现有四张表

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)

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)

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)

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

#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;

#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；

#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；

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；

#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；

#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 BLEFT 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；

#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；

#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； 

#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；

#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；

#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 studenton score.s_id = student.s_idwhere 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；

# 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；

#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；

#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；

#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

#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

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

#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

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_idGROUP 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] 

#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.

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

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

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

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

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

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

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

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_idwhere 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

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；

#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_idGROUP 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

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

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

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

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

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

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

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

#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

#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

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

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

DELETE from score where score.s_id = '2' and score.c_id ='1'