MySQL练习-主外键多表查询

练习:

1.建立表关系:

请创建如下表，并创建相关约束

 USE db1;
 CREATE TABLE class(
     cid INT AUTO_INCREMENT PRIMARY KEY,
     caption VARCHAR(10)
 );
 CREATE TABLE teacher(
     tid INT AUTO_INCREMENT PRIMARY KEY,
     tname VARCHAR(10)
 );
 CREATE TABLE course(
     cid INT AUTO_INCREMENT PRIMARY KEY,
     cname VARCHAR(10),
     teach_id INT,
     CONSTRAINT fk_course_teacher FOREIGN KEY(teach_id) REFERENCES teacher(tid)
 );
 CREATE TABLE student(
     sid INT AUTO_INCREMENT PRIMARY KEY,
     sname VARCHAR(10),
     gender VARCHAR(10),
     class_id INT,
     CONSTRAINT fk_student_class FOREIGN KEY(class_id) REFERENCES class(cid)
 );
 CREATE TABLE score(
     sid INT AUTO_INCREMENT PRIMARY KEY,
     student_id INT,
     course_id INT,
     number INT,
     CONSTRAINT fk_score_student FOREIGN KEY(student_id) REFERENCES student(sid),
     CONSTRAINT fk_score_course FOREIGN KEY(course_id) REFERENCES course(cid)
 );

创建表并建立表关系

 INSERT INTO class(caption) VALUES("三年二班");
 INSERT INTO class(caption) VALUES("一年三班");
 INSERT INTO class(caption) VALUES("三年一班");

 INSERT INTO teacher(tname) VALUES("波多");
 INSERT INTO teacher(tname) VALUES("苍空");
 INSERT INTO teacher(tname) VALUES("饭岛");

 INSERT INTO student(sname,gender,class_id) VALUES("钢蛋","女",1);
 INSERT INTO student(sname,gender,class_id) VALUES("铁锤","女",1);
 INSERT INTO student(sname,gender,class_id) VALUES("山炮","男",2);

 INSERT INTO course(cname,teach_id) VALUES("生物",1);
 INSERT INTO course(cname,teach_id) VALUES("体育",1);
 INSERT INTO course(cname,teach_id) VALUES("物理",2);

 INSERT INTO score(student_id,course_id,number) VALUES(1,1,60);
 INSERT INTO score(student_id,course_id,number) VALUES(1,2,59);
 INSERT INTO score(student_id,course_id,number) VALUES(2,2,100);

表数据

 ALTER TABLE score ADD UNIQUE sco_que(student_id,course_id);

补充联合唯一

2.操作练习:

1、自行创建测试数据

 INSERT INTO student(sname,gender,class_id) VALUES("钢铁侠","男",4);
 -- Cannot add or update a child row: a foreign key constraint fails (`db1`.`student`, CONSTRAINT `fk_student_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`))

测试失败数据

2、查询“生物”课程比“物理”课程成绩高的所有学生的学号；

 SELECT s1.student_id FROM (SELECT a.student_id,a.`number` FROM score a,course b
 WHERE a.`course_id` = b.`cid` AND b.`cname` ="生物") s1 , (SELECT a.student_id,a.`number` FROM score a,course b
 WHERE a.`course_id` = b.`cid` AND b.`cname`="体育") s2
 WHERE s1.`student_id` = s2.student_id AND s1.number >s2.number

多表嵌套查询

3、查询平均成绩大于60分的同学的学号和平均成绩；

 SELECT student_id,AVG(number)
 FROM score
 GROUP BY student_id
 HAVING AVG(number) > 60

HAVING用法

4、查询所有同学的学号、姓名、选课数、总成绩；

 SELECT a.`sid`,a.`sname`,b.`cname`,c.`number`,c.`course_id`,a.`class_id`
 FROM student a,course b,score c
 WHERE a.`sid` = c.`student_id` AND c.`course_id` = b.`cid`

多表查询

5、查询姓“李”的老师的个数；

 SELECT count(*) as a from teacher where tname LIKE '李%'

6、查询没学过“叶平”老师课的同学的学号、姓名；

 SELECT a.sid,a.sname FROM student a,teacher b,course c
 WHERE a.class_id = c.cid and b.tid = c.teach_id AND b.tname != '叶平'

多表查询

7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；

 SELECT a.`student_id`,c.`sname` FROM score a , score b ,student c
 WHERE a.`student_id`=b.`student_id` AND a.`course_id` =''
 AND b.`course_id`='' AND c.`sid` = a.`student_id`

多表查询

8、查询学过“叶平”老师所教的所有课的同学的学号、姓名；

 SELECT c.`sid`,c.`sname` FROM student c
 LEFT JOIN score d ON c.`sid` = d.`student_id`
 LEFT JOIN course b ON d.`course_id` = b.`cid`
 LEFT JOIN teacher a ON a.`tid` = b.`teach_id`
 WHERE a.`tname`='波多'
 GROUP BY c.`sname`
 HAVING COUNT(c.sname)
 <(SELECT COUNT(*) FROM teacher a
 LEFT JOIN course b ON a.`tid` = b.`teach_id`
 WHERE a.`tname`='波多')

乱七八糟一大堆

9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；

 SELECT s1.student_id FROM (SELECT a.student_id,a.`number` FROM score a,course b
 WHERE a.`course_id` = b.`cid` AND b.`cid` ='') s1 , (SELECT a.student_id,a.`number` FROM score a,course b
 WHERE a.`course_id` = b.`cid` AND b.`cid`='') s2
 WHERE s1.`student_id` = s2.student_id AND s1.number >s2.number

多表嵌套查询

10、查询有课程成绩小于60分的同学的学号、姓名；

 SELECT a.`sid`,a.`sname` FROM student a , score    b
 WHERE a.sid = b.student_id AND number < 60

11、查询没有学全所有课的同学的学号、姓名；

 SELECT a.`sid`,a.`sname` FROM student a
 LEFT JOIN score b ON  a.`sid` = b.`student_id`
 GROUP BY a.`sid`
 HAVING COUNT(b.`student_id`) < (SELECT COUNT(*) FROM course)

Having 子查询

12、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名；

 SELECT a.`sid`,a.`sname` FROM student a LEFT JOIN score b ON a.`sid` = b.`student_id`
 LEFT JOIN course c ON c.`cid` = b.`course_id`
 WHERE c.`cid` IN (
 SELECT cc.`cid` FROM student aa LEFT JOIN score bb ON aa.`sid` = bb.`student_id`
 LEFT JOIN course cc ON cc.`cid` = bb.`course_id` WHERE aa.`sid`=''
 ) AND a.`sid` !=''

又是乱七八糟一大堆

13、查询至少学过学号为“001”同学所选课程中任意一门课的其他同学学号和姓名；

SELECT a.`sid`,a.`sname` FROM student a LEFT JOIN score b ON a.`sid` = b.`student_id`
LEFT JOIN course c ON c.`cid` = b.`course_id`
WHERE c.`cid` IN (
SELECT cc.`cid` FROM student aa LEFT JOIN score bb ON aa.`sid` = bb.`student_id`
LEFT JOIN course cc ON cc.`cid` = bb.`course_id` WHERE aa.`sid`=''
) AND a.`sid` !=''

完全一模一样啊

14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名；

 SELECT a.`sid`,a.`sname` FROM student a LEFT JOIN score b ON a.`sid` = b.`student_id`
 LEFT JOIN course ac ON b.`course_id`=ac.`cid`
 WHERE
  ac.`cid` IN (SELECT c.`cid` FROM student a LEFT JOIN score b ON a.`sid` = b.`student_id`
 LEFT JOIN course c ON b.`course_id`=c.`cid` WHERE a.`sid`='')
 AND a.`sid` != ''
 GROUP BY a.`sid`,a.`sname`
 HAVING
 (SELECT COUNT(ac.`cid`) AS countCid FROM student aa LEFT JOIN score b ON aa.`sid` = b.`student_id`
 LEFT JOIN course ac ON b.`course_id`=ac.`cid` WHERE aa.`sid`=a.`sid`)
  = (SELECT COUNT(c.`cid`) AS cc FROM student a LEFT JOIN score b ON a.`sid` = b.`student_id`
 LEFT JOIN course c ON b.`course_id`=c.`cid` WHERE a.`sid`='')

抠抠抠抠抠一下午,抠出来了

15、删除学习“叶平”老师课的SC表记录；

 DELETE FROM score WHERE course_id IN (
 SELECT a.`cid` FROM course a LEFT JOIN teacher b ON a.`teach_id`=b.`tid`
 WHERE b.`tname`='叶平'
 )

这个这么简单?

16、向SC表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“002”课程的同学学号；②插入“002”号课程的平均成绩；
17、按平均成绩从低到高显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示： 学生ID,语文,数学,英语,有效课程数,有效平均分；
18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分；
19、按各科平均成绩从低到高和及格率的百分数从高到低顺序；
20、课程平均分从高到低显示（现实任课老师）；
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
22、查询每门课程被选修的学生数；
23、查询出只选修了一门课程的全部学生的学号和姓名；
24、查询男生、女生的人数；
25、查询姓“张”的学生名单；
26、查询同名同姓学生名单，并统计同名人数；
27、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；
28、查询平均成绩大于85的所有学生的学号、姓名和平均成绩；
29、查询课程名称为“数学”，且分数低于60的学生姓名和分数；
30、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名；
31、求选了课程的学生人数
32、查询选修“杨艳”老师所授课程的学生中，成绩最高的学生姓名及其成绩；
33、查询各个课程及相应的选修人数；
34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩；

35、查询每门课程成绩最好的前两名；

 SELECT student_id FROM score
 ORDER BY number DESC LIMIT 2

这个肯定不对

36、检索至少选修两门课程的学生学号；

 SELECT a.`sid` FROM student a LEFT JOIN score b ON a.`sid` = b.`student_id`
 GROUP BY a.`sid`
 HAVING COUNT(a.`sid`)>=2

完美~

37、查询全部学生都选修的课程的课程号和课程名；

 SELECT DISTINCT c.`cid`,c.`cname` FROM student a
 LEFT JOIN score b ON a.`sid` = b.`student_id`
 LEFT JOIN course c ON b.`course_id`=c.`cid`

应该对了

38、查询没学过“叶平”老师讲授的任一门课程的学生姓名；

 SELECT a.`sname` FROM student a
 LEFT JOIN score b ON a.`sid` = b.`student_id`
 LEFT JOIN course c ON c.`cid` = b.`course_id`
 WHERE c.`cid` NOT IN (
 SELECT cour.`cid` FROM teacher teac
 LEFT JOIN course cour ON teac.`tid`=cour.`teach_id`
 WHERE teac.`tname` = '波多'
 )

差不多吧

39、查询两门以上不及格课程的同学的学号及其平均成绩；

 SELECT student_id,COUNT(sid) AS co,AVG(number)AS avgg  FROM score
 WHERE number < 60
 GROUP BY student_id
 HAVING co >=2

so anyway

40、检索“004”课程分数小于60，按分数降序排列的同学学号；

 SELECT b.`sid`,b.`sname`,a.`number` FROM score a LEFT JOIN student b ON a.`student_id` = b.`sid`
 WHERE a.`course_id` = '' AND a.`number` < ''
 ORDER BY a.`number` DESC

恩不难

41、删除“002”同学的“001”课程的成绩；

 DELETE FROM score WHERE student_id ='' AND course_id = ''