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  • 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
    

      

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  • 原文地址:https://www.cnblogs.com/nulige/p/6544052.html
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