• MySQL练习题


    MySQL测试题

    一、表关系

    请创建如下表,并创建相关约束

    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;

    3、查询平均成绩大于60分的同学的学号和平均成绩; 

    4、查询所有同学的学号、姓名、选课数、总成绩;

    5、查询姓“李”的老师的个数;

    6、查询没学过“叶平”老师课的同学的学号、姓名;

    7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;

    8、查询学过“叶平”老师所教的所有课的同学的学号、姓名;

    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;

    10、查询有课程成绩小于60分的同学的学号、姓名;

    11、查询没有学全所有课的同学的学号、姓名;

    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”同学所选课程中任意一门课的其他同学学号和姓名;

    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;

    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;

    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;

    36、检索至少选修两门课程的学生学号;

    37、查询全部学生都选修的课程的课程号和课程名;

    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;

    40、检索“004”课程分数小于60,按分数降序排列的同学学号;

    41、删除“002”同学的“001”课程的成绩;

    MySQL练习题参考答案

     

    导出现有数据库数据:

    • mysqldump -u用户名 -p密码 数据库名称 >导出文件路径           # 结构+数据
    • mysqldump -u用户名 -p密码 -d 数据库名称 >导出文件路径       # 结构 

    导入现有数据库数据:

    • mysqldump -uroot -p密码  数据库名称 < 文件路径  
    Navicat Premium Data Transfer
    
     Source Server         : localhost
     Source Server Type    : MySQL
     Source Server Version : 50624
     Source Host           : localhost
     Source Database       : sqlexam
    
     Target Server Type    : MySQL
     Target Server Version : 50624
     File Encoding         : utf-8
    
     Date: 10/21/2016 06:46:46 AM
    */
    
    SET NAMES utf8;
    SET FOREIGN_KEY_CHECKS = 0;
    
    -- ----------------------------
    --  Table structure for `class`
    -- ----------------------------
    DROP TABLE IF EXISTS `class`;
    CREATE TABLE `class` (
      `cid` int(11) NOT NULL AUTO_INCREMENT,
      `caption` varchar(32) NOT NULL,
      PRIMARY KEY (`cid`)
    ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;
    
    -- ----------------------------
    --  Records of `class`
    -- ----------------------------
    BEGIN;
    INSERT INTO `class` VALUES ('1', '三年二班'), ('2', '三年三班'), ('3', '一年二班'), ('4', '二年九班');
    COMMIT;
    
    -- ----------------------------
    --  Table structure for `course`
    -- ----------------------------
    DROP TABLE IF EXISTS `course`;
    CREATE TABLE `course` (
      `cid` int(11) NOT NULL AUTO_INCREMENT,
      `cname` varchar(32) NOT NULL,
      `teacher_id` int(11) NOT NULL,
      PRIMARY KEY (`cid`),
      KEY `fk_course_teacher` (`teacher_id`),
      CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`)
    ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;
    
    -- ----------------------------
    --  Records of `course`
    -- ----------------------------
    BEGIN;
    INSERT INTO `course` VALUES ('1', '生物', '1'), ('2', '物理', '2'), ('3', '体育', '3'), ('4', '美术', '2');
    COMMIT;
    
    -- ----------------------------
    --  Table structure for `score`
    -- ----------------------------
    DROP TABLE IF EXISTS `score`;
    CREATE TABLE `score` (
      `sid` int(11) NOT NULL AUTO_INCREMENT,
      `student_id` int(11) NOT NULL,
      `course_id` int(11) NOT NULL,
      `num` int(11) NOT NULL,
      PRIMARY KEY (`sid`),
      KEY `fk_score_student` (`student_id`),
      KEY `fk_score_course` (`course_id`),
      CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`),
      CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`)
    ) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;
    
    -- ----------------------------
    --  Records of `score`
    -- ----------------------------
    BEGIN;
    INSERT INTO `score` VALUES ('1', '1', '1', '10'), ('2', '1', '2', '9'), ('5', '1', '4', '66'), ('6', '2', '1', '8'), ('8', '2', '3', '68'), ('9', '2', '4', '99'), ('10', '3', '1', '77'), ('11', '3', '2', '66'), ('12', '3', '3', '87'), ('13', '3', '4', '99'), ('14', '4', '1', '79'), ('15', '4', '2', '11'), ('16', '4', '3', '67'), ('17', '4', '4', '100'), ('18', '5', '1', '79'), ('19', '5', '2', '11'), ('20', '5', '3', '67'), ('21', '5', '4', '100'), ('22', '6', '1', '9'), ('23', '6', '2', '100'), ('24', '6', '3', '67'), ('25', '6', '4', '100'), ('26', '7', '1', '9'), ('27', '7', '2', '100'), ('28', '7', '3', '67'), ('29', '7', '4', '88'), ('30', '8', '1', '9'), ('31', '8', '2', '100'), ('32', '8', '3', '67'), ('33', '8', '4', '88'), ('34', '9', '1', '91'), ('35', '9', '2', '88'), ('36', '9', '3', '67'), ('37', '9', '4', '22'), ('38', '10', '1', '90'), ('39', '10', '2', '77'), ('40', '10', '3', '43'), ('41', '10', '4', '87'), ('42', '11', '1', '90'), ('43', '11', '2', '77'), ('44', '11', '3', '43'), ('45', '11', '4', '87'), ('46', '12', '1', '90'), ('47', '12', '2', '77'), ('48', '12', '3', '43'), ('49', '12', '4', '87'), ('52', '13', '3', '87');
    COMMIT;
    
    -- ----------------------------
    --  Table structure for `student`
    -- ----------------------------
    DROP TABLE IF EXISTS `student`;
    CREATE TABLE `student` (
      `sid` int(11) NOT NULL AUTO_INCREMENT,
      `gender` char(1) NOT NULL,
      `class_id` int(11) NOT NULL,
      `sname` varchar(32) NOT NULL,
      PRIMARY KEY (`sid`),
      KEY `fk_class` (`class_id`),
      CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`)
    ) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;
    
    -- ----------------------------
    --  Records of `student`
    -- ----------------------------
    BEGIN;
    INSERT INTO `student` VALUES ('1', '', '1', '理解'), ('2', '', '1', '钢蛋'), ('3', '', '1', '张三'), ('4', '', '1', '张一'), ('5', '', '1', '张二'), ('6', '', '1', '张四'), ('7', '', '2', '铁锤'), ('8', '', '2', '李三'), ('9', '', '2', '李一'), ('10', '', '2', '李二'), ('11', '', '2', '李四'), ('12', '', '3', '如花'), ('13', '', '3', '刘三'), ('14', '', '3', '刘一'), ('15', '', '3', '刘二'), ('16', '', '3', '刘四');
    COMMIT;
    
    -- ----------------------------
    --  Table structure for `teacher`
    -- ----------------------------
    DROP TABLE IF EXISTS `teacher`;
    CREATE TABLE `teacher` (
      `tid` int(11) NOT NULL AUTO_INCREMENT,
      `tname` varchar(32) NOT NULL,
      PRIMARY KEY (`tid`)
    ) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;
    
    -- ----------------------------
    --  Records of `teacher`
    -- ----------------------------
    BEGIN;
    INSERT INTO `teacher` VALUES ('1', '张磊老师'), ('2', '李平老师'), ('3', '刘海燕老师'), ('4', '朱云海老师'), ('5', '李杰老师');
    COMMIT;
    
    SET FOREIGN_KEY_CHECKS = 1;
    表数据结构
      1 2、查询“生物”课程比“物理”课程成绩高的所有学生的学号;
      2 思路:
      3     获取所有有生物课程的人(学号,成绩) - 临时表
      4     获取所有有物理课程的人(学号,成绩) - 临时表
      5     根据【学号】连接两个临时表:
      6         学号  物理成绩   生物成绩
      7  
      8     然后再进行筛选
      9  
     10         select A.student_id,sw,ty from
     11  
     12         (select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = '生物') as A
     13  
     14         left join
     15  
     16         (select student_id,num  as ty from score left join course on score.course_id = course.cid where course.cname = '体育') as B
     17  
     18         on A.student_id = B.student_id where sw > if(isnull(ty),0,ty);
     19  
     20 3、查询平均成绩大于60分的同学的学号和平均成绩; 
     21     思路:
     22         根据学生分组,使用avg获取平均值,通过having对avg进行筛选
     23  
     24         select student_id,avg(num) from score group by student_id having avg(num) > 60
     25  
     26 4、查询所有同学的学号、姓名、选课数、总成绩;
     27  
     28     select score.student_id,sum(score.num),count(score.student_id),student.sname 
     29     from
     30         score left join student on score.student_id = student.sid   
     31     group by score.student_id
     32  
     33 5、查询姓“李”的老师的个数;
     34     select count(tid) from teacher where tname like '李%'
     35  
     36     select count(1) from (select tid from teacher where tname like '李%') as B
     37  
     38 6、查询没学过“叶平”老师课的同学的学号、姓名;
     39     思路:
     40         先查到“李平老师”老师教的所有课ID
     41         获取选过课的所有学生ID
     42         学生表中筛选
     43     select * from student where sid not in (
     44         select DISTINCT student_id from score where score.course_id in (
     45             select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老师'
     46         )
     47     )
     48  
     49 7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
     50     思路:
     51         先查到既选择001又选择002课程的所有同学
     52         根据学生进行分组,如果学生数量等于2表示,两门均已选择
     53  
     54     select student_id,sname from
     55  
     56     (select student_id,course_id from score where course_id = 1 or course_id = 2) as B
     57       
     58     left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1
     59  
     60  
     61 8、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
     62  
     63     同上,只不过将001和002变成 in (叶平老师的所有课)
     64  
     65 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;
     66     同第1题
     67  
     68  
     69 10、查询有课程成绩小于60分的同学的学号、姓名;
     70          
     71     select sid,sname from student where sid in (
     72         select distinct student_id from score where num < 60
     73     )
     74  
     75 11、查询没有学全所有课的同学的学号、姓名;
     76     思路:
     77         在分数表中根据学生进行分组,获取每一个学生选课数量
     78         如果数量 == 总课程数量,表示已经选择了所有课程
     79  
     80         select student_id,sname 
     81         from score left join student on score.student_id = student.sid 
     82         group by student_id HAVING count(course_id) = (select count(1) from course)
     83  
     84  
     85 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;
     86     思路:
     87         获取 001 同学选择的所有课程
     88         获取课程在其中的所有人以及所有课程
     89         根据学生筛选,获取所有学生信息
     90         再与学生表连接,获取姓名
     91  
     92         select student_id,sname, count(course_id) 
     93         from score left join student on score.student_id = student.sid
     94         where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id
     95  
     96 13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名;
     97         先找到和001的学过的所有人
     98         然后个数 = 001所有学科     ==》 其他人可能选择的更多
     99  
    100         select student_id,sname, count(course_id) 
    101         from score left join student on score.student_id = student.sid
    102         where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) = (select count(course_id) from score where student_id = 1)
    103  
    104 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;
    105          
    106         个数相同
    107         002学过的也学过
    108  
    109         select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
    110             select student_id from score  where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
    111         ) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
    112  
    113  
    114 15、删除学习“叶平”老师课的score表记录;
    115  
    116     delete from score where course_id in (
    117         select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = '叶平'
    118     )
    119  
    120 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; 
    121     思路:
    122         由于insert 支持  
    123                 inset into tb1(xx,xx) select x1,x2 from tb2;
    124         所有,获取所有没上过002课的所有人,获取002的平均成绩
    125  
    126     insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2) 
    127     from student where sid not in (
    128         select student_id from score where course_id = 2
    129     )
    130      
    131 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;
    132     select sc.student_id,
    133         (select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,
    134         (select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,
    135         (select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,
    136         count(sc.course_id),
    137         avg(sc.num)
    138     from score as sc
    139     group by student_id desc        
    140  
    141 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;
    142      
    143     select course_id, max(num) as max_num, min(num) as min_num from score group by course_id;
    144  
    145 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;
    146     思路:case when .. then
    147     select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;
    148  
    149 20、课程平均分从高到低显示(现实任课老师);
    150  
    151     select avg(if(isnull(score.num),0,score.num)),teacher.tname from course 
    152     left join score on course.cid = score.course_id 
    153     left join teacher on course.teacher_id = teacher.tid
    154  
    155     group by score.course_id
    156  
    157  
    158 21、查询各科成绩前三名的记录:(不考虑成绩并列情况) 
    159     select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
    160     (
    161     select
    162         sid,
    163         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
    164         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 3,1) as second_num
    165     from
    166         score as s1
    167     ) as T
    168     on score.sid =T.sid
    169     where score.num <= T.first_num and score.num >= T.second_num
    170  
    171 22、查询每门课程被选修的学生数;
    172      
    173     select course_id, count(1) from score group by course_id;
    174  
    175 23、查询出只选修了一门课程的全部学生的学号和姓名;
    176     select student.sid, student.sname, count(1) from score
    177  
    178     left join student on score.student_id  = student.sid
    179  
    180      group by course_id having count(1) = 1
    181  
    182  
    183 24、查询男生、女生的人数;
    184     select * from
    185     (select count(1) as man from student where gender='') as A ,
    186     (select count(1) as feman from student where gender='') as B 
    187  
    188 25、查询姓“张”的学生名单;
    189     select sname from student where sname like '张%';
    190  
    191 26、查询同名同姓学生名单,并统计同名人数;
    192  
    193     select sname,count(1) as count from student group by sname;
    194  
    195 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;
    196     select course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg     asc,course_id desc;
    197  
    198 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;
    199  
    200     select student_id,sname, avg(if(isnull(num), 0 ,num)) from score left join student on score.student_id = student.sid group by student_id;
    201  
    202 29、查询课程名称为“数学”,且分数低于60的学生姓名和分数;
    203  
    204     select student.sname,score.num from score 
    205     left join course on score.course_id = course.cid
    206     left join student on score.student_id = student.sid
    207     where score.num < 60 and course.cname = '生物'
    208  
    209 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; 
    210     select * from score where score.student_id = 3 and score.num > 80
    211  
    212 31、求选了课程的学生人数
    213  
    214     select count(distinct student_id) from score
    215  
    216     select count(c) from (
    217         select count(student_id) as c from score group by student_id) as A
    218  
    219 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;
    220      
    221     select sname,num from score 
    222     left join student on score.student_id = student.sid
    223     where score.course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname='张磊老师') order by num desc limit 1;
    224  
    225 33、查询各个课程及相应的选修人数;
    226     select course.cname,count(1) from score
    227     left join course on score.course_id = course.cid
    228     group by course_id;
    229  
    230  
    231 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;
    232     select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;
    233  
    234 35、查询每门课程成绩最好的前两名;
    235  
    236     select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
    237     (
    238     select
    239         sid,
    240         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
    241         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1,1) as second_num
    242     from
    243         score as s1
    244     ) as T
    245     on score.sid =T.sid
    246     where score.num <= T.first_num and score.num >= T.second_num
    247  
    248 36、检索至少选修两门课程的学生学号;
    249     select student_id from score group by student_id having count(student_id) > 1
    250  
    251 37、查询全部学生都选修的课程的课程号和课程名;
    252     select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student);
    253  
    254 38、查询没学过“叶平”老师讲授的任一门课程的学生姓名;
    255     select student_id,student.sname from score 
    256     left join student on score.student_id = student.sid
    257     where score.course_id not in (
    258         select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '张磊老师'
    259     ) 
    260     group by student_id
    261  
    262 39、查询两门以上不及格课程的同学的学号及其平均成绩;
    263  
    264     select student_id,count(1) from score where num < 60 group by student_id having count(1) > 2
    265  
    266 40、检索“004”课程分数小于60,按分数降序排列的同学学号;
    267     select student_id from score where num< 60 and course_id = 4 order by num desc;
    268  
    269 41、删除“002”同学的“001”课程的成绩;
    270     delete from score where course_id = 1 and student_id = 2

  • 相关阅读:
    [Leetcode] Convert Sorted List to Binary Search Tree
    [Leetcode] Sqrt(x)
    [Leetcode] Pow(x, n)
    [Leetcode] Balanced Binary Tree
    [Leetcode] Convert Sorted Array to Binary Search Tree
    [Leetcode] Construct Binary Tree from Preorder and Inorder Traversal
    [Leetcode] Remove Element
    [Leetcode] Letter Combinations of a Phone Number
    [Leetcode] Generate Parentheses
    [Leetcode] Valid Parentheses
  • 原文地址:https://www.cnblogs.com/shiluoliming/p/6625016.html
Copyright © 2020-2023  润新知