现有四张表
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mysql> select * from student; +------+--------+-------+-------+ | s_id | s_name | s_age | s_sex | +------+--------+-------+-------+ | 1 | 鲁班 | 12 | 男 | | 2 | 貂蝉 | 20 | 女 | | 3 | 刘备 | 35 | 男 | | 4 | 关羽 | 34 | 男 | | 5 | 张飞 | 33 | 女 | +------+--------+-------+-------+ 5 rows in set (0.00 sec)
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mysql> select * from teacher; +------+-----------+ | t_id | t_name | +------+-----------+ | 1 | 张雪峰 | | 2 | 老子 | | 3 | 墨子 | +------+-----------+ 3 rows in set (0.00 sec)
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mysql> select * from course; +------+--------+------+ | c_id | c_name | t_id | +------+--------+------+ | 1 | python | 1 | | 2 | java | 1 | | 3 | linux | 3 | | 4 | web | 2 | +------+--------+------+ 4 rows in set (0.00 sec)
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mysql> select * from score; +-------+------+------+---------+ | sc_id | s_id | c_id | s_score | +-------+------+------+---------+ | 1 | 1 | 1 | 79 | | 2 | 1 | 2 | 78 | | 3 | 1 | 3 | 35 | | 4 | 2 | 2 | 32 | | 5 | 3 | 1 | 66 | | 6 | 4 | 2 | 77 | | 7 | 4 | 1 | 68 | | 8 | 5 | 1 | 66 | | 9 | 2 | 1 | 69 | | 10 | 4 | 4 | 75 | | 11 | 5 | 4 | 75 | +-------+------+------+---------+ 11 rows in set (0.00 sec)
有以下需求:
1、查询课程编号“001”比课程编号“002” 成绩高的所有学生的学号
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#1.先查询001课程和"002"课程的学生成绩,临时表 #2.让两个临时表进行比较 select a.s_id from (select * from score where c_id ='1') a, (select * from score where c_id ='2') b where a.s_id = b.s_id and a.s_score > b.s_score;
2、查询平均成绩大于60分的同学的学号和平均成绩;
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#1.先查询学生的学号和平均成绩 #2.再进行条件过滤 select s_id, avg(s_score) as sc from score GROUP BY s_id having sc>60;
3、查询所有同学的学号、姓名、选课数、总成绩;
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#1.先查学生表中的字段 #2.然后再连表查询成绩表中的字段 select s.s_id,s.s_name,COUNT(sc.c_id)AS'选课数',sum(sc.s_score) from student s LEFT JOIN score sc on s.s_id = sc.s_id GROUP BY s.s_id
4、查询含有"子"的老师的个数;
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select count(t_id) from teacher where t_name like'%子%'
5、查询没学过“老子”老师课的同学的学号、姓名;
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#1.先查询"老子"老师教什么课程 #2.再查询学过该老师课程的学生有哪些 #3.排除学过该老师课的学生,剩下的就是没有学过的学生 select s_id,s_name from student where s_id not in( select s_id FROM score where c_id = (select c_id from teacher,course where teacher.t_id = course.t_id and t_name ='老子') )
6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
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#1.注意:是即学过001也学过002课程的学生 #2.思路:先查询有哪些学生学了'001'或者'002'课程 #3.然后进行分组,学科数 = 2 表示学了两门学科 select student.s_id,student.s_name FROM (select s_id from score se where se.c_id='1' or se.c_id ='2' GROUP BY se.s_id HAVING count(c_id)>1) as B LEFT JOIN student on student.s_id = B.s_id;
7、查询学过“老子”老师所教的所有课的同学的学号、姓名;
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#1.先查询"老子"老师教哪些课程 #2.再查询哪些学生学习了这些课程 #3.再根据学生编号分组,如果分组后的个数 ="老子"老师所教授课程的个数,则表示学过该老师所有课程. select s_id,s_name from student where s_id in( select s_id FROM score where c_id in( select c_id from teacher,course where teacher.t_id = course.t_id and t_name ='老子' ) group by s_id having count(s_id) =( select count(c_id) from teacher,course where teacher.t_id = course.t_id and t_name ='老子') )
8、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;
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#同第一题 select a.s_id from (select * from score where c_id ='1') a, (select * from score where c_id ='2') b where a.s_id = b.s_id and a.s_score < b.s_score;
9、查询有课程成绩小于60分的同学的学号、姓名;
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#1.查询所有成绩分数小于60分的同学 #2.关联学生表,去重复 select DISTINCT student.s_id,student.s_name from score,student where score.s_id=student.s_id and s_score < 60
10、查询没有学全所有课的同学;
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#1.分数表分组得到学生选课数量 #2.选课数量 = 课程表总课程 select student.* from score LEFT JOIN student on score.s_id = student.s_id GROUP BY score.s_id HAVING count(score.s_id) = (select count(c_id) from course);
11、查询至少有一门课与学号为“002”的同学所学相同的同学的学号和姓名;
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#1 002学生学了什么课程 #2.其他学过002学生其中一门课程的学生 #3.关联学生表 select student.s_id,student.s_name from score LEFT JOIN student on score.s_id = student.s_id where score.c_id in(select c_id from score where s_id = '2') and score.s_id !='2' GROUP BY score.s_id
12、查询学过 学号为“002”同学全部课程 的其他同学的学号和姓名;
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# 1先找到学过002同学课程的人 # 2.课程个数 = 002学生课程个数 # 3.关联学生表,如果不显示自身就去掉 select student.s_id,student.s_name from score LEFT JOIN student on score.s_id = student.s_id where score.c_id in(select c_id from score where score.s_id ='2') and score.s_id !='2' GROUP BY score.s_id having count(score.s_id) =(select count(c_id) from score where score.s_id ='2')
13、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;
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#1.找出与002学生学习课程数相同的学生(你学两门,我也学两门) #2.然后再找出学过'002'学生课程的学生,剩下的一定是至少学过一门002课程的学生 #3.再根据学生ID进行分组,剩下学生数count(1) = 002学生所学课程数 SELECT * FROM score where score.s_id in( select score.s_id from score GROUP BY s_id HAVING count(1) =(select count(1) from score where score.s_id = '2') ) and score.c_id in (select c_id from score where score.s_id = '2') and score.s_id!='2' GROUP BY score.s_id HAVING count(1) = (select count(1) from score where score.s_id = '2')
14、把“score”表中“老子”老师教的课的成绩都更改为此课程的平均成绩;
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#1.获得"老子"老师所教的课程号 -- select c_id from course LEFT JOIN teacher on teacher.t_id = course.t_id and teacher.t_name ='老子'; #2. 获得"老子"老师课程的平均成绩 -- select AVG(score.s_score) s_score from score where score.c_id -- in(select c_id from course,teacher where teacher.t_id = course.t_id and teacher.t_name ='老子') -- #3.注意:如果直接把上面的查询结果作为更新字段,则会报错(不能先select出同一表中的某些值,再update这个表(在同一语句中)) #所以 需要将查询结果集包装(加一层查询)变为临时表.则可以作为更新字段 update score SET s_score = ( select bb.s_score from ( select AVG(s_score) s_score from score where score.c_id in(select c_id from course,teacher where teacher.t_id = course.t_id and teacher.t_name ='老子') )as bb) where score.c_id in (select c_id from course,teacher where teacher.t_id = course.t_id and teacher.t_name ='老子')
15、删除学习“墨子”老师课的score表记录;
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#1.找到墨子老师教的课程 #2.根据课程号直接删除 DELETE from score where c_id in(select c_id from course INNER JOIN teacher on teacher.t_id = course.t_id and teacher.t_name = '墨子')
16、按平均成绩从高到低显示所有学生的“python”、“java”、“linux”三门的课程成绩,按如下形式显示: 学生ID,python,java,linux,有效课程数,有效平均分
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#1.学生python课程的平均成绩是多少? select sc1.s_score from course c,score sc1 where c.c_id = sc1.c_id and c.c_name = 'python' and sc1.s_id = 1 ORDER BY sc1.s_score desc; select sc.s_score from course c,score sc where c.c_id = sc.c_id and c.c_name = 'java' and sc.s_id = 1 ORDER BY sc.s_score desc; select sc.s_score from course c,score sc where c.c_id = sc.c_id and c.c_name = 'linux' and sc.s_id = 1 ORDER BY sc.s_score desc #2.学生id,有效课程数,有效平均分如何查询? select sc.s_id, count(*), AVG(sc.s_score) from score sc,course c where sc.c_id = c.c_id GROUP BY sc.s_id; #3.组合SQL:按平均分排序 select sc.s_id, (select sc1.s_score from course c,score sc1 where c.c_id = sc1.c_id and c.c_name = 'python' and sc1.s_id = sc.s_id ORDER BY sc1.s_score desc)as 'python', (select sc1.s_score from course c,score sc1 where c.c_id = sc1.c_id and c.c_name = 'java' and sc1.s_id = sc.s_id ORDER BY sc1.s_score desc)as 'java', (select sc1.s_score from course c,score sc1 where c.c_id = sc1.c_id and c.c_name = 'linux' and sc1.s_id = sc.s_id ORDER BY sc1.s_score desc)as 'linux', count(*) as '课程数', AVG(sc.s_score) as '平均分' from score sc,course c where sc.c_id = c.c_id GROUP BY sc.s_id order by AVG(sc.s_score) desc;
17、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
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select c_id,MAX(s_score),MIN(s_score) from score GROUP BY c_id
18、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
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#1. case when .... then ...else ... end #2.先获得学生ID,各科平均成绩 #3.计算及格率. 规则:及格课数/总科数 *100 select sc.c_id as '学生号', avg(sc.s_score) as '平均成绩', sum(case when sc.s_score >=60 then 1 ELSE 0 end)/count(1) * 100 as '及格率' from score sc GROUP BY sc.c_id order by avg(sc.s_score) asc , sum(case when sc.s_score >=60 then 1 ELSE 0 end)/count(1) * 100 desc;
19、查询老师所教课程平均分从高到低显示,并显示老师的名称及课程名称
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select teacher.t_name,avg(score.s_score),course.c_name from teacher LEFT JOIN course on course.t_id = teacher.t_id LEFT join score on score.c_id = course.c_id GROUP BY score.c_id
20、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
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#1.先统计出课程ID和课程名称,可以利用分组 #2.通过判断语句计算和的方式获得分数段人数 select score.c_id, course.c_name, sum(case when score.s_score between 85 and 100 THEN 1 ELSE 0 END) as '[100-85]', sum(case when score.s_score between 70 and 85 THEN 1 ELSE 0 END) as '[85-70]', sum(case when score.s_score between 60 and 70 THEN 1 ELSE 0 END) as '[70-60]', sum(case when score.s_score < 60 THEN 1 ELSE 0 END) as '[<60]' from score,course where score.c_id = course.c_id group by score.c_id
21、查询每门课程被选修的学生数.
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select c_id,count(s_id) from score GROUP BY c_id
22、查询出只选修了一门课程的学生的学号和姓名
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select student.s_id,student.s_name from score LEFT JOIN student on score.s_id = student.s_id group by s_id HAVING count(1)='1';
23、查询学生表中男生、女生人数
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select sum(case when s_sex ='男' then 1 ELSE 0 end )as '男', sum(case when s_sex ='女' then 1 ELSE 0 end )as '女' from student
24、查询姓“张”的学生名单
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select * from student where student.s_name like '张%'
25、查询同名学生名单,并统计同名人数
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select s_name,count(1) from student group by s_name;
26、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
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select AVG(IFNULL(s1.s_score,0)) from score s1 GROUP BY s1.c_id ORDER BY AVG(IFNULL(s1.s_score,0)) asc,s1.c_id DESC
27、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
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select st1.*,avg(sc2.s_score) from student st1,score sc2 where st1.s_id = sc2.s_id GROUP BY sc2.s_id HAVING avg(sc2.s_score)>65
28、查询课程名称为“python”,且分数低于60的学生姓名和分数
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select student.s_name,score.s_score from score LEFT JOIN course on score.c_id = course.c_id left join student on student.s_id = score.s_id where course.c_name ='python' AND score.s_score <60
29、查询所有学生的选课情况,显示 学生编号,学生姓名,所选课程名称
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select score.s_id,student.s_name,c_name from score LEFT JOIN student on student.s_id = score.s_id LEFT join course on course.c_id = score.c_id
30、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
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#方式一 #学生编号分组,获得最小分数 select st.s_name,c.c_name,sc.s_score from score sc,student st,course c where sc.s_id = st.s_id and sc.c_id = c.c_id GROUP BY sc.s_id HAVING MIN(sc.s_score) > 70 #方式二 select s2.s_name,c3.c_name,s1.s_score from score s1,student s2,course c3 where s1.s_id = s2.s_id and s1.c_id = c3.c_id GROUP BY s2.s_id HAVING sum(case when s1.s_score>60 THEN 1 ELSE 0 end) = (select count(c_id) from score where s2.s_id= score.s_id GROUP BY score.s_id)
31、查询不及格的课程,并按课程号从大到小排列
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select c_id from score where score.s_score < 60 ORDER BY c_id DESC
32、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
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select student.s_id,student.s_name from score LEFT JOIN student on score.s_id = student.s_id where score.s_score>60 AND score.c_id = 2
33、求 已选课程的学生人数
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select count( DISTINCT s_id) as '人数' from score
34、查询选修“老子”老师所授课程的学生中,成绩最高的学生姓名及其成绩
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select student.s_name,score.s_score from teacher LEFT JOIN course on teacher.t_id = course.t_id LEFT JOIN score on score.c_id = course.c_id LEFT JOIN student on score.s_id = student.s_id where teacher.t_name = '老子' order BY score.s_score desc LIMIT 1
35、查询各个课程及相应的选修人数
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select score.c_id,course.c_name,count(score.s_id) as'选修人数' from score LEFT JOIN course on course.c_id = score.c_id GROUP BY score.c_id
36、查询不同课程但成绩相同的学生的学号、课程号、学生成绩
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select s1.s_id, s1.c_id, s1.s_score from score s1,score s2 where s1.s_score = s2.s_score and s1.c_id != s2.c_id
37、检索至少选修两门课程的学生学号
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select s_id from score GROUP BY s_id HAVING COUNT(s_id)>1
38、查询全部学生都选修的课程的课程号和课程名
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#1.学生数量 = 分组的课程数量 select score.c_id,course.c_name from score LEFT JOIN course ON score.c_id = course.c_id GROUP BY score.c_id HAVING count(score.c_id) = (select count(1) from student)
39、查询没学过“老子”老师讲授的任一门课程的学生姓名
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#1.先查询"老子"老师教什么课程 #2.再查询学过该老师课程的学生有哪些 #3.排除学过该老师课的学生,剩下的就是没有学过的学生 select s_id,s_name from student where s_id not in( select s_id FROM score where c_id = (select c_id from teacher,course where teacher.t_id = course.t_id and t_name ='老子') )
40、查询两门以上不及格课程的同学的学号及其平均成绩
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select score.s_id,avg(score.s_score),COUNT(1) from score where score.s_score <60 GROUP BY score.s_id HAVING COUNT(1)>1
41、检索“003”课程分数小于60,按分数降序排列的同学学号
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select s_id from score where score.c_id='1' and score.s_score < 160 ORDER BY score.s_score desc
42、删除“002”同学的“001”课程的成绩
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DELETE from score where score.s_id = '2' and score.c_id ='1'