mysql经典4张表问题详细讲解

/*
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 Source Server         : 127.0.0.1
 Source Server Type    : MySQL
 Source Server Version : 50720
 Source Host           : localhost:3306
 Source Schema         : work

 Target Server Type    : MySQL
 Target Server Version : 50720
 File Encoding         : 65001

 Date: 16/02/2022 16:39:35
*/

SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course`  (
  `cid` int(11) NOT NULL AUTO_INCREMENT COMMENT '课程编号',
  `cname` varchar(20) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '课程名称',
  `tid` int(11) NULL DEFAULT NULL COMMENT '教师编号',
  PRIMARY KEY (`cid`) USING BTREE,
  INDEX `tid`(`tid`) USING BTREE,
  CONSTRAINT `course_ibfk_1` FOREIGN KEY (`tid`) REFERENCES `teacher` (`tid`) ON DELETE RESTRICT ON UPDATE RESTRICT
) ENGINE = InnoDB AUTO_INCREMENT = 4 CHARACTER SET = utf8 COLLATE = utf8_general_ci COMMENT = '课程表' ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of course
-- ----------------------------
INSERT INTO `course` VALUES (1, '语文', 2);
INSERT INTO `course` VALUES (2, '数学', 1);
INSERT INTO `course` VALUES (3, '英语', 3);

-- ----------------------------
-- Table structure for sc
-- ----------------------------
DROP TABLE IF EXISTS `sc`;
CREATE TABLE `sc`  (
  `sid` int(11) NOT NULL COMMENT '学生编号',
  `cid` int(11) NULL DEFAULT NULL COMMENT '课程编号',
  `score` int(11) NULL DEFAULT NULL COMMENT '分数'
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci COMMENT = '成绩表' ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of sc
-- ----------------------------
INSERT INTO `sc` VALUES (1, 1, 90);
INSERT INTO `sc` VALUES (1, 2, 80);
INSERT INTO `sc` VALUES (1, 3, 90);
INSERT INTO `sc` VALUES (2, 1, 70);
INSERT INTO `sc` VALUES (2, 2, 60);
INSERT INTO `sc` VALUES (2, 3, 80);
INSERT INTO `sc` VALUES (3, 1, 80);
INSERT INTO `sc` VALUES (3, 2, 80);
INSERT INTO `sc` VALUES (3, 3, 80);
INSERT INTO `sc` VALUES (4, 1, 50);
INSERT INTO `sc` VALUES (4, 2, 30);
INSERT INTO `sc` VALUES (4, 3, 20);
INSERT INTO `sc` VALUES (5, 1, 76);
INSERT INTO `sc` VALUES (5, 2, 87);
INSERT INTO `sc` VALUES (6, 1, 31);
INSERT INTO `sc` VALUES (6, 3, 34);
INSERT INTO `sc` VALUES (7, 2, 89);
INSERT INTO `sc` VALUES (7, 3, 98);

-- ----------------------------
-- Table structure for student
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student`  (
  `sid` int(11) NOT NULL AUTO_INCREMENT COMMENT '学生编号',
  `sname` varchar(20) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '学生姓名',
  `sage` date NULL DEFAULT NULL COMMENT '出生年月',
  `ssex` char(4) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '学生性别',
  PRIMARY KEY (`sid`) USING BTREE
) ENGINE = InnoDB AUTO_INCREMENT = 9 CHARACTER SET = utf8 COLLATE = utf8_general_ci COMMENT = '学生表' ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES (1, '赵雷', '1990-01-01', '男');
INSERT INTO `student` VALUES (2, '钱电', '1990-12-21', '男');
INSERT INTO `student` VALUES (3, '孙风', '1990-05-20', '男');
INSERT INTO `student` VALUES (4, '李云', '1990-08-06', '男');
INSERT INTO `student` VALUES (5, '周梅', '1991-12-01', '女');
INSERT INTO `student` VALUES (6, '吴兰', '1992-03-01', '女');
INSERT INTO `student` VALUES (7, '郑竹', '1989-07-01', '女');
INSERT INTO `student` VALUES (8, '王菊', '1990-01-20', '女');

-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher`  (
  `tid` int(11) NOT NULL AUTO_INCREMENT COMMENT '教师编号',
  `tname` varchar(20) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '教师姓名',
  PRIMARY KEY (`tid`) USING BTREE
) ENGINE = InnoDB AUTO_INCREMENT = 4 CHARACTER SET = utf8 COLLATE = utf8_general_ci COMMENT = '教师表' ROW_FORMAT = Dynamic;

-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES (1, '张三');
INSERT INTO `teacher` VALUES (2, '李四');
INSERT INTO `teacher` VALUES (3, '王五');

SET FOREIGN_KEY_CHECKS = 1;

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.sid,s.sname,s.sage,s.ssex,sc1.score,sc2.score from student s,sc sc1,sc sc2 where sc1.cid=1 and sc2.cid=2 and sc1.score>sc2.score and sc1.sid=sc2.sid and s.sid=sc1.sid;

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 s.sid,s.sname,avg(sc.score) from student s,sc group by s.sid having avg(sc.score)>=60;

4、查询名字中含有"风"字的学生信息
select * from student where sname like ‘%风%';

5、查询课程名称为"数学"，且分数低于60的学生姓名和分数
select s.sname,score from student s,sc where s.sid=sc.sid and cid=2 and score<60;

6、查询所有学生的课程及分数情况；
select cname,score from sc,course where sc.cid=course.cid;

7、查询没学过"张三"老师授课的同学的信息
select s.* from student s where s.sid not in(select sc1.sid from sc sc1,course c,teacher t where t.tid=c.tid and sc1.cid=c.cid and t.tname=‘张三');

8.查询学过"张三"老师授课的同学的信息
select s.* from student s ,sc sc1,course c,teacher t where s.sid=sc1.sid and sc1.cid=c.cid and c.tid=t.tid and t.tname=‘张三';

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
student(sid) sc(sid cid tid) sc2(sid cid tid) course(cid tid cname)
select s.* from student s,sc sc1,sc sc2 where s.sid=sc1.sid and sc1.sid=sc2.sid and sc1.cid=1 and sc2.cid=2;

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 distinct s.* 
	from student s,sc sc1,sc sc2,sc sc3 
		where s.sid=sc1.sid and sc1.sid=sc2.sid and sc1.cid=1 and sc2.cid!=2;

11、查询没有学全所有课程的同学的信息
select s.* from student s where s.sid not in(select sc1.sid from sc sc1,sc sc2,sc sc3 where sc1.cid=1 and sc2.cid=2 and sc3.cid =3 and sc1.sid=sc2.sid and sc1.sid=sc3.sid) group by s.sid;

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 distinct s.* from student s,sc sc1 where s.sid=sc1.sid and sc1.cid in(select cid from sc where sid=1) and s.sid<> 1;

13、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
select s.* from student s where s.sid in(select distinct sc.sid from sc where sid<>1 and sc.cid in(select distinct cid from sc where sid=1)group by sc.sid having count(1)=(select count(1) from sc where s.sid=1));

14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select s.* from student s where s.sid not in(select sc1.sid from sc sc1,course c,teacher t where sc1.cid=c.cid and c.tid=t.tid and t.tname=‘张三');

15、查询出只有两门课程的全部学生的学号和姓名
select s.* from student s,sc group by sc.sid having count(sc.sid)=2 and s.sid=sc.sid;

16、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)
select * from student where sage>=‘1900-01-01' and sage<=‘1900-12-31';
select s.* from student s where s.sage like ‘1900-%';（方法2）

17、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 sc.cid,avg(score) from sc group by sc.cid order by avg(score) DESC , sc.cid;

18、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；
select s.sname,c.cname,score from student s,sc,course c where s.sid=sc.sid and sc.cid=c.cid and score>70;

19、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 s.sname,avg(score) from sc,student s where s.sid=sc.sid group by sc.sid having avg(score)>=85;

20、查询不及格的课程
select s.sname,c.cname,score from student s,sc,course c where s.sid=sc.sid and sc.cid=c.cid and score<60;

21、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名；
select s.sid,s.sname from student s,sc where sc.sid=s.sid and sc.cid=1 and score>80;

22、求每门课程的学生人数
select cid,count(sid) from sc group by sc.cid;

23、统计每门课程的学生选修人数（超过5人的课程才统计）。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 cid,count(sid) from sc group by cid having count(sid)>5 order by count(sid),cid ASC;

24、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
select s1.sid,s2.sid,sc1.cid,sc1.score,sc2.score from student s1,student s2,sc sc1,sc sc2 where s1.sid!=s2.sid and s1.sid=sc1.sid and s2.sid=sc2.sid and sc1.cid!=sc2.cid and sc1.score=sc2.score;

25、检索至少选修两门课程的学生学号
select sid from sc group by sid having count(cid)>=2;

26、查询选修了全部课程的学生信息
select s.* from sc,student s where s.sid=sc.sid group by sid having count
(cid)=3;

27、查询各学生的年龄
select s.sname,(TO_DAYS(‘2017-09-07')-TO_DAYS(s.sage))/365 as age from student s;

28、查询本月过生日的学生
select s.sname from student s where s.sage like ‘_____07%';

29、查询下月过生日的学生
select s.sname from student s where s.sage like ‘_____08%';

30、查询学全所有课程的同学的信息
select s.* from student s,sc sc1,sc sc2,sc sc3 where sc1.cid=1 and sc2.cid=2 and sc3.cid=3 and sc1.sid=sc2.sid and sc1.sid=sc3.cid and s.sid =sc1.sid group by s.sid;