Cid`),
CONSTRAINT `course_ibfk_1` FOREIGN KEY (`Tid`) REFERENCES `teacher` (`Tid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Table structure for sc
-- ----------------------------
DROP TABLE IF EXISTS `sc`;
CREATE TABLE `sc` (
`Sid` varchar(10) NOT NULL DEFAULT '',
`Cid` varchar(10) NOT NULL DEFAULT '',
`score` decimal(18,1) DEFAULT NULL,
CONSTRAINT `sc_ibfk_1` FOREIGN KEY (`Sid`) REFERENCES `student` (`Sid`),
CONSTRAINT `sc_ibfk_2` FOREIGN KEY (`Cid`) REFERENCES `course` (`Cid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
1. 查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
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
3. 查询在 SC 表存在成绩的学生信息
4. 查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )
4.1 查有成绩的学生信息
5. 查询「李」姓老师的数量
6. 查询学过「张三」老师授课的同学的信息
7. 查询没有学全所有课程的同学的信息
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
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
10. 查询没学过"张三"老师讲授的任一门课程的学生姓名
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
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
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）。
15. 按平均成绩进行排序，显示总排名和各科排名，Score 重复时保留名次空缺
15.1 按平均成绩进行排序，显示总排名和各科排名，Score 重复时合并名次
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
18. 查询各科成绩前三名的记录
20. 查询出只选修两门课程的学生学号和姓名
22. 查询名字中含有「风」字的学生信息
24. 查询 1990 年出生的学生名单
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
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
40. 查询各学生的年龄，只按年份来算
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
42. 查询本周过生日的学生
43. 查询下周过生日的学生
44. 查询本月过生日的学生
45. 查询下月过生日的学生
46.查询所有课程成绩都大于80分的学生名字
47.查询所有课程成绩都大于80分的学生名字，不显示重复名字。
查询学生的最高成绩。
SELECT sid,score from sc group by sid ;
#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 s.* ,a.score as score_01,b.score as score_02 FROM
student s,
(SELECT Sid,score from sc WHERE sc.Cid = "01") a,
(SELECT Sid,score from sc WHERE sc.Cid = "02") b
where a.Sid = b.Sid and a.score> b.score and s.Sid = a.Sid;
SELECT a.*,b.score score1 from
(SELECT Sid,score from sc WHERE sc.Cid = "01") a,
(SELECT Sid,score from sc WHERE sc.Cid = "02") b
where a.sid = b.sid and a.score > b.score;
#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 sid,AVG(score) as avgg from sc GROUP BY sid having AVG(score) >60;
SELECT Sname,sc.sid,AVG(score) as avgg from sc,student s WHERE sc.Sid = s.Sid GROUP BY sid having AVG(score) >60;
#3. 查询在 SC 表存在成绩的学生信息
SELECT * from student where sid in(SELECT Sid from sc WHERE score is not null);
SELECT s.*,sc.score FROM student s,sc WHERE s.Sid = sc.sid and sc.score is not null group BY sc.sid;
#4. 查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )
这道题得用到left join或者right join，不能用where连接，因为题目说了要求有显示为null的，where是inner join，不会出现null，在这道题里会查不出第08号学生。
select count(sid) as xuankezongshu, sum(score) as xuankezongfen from sc GROUP BY Sid;
select s.*,count(sc.sid) as xuankezongshu, sum(score) as xuankezongshu from student s, sc WHERE s.Sid = sc.sid GROUP BY sc.Sid;
select s.*,count(sc.sid) as xuankezongshu, sum(score) as xuankezongshu from student s LEFT JOIN sc on s.Sid = sc.sid GROUP BY sc.Sid;
4.1 查有成绩的学生信息 ***************************
select s.*,a.xuankezongshu,a.xuankezongfen,a.score from student s,
(SELECT sid,score,count(sid) as xuankezongshu, sum(score) as xuankezongfen from sc where score is not NULL GROUP BY sid) a
WHERE s.sid = a.sid
select s.sid, s.sname, count(*) as 选课总数, sum(score) as 总成绩,
sum(case when cid = 01 then score else null end) as score_01,
sum(case when cid = 02 then score else null end) as score_02,
sum(case when cid = 03 then score else null end) as score_03
from student as s, sc
where s.sid = sc.sid
group by s.sid
5. 查询「李」姓老师的数量
SELECT count(tid) 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 WHERE sid in (SELECT sid FROM sc where cid in (SELECT cid from course WHERE tid in ( select tid from teacher t WHERE t.Tname = "张三" )));
select * from student where sid in (
select sid from sc, course, teacher
where sc.cid = course.cid
and course.tid = teacher.tid
and 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 in (select sid from sc group by sid having count(cid) < 3)
SELECT s.*,b.xuankeshu from student s,
(
select sid,count(*) as xuankeshu from sc group by sid HAVING xuankeshu <
(select count(*) coursecount from (SELECT count(cid) from sc group by cid) a)
) b
where s.sid = b.sid;
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
参考地址：https://blog.csdn.net/u010452388/article/details/80150985
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