MySQL习题
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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
MySQL习题
标签：let 表连接 like max 名称 大于 筛选 生成 font
本文系统来源：https://www.cnblogs.com/folishJ/p/11137561.html
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