MySQL数据库练习题
Posted 梦想一步登天
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了MySQL数据库练习题相关的知识,希望对你有一定的参考价值。
二、练习操作
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=\'男\') as A , 186 (select count(1) as feman from student where gender=\'女\') 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
附注参考:
http://www.cnblogs.com/wupeiqi/articles/5729934.html
http://www.cnblogs.com/wupeiqi/articles/5748496.html
以上是关于MySQL数据库练习题的主要内容,如果未能解决你的问题,请参考以下文章
Python练习册 第 0013 题: 用 Python 写一个爬图片的程序,爬 这个链接里的日本妹子图片 :-),(http://tieba.baidu.com/p/2166231880)(代码片段
spring练习,在Eclipse搭建的Spring开发环境中,使用set注入方式,实现对象的依赖关系,通过ClassPathXmlApplicationContext实体类获取Bean对象(代码片段