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MySQL数据库练习题

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标签:形式   having   rom   思路   情况   连接   article   筛选   获取   

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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=) 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

附注参考:

http://www.cnblogs.com/wupeiqi/articles/5729934.html

http://www.cnblogs.com/wupeiqi/articles/5748496.html

MySQL数据库练习题

标签:形式   having   rom   思路   情况   连接   article   筛选   获取   

原文地址:https://www.cnblogs.com/june-L/p/12173661.html

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