mysql考试总结

USE school;

-- 班级表

CREATE TABLE class(

    cid TINYINT PRIMARY KEY AUTO_INCREMENT,

    caption VARCHAR(20)

);

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

SELECT * FROM class;

-- 老师表

CREATE TABLE teacher(

    tid TINYINT PRIMARY KEY AUTO_INCREMENT,

    tname VARCHAR(20)

);

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

SELECT * FROM teacher;

-- 学生表

CREATE TABLE student(

    sid TINYINT PRIMARY KEY AUTO_INCREMENT,

    sname VARCHAR(20),

    gender VARCHAR(10),

    class_id TINYINT,

    FOREIGN KEY (class_id) REFERENCES class(cid)

);

INSERT INTO student(sname,gender,class_id) VALUES

    ("钢蛋","女",1),

    ("铁锤","女",1),

    ("山炮","男",2);

SELECT * FROM student;

-- 课程表

CREATE TABLE course(

    cid TINYINT PRIMARY KEY AUTO_INCREMENT,

    cname VARCHAR(20),

    teacher_id TINYINT,

    FOREIGN KEY (teacher_id) REFERENCES teacher(tid)

);

ALTER TABLE course MODIFY cid TINYINT, DROP PRIMARY KEY;

ALTER TABLE course ADD CONSTRAINT xx FOREIGN KEY (cid) REFERENCES class(cid);

DESC course;

SHOW CREATE TABLE course;

INSERT INTO course(cname,teacher_id) VALUES

    ("生物",1),

    ("体育",1),

    ("物理",2);

SELECT * FROM course;

-- 成绩表

CREATE TABLE score(

    sid TINYINT PRIMARY KEY AUTO_INCREMENT,

    student_id TINYINT,

    course_id TINYINT,

    number INT,

    FOREIGN KEY (student_id) REFERENCES student(sid),

    FOREIGN KEY (course_id) REFERENCES course(cid)

);

INSERT INTO score(student_id, course_id, number) VALUES

    (1,1,60),

    (1,2,59),

    (2,2,100);

SELECT * FROM score;

DELETE FROM score WHERE sid=6;

# 二、操作表

#

# 1、自行创建测试数据

#

# 2、查询“生物”课程比“物理”课程成绩高的所有学生的学号；

SELECT B.student_id FROM

(SELECT score.student_id,score.number FROM score INNER JOIN course ON score.course_id = course.cid WHERE course.cname="生物") as A

    INNER JOIN

(SELECT score.student_id,score.number FROM score INNER JOIN course ON score.course_id = course.cid WHERE course.cname="物理") as B

        ON A.student_id=B.student_id AND A.number>B.number;

# 3、查询平均成绩大于60分的同学的学号和平均成绩；

INSERT INTO score(student_id, course_id, number) VALUES(2,3,56),

    (3,1,46),(3,2,59),(3,3,71),(4,1,90),(4,2,27);

SELECT student_id,avg(number) 平均成绩 FROM score GROUP BY student_id HAVING avg(number)>60;

# 4、查询所有同学的学号、姓名、选课数、总成绩；

SELECT student.sid,student.sname,count(student.class_id) 选课数,sum(number) 总成绩

    FROM student INNER JOIN score ON student.sid = score.student_id GROUP BY sname ORDER BY 总成绩;

# 5、查询姓“波”的老师的个数；

SELECT count(*) 波老师个数 FROM teacher WHERE tname LIKE "波%";

INSERT INTO teacher(tname) VALUES ("波大");

# 6、查询没学过“叶平”老师课的同学的学号、姓名；

  -- 得到所有同学学过的课程及其对应的老师,然后对应筛选

INSERT INTO teacher(tname) VALUES("叶平");

INSERT INTO course(cname, teacher_id) VALUES("历史",5);

INSERT INTO score(student_id, course_id, number) VALUES(1,5,81);

SELECT student.sid,student.sname,A.tname FROM score

    INNER JOIN student ON score.student_id = student.sid

    INNER JOIN (SELECT DISTINCT course.cid,course.teacher_id,course.cname,teacher.tname FROM

    course INNER JOIN teacher ON course.teacher_id = teacher.tid) AS A ON score.course_id = A.cid

    GROUP BY A.tname HAVING A.tname!="叶平";

# 7、查询学过“1”并且也学过编号“2”课程的同学的学号、姓名；

INSERT INTO student(sname,gender,class_id) VALUES("张三","男",3);

INSERT INTO score(student_id, course_id, number) VALUES(5,2,63);

SELECT B.sid 学号,B.sname 姓名 FROM

    (SELECT student.sid,student.sname FROM score INNER JOIN student ON score.student_id = student.sid

WHERE course_id=1) AS A

    INNER JOIN

    (SELECT student.sid,student.sname FROM score INNER JOIN student ON score.student_id = student.sid

WHERE course_id=2) AS B ON A.sid = B.sid;

# 8、查询学过“叶平”老师所教的所有课的同学的学号、姓名；

SELECT * FROM student;    SELECT * FROM score;

INSERT INTO student(sname, gender, class_id) VALUES("王五","男",3);

INSERT INTO score(student_id, course_id, number) VALUES(8,5,93);

SELECT student.sid 学号,student.sname 姓名 FROM score INNER JOIN student

        ON score.student_id = student.sid AND course_id=5;

# 9、查询课程编号“2”的成绩比课程编号“1”课程低的所有同学的学号、姓名；

INSERT INTO student(sname, gender, class_id) VALUES("赵六","女",2);

INSERT INTO score(student_id, course_id, number) VALUES(9,2,65),(9,1,70);

UPDATE score SET number = 65 WHERE sid=21;

SELECT * FROM student WHERE

    (SELECT number FROM score WHERE student.sid=score.student_id AND score.course_id=2)<

    (SELECT number FROM score WHERE student.sid=score.student_id AND score.course_id=1);

# 10、查询有课程成绩小于60分的同学的学号、姓名；

SELECT student.sid,student.sname FROM student

    INNER JOIN score ON student.sid = score.student_id WHERE score.number<60 GROUP BY student.sname;

INSERT INTO student(sname, gender, class_id) VALUES("钢镚","男",2);

INSERT INTO score(student_id, course_id, number) VALUES(4,3,48);

# 11、查询没有学全所有课的同学的学号、姓名；

-- 测试

SELECT count(cid) 总课程数 FROM course;

SELECT count(course_id) study_course FROM score GROUP BY student_id;

SELECT * FROM (SELECT count(cid) 总课程数 FROM course) AS A

    INNER JOIN

    (SELECT count(course_id) study_course FROM score GROUP BY student_id) AS B

    ON A.总课程数 = B.study_course;

-- 正确答案

SELECT student.sid,student.sname,count(course_id) 学习课程数 FROM score INNER JOIN student

    ON score.student_id = student.sid GROUP BY student_id

    HAVING count(course_id)=(SELECT count(cid) FROM course);

# 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；

#

# 13、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名；

#

# 14、查询和“2”号的同学学习的课程完全相同的其他同学学号和姓名；

# 15、删除学习“叶平”老师课的SC表记录；

#

# 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；

SELECT course_id,max(number) 最高分,min(number) 最低分 FROM score GROUP BY course_id;

INSERT INTO score(student_id, course_id, number) VALUES(2,1,76);

# 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；

    -- 分析求出平均成绩并计算及格率

SELECT avg(number) 课程平均成绩 FROM score GROUP BY course_id;

SELECT count(student_id) 各科不及格人数 FROM score WHERE number>60 GROUP BY course_id;

SELECT count(student_id) 各科总人数 FROM score GROUP BY course_id;

    -- 答案如下

SELECT A.课程平均成绩,B.各科不及格人数/C.各科总人数 AS 及格率 FROM

    (SELECT course_id,avg(number) 课程平均成绩 FROM score GROUP BY course_id) AS A

        INNER JOIN

    (SELECT course_id,count(student_id) 各科不及格人数 FROM score WHERE number>60 GROUP BY course_id) AS B

        INNER JOIN

    (SELECT course_id,count(student_id) 各科总人数 FROM score GROUP BY course_id) AS C

    ON A.course_id = B.course_id AND A.course_id = C.course_id ORDER BY A.课程平均成绩;

# 20、课程平均分从高到低显示（显示任课老师）；

SELECT teacher.tname,course.cname FROM teacher,course WHERE course.teacher_id = teacher.tid;

SELECT avg(number) 课程平均分 FROM teacher,score GROUP BY course_id;

-- 答案如下

SELECT A.tname 任课老师,B.课程平均分 FROM

    (SELECT teacher.tname,course.cname,course.cid FROM teacher,course WHERE course.teacher_id = teacher.tid) AS A

    INNER JOIN

    (SELECT avg(number) 课程平均分,teacher.tname,score.course_id FROM teacher,score GROUP BY course_id) AS B

    ON A.cid=B.course_id ORDER BY B.课程平均分 DESC;

# 21、查询各科成绩前三名的记录:(不考虑成绩并列情况)

SELECT * FROM score ORDER BY course_id;

# 22、查询每门课程被选修的学生数；

    -- 分析  根据成绩根据课程进行分类,按照学生ID统计次数

SELECT score.course_id 课程ID,count(student_id) 课程人次 FROM score GROUP BY course_id;

# 23、查询出只选修了一门课程的全部学生的学号和姓名；

SELECT student.sid,student.sname FROM

    (SELECT * FROM score GROUP BY student_id HAVING count(student_id)=1) AS A

    INNER JOIN student ON A.student_id = student.sid;

# 24、查询男生、女生的人数；

SELECT boy.男,girl.女 FROM

    (SELECT count(gender) 男 FROM student WHERE gender="男") as boy,

    (SELECT count(gender) 女 FROM student WHERE gender="女") as girl;

# 25、查询姓“张”的学生名单；

SELECT sid,sname FROM student WHERE sname LIKE "张%";

# 26、查询同名同姓学生名单，并统计同名人数；

SELECT sname 姓名,count(sname) 同名人数 FROM student GROUP BY sname;

# 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；

SELECT A.cid 课程号,B.课程平均分 FROM

    (SELECT teacher.tname,course.cname,course.cid FROM teacher,course WHERE course.teacher_id = teacher.tid) AS A

    INNER JOIN

    (SELECT avg(number) 课程平均分,teacher.tname,score.course_id FROM teacher,score GROUP BY course_id) AS B

    ON A.cid=B.course_id ORDER BY B.课程平均分 DESC;

# 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；

SELECT student.sid 学号,student.sname 姓名,avg(score.number) 平均成绩 FROM

    student INNER JOIN score ON student.sid = score.student_id GROUP BY sname;

# 29、查询课程名称为“生物”，且分数低于60的学生姓名和分数；

SELECT course.cname 课程,student.sname 姓名,score.number 分数 FROM score INNER JOIN student INNER JOIN course

    ON score.student_id = student.sid AND score.course_id=course.cid

    WHERE course.cname="生物" AND score.number<60;

# 30、查询课程编号为3且课程成绩在80分以上的学生的学号和姓名；

SELECT student.sid 学号,student.sname 姓名 FROM score INNER JOIN student

    ON score.student_id = student.sid AND score.course_id=3 AND score.number>80;

# 31、求选了课程的学生人数

INSERT INTO student(sname, gender, class_id) VALUES("李四","男",3);

INSERT INTO student(sname, gender, class_id) VALUES("胜七","女",2);

INSERT INTO score(student_id, course_id, number) VALUES(7,1,74);

-- 答案如下

SELECT count(A.student_id) 选课人数 FROM (SELECT DISTINCT student_id FROM score) AS A;

# 32、查询选修“苍空”老师所授课程的学生中，成绩最高的学生姓名及其成绩；

SELECT student.sname 姓名,max(number) 成绩 FROM score INNER JOIN student INNER JOIN course INNER JOIN teacher

    ON score.student_id=student.sid AND score.course_id=course.cid AND course.teacher_id=teacher.tid

    WHERE teacher.tname="苍空";

# 33、查询各个课程及相应的选修人数；

SELECT score.course_id 课程ID,count(student_id) 课程人次 FROM score GROUP BY course_id;

# 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；

#

# 36、检索至少选修两门课程的学生学号；

    -- 分析 根据学生ID进行分组，统计课程出现次数,筛选课程次数大于等于2的学生

SELECT student_id FROM score GROUP BY student_id HAVING count(course_id)>=2;

# 37、查询全部学生都选修的课程的课程号和课程名；

SELECT score.student_id 学生ID,course.cname 所选课程,course.cid 课程ID FROM score INNER JOIN course

    ON score.course_id = course.cid ORDER BY 学生ID;

# 38、查询没学过“叶平”老师讲授的任一门课程的学生姓名；

#

# 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 A.student_id 学号,B.平均成绩 FROM

    (SELECT student_id FROM score WHERE number<60 GROUP BY student_id HAVING count(number)>=2) AS A

INNER JOIN

    (SELECT student_id,avg(number) 平均成绩 FROM score GROUP BY student_id) AS B ON A.student_id=B.student_id;

# 40、检索“1”课程分数小于60，按分数降序排列的同学学号；

SELECT student_id FROM score WHERE course_id=1 AND number<60;

# 41、删除“2”同学的“1”课程的成绩；

    -- 此题已答
巴特西

mysql考试总结

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