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t_tudent(sid,sname,sage,ssex,sdept) 学生表

t_course(cid,cname,tid) 课程表

t_score( scid,sid,cid,grade) 成绩表

t_teacher(tid,tname) 教师表

问题：

1、查询“001”课程比“002”课程成绩高的所有学生的学号;

select t1.sid from

(select sid,grade from t_score where cid = ‘001‘) t1,

(select sid,grade from t_score where cid = ‘002‘) t2

where t1.grade >t2.grade and t1.sid = t2.sid;

select t2.sid,t2.ca,t2.cb(

select t1.sid,sum(t1.ca) ca,sum(t1.cb) cb

from

(

select sid,

CASE WHEN cid = ‘001‘ THEN grade ELSE 0 END ca,

CASE WHEN cid = ‘002‘ THEN grade ELSE 0 END cb

from t_score

)t1

GROUP BY sid

)t2

where t2.ca > t2.cb

2、查询平均成绩大于60分的同学的学号和平均成绩;

select sid,avg(grade ) from t_score

group by sid

having avg(grade ) >60;

3、查询所有同学的学号、姓名、选课数、总成绩;

select t1.sid,t1.sname,count(t2.cid),sum(grade ) from t_student t1

left join t_score t2

on t1.sid = t2sid

group by t1sid,t1.sname;

4、查询姓“李”的老师的个数;

select count(distinct(tname)) from t_teacher where tname like ‘李%‘;

5、查询没学过“叶平”老师课的同学的学号、姓名;

select sid, sname from t_student

where sid not in

(

select distinct(t1.sid) from t_score t1,t_course t2,t_teacher t3

where t1.cid = t2.cid and t3.tid = t2.tid and t3.tname = ‘叶平‘

);

6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;

select t0.sid,t0.sname from t_student t0, t_score t1

where t0.sid = t1.sid and t1.cid = ‘001‘

and exists

(

select * from t_score t2 where t2.sid = t1.sid and t2.cid = ‘002‘

);

7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;

select sid,sname from t_student

where sid in

(

select sid from t_score t1 ,t_course t2,t_teacher t3

where t1.cid=t2.cid

and t3.tid=t2.tid

and t3.tname = ‘叶平‘

group by sid

having count(t1.cid)=(

select count(cid) from t_course t1,t_teacher t2

where t2.tid=t1.tid and t2.tname = ‘叶平‘

)

);

8、查询“001”课程比“002”课程成绩高的所有同学的学号、姓名;

Select sid,sname from

(

select t1.sid,t1.sname,t2.grade ,

(

select grade from t_score t3

where t3.sid = t1.sid and t3.cid = ‘002‘

) grade2

from t_student t1, t_score t2

where t1sid = t2.sid and t2.cid = ‘001‘

) S_2

where grade 2 < grade ;

9、查询所有课程成绩小于60分的同学的学号、姓名;

select sid,sname from t_student

where sid not in

(

select t1.sid from t_student t1, t_score t2

where t1.sid = t2.sid and t2.grade >60

);

10、查询没有学全所有课的同学的学号、姓名;

select t1.sid,t1.sname from Student t1, t_score t2

where t1.sid = t2.sid

group by t1.sid,t1.sname

having count(t2.cid) < (select count(cid) from t_course);

11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名;

select sid,sname from t_student t1, t_score t2

where t1.sid = t2.sid

and cid in (select cid from t_score where sid = ‘1001‘);

12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名;

select distinct t2.sid,t1.sname from t_student t1, t_score t2

where t1.sid = t2.sid

and cid in (select cid from t_score where sid = ‘001‘);

13、把“ t_score”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩;

update t_score t4 set grade =(select avg( t3.grade ) from t_score t3 where t3.cid= t4.cid )

from t_course t1,t_teacher t2

where t1.cid = t4.cid

and t1.tid = t2.tid

and t2.tname = ‘叶平‘);

14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名;

select sid from t_score

where cid in (select cid from t_score where sid = ‘1002‘)

group by sid

having count(*)=(select count(*) from t_score where sid = ‘1002‘);

15、删除学习“叶平”老师课的 t_score表记录;

delect t_score from t_course t1 ,t_teacher t2

Where t1.cid = t2.cid

and t1.tid = t2.tid

and t2.tname = ‘叶平‘;

16、向 t_score表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2、

号课的平均成绩;

insert t_score

select sid,‘002‘,(Select avg(grade ) from t_score where cid = ‘002‘)

from t_student

where sid not in (Select sid from t_score where cid = ‘002‘);

17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示： 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分

SELECT sid as 学生ID,

(SELECT grade FROM t_score WHERE sid=t.sid AND cid=‘004‘) AS 数据库,

(SELECT grade FROM t_score WHERE sid=t.sid AND cid=‘001‘) AS 企业管理,

(SELECT grade FROM t_score WHERE sid=t.sid AND cid=‘006‘) AS 英语,

COUNT(*) AS 有效课程数,

AVG(t.grade ) AS 平均成绩

FROM t_score AS t

GROUP BY sid

ORDER BY avg(t.grade )

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

SELECT t1.cid As 课程ID,t1.grade AS 最高分,t2.grade AS 最低分

FROM t_score t1 , t_score t2

WHERE t1.cid = t2.cid

and t1.grade = (

SELECT MAX(tt1.grade ) FROM t_score tt1,t_student tt2

WHERE t1.cid = tt1.cid and tt2.sid = tt1.sid

GROUP BY tt1.cid

)

AND t2.grade = (

SELECT MIN(tt3.grade ) FROM t_score AS tt3

WHERE t2.cid = tt3.cid GROUP BY tt2.cid

);

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

SELECT t.cid AS 课程号,max(course.cname)AS 课程名,isnull(AVG(grade ),0) AS 平均成绩 ,

100 * SUM(CASE WHEN isnull(grade ,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数

FROM t_score t1,t_course t2

where t1.cid = t2.cid

GROUP BY t1.cid

ORDER BY 100 * SUM(CASE WHEN isnull(grade ,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC

20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理(001)，马克思(002)，OO&UML (003)，数据库(004)

SELECT

SUM(

CASE WHEN cid = ‘001‘

THEN grade ELSE 0 END)/SUM(CASE cid WHEN ‘001‘

THEN 1 ELSE 0 END

) AS 企业管理平均分 ,

100 * SUM(

CASE WHEN cid = ‘001‘ AND grade >= 60

THEN 1 ELSE 0 END)/SUM(CASE WHEN cid = ‘001‘

THEN 1 ELSE 0 END

) AS 企业管理及格百分数 FROM t_score

21、查询不同老师所教不同课程平均分从高到低显示

SELECT max(t3.tid) AS 教师ID,MAX(t3.tname) AS 教师姓名,

T2.cid AS 课程ID,MAX(t2.cname) AS 课程名称,AVG(grade ) AS 平均成绩

FROM t_score AS t1,t_course AS t2 ,t_teacher AS t3

where t1.cid = t2.cid and t2.tid = t3.tid

GROUP BY t2.cid

ORDER BY AVG(grade ) DESC ;

22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理(001)，马克思(002)，UML (003)，数据库(004)

[学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩

SELECT DISTINCT top 3 t_score.sid As 学生学号,Student.Sname AS 学生姓名 ,

T1.grade AS 企业管理, T2.grade AS 马克思, T3.grade AS UML,

T4.grade AS 数据库,ISNULL(T1.grade ,0) + ISNULL(T2.grade ,0) + ISNULL(T3.grade ,0) + ISNULL(T4.grade ,0) as 总分

FROM Student, t_score

LEFT JOIN t_score AS T1

ON t_score.sid = T1.sid AND T1.cid = ‘001‘

LEFT JOIN t_score AS T2

ON t_score.sid = T2.sid AND T2.cid = ‘002‘

LEFT JOIN t_score AS T3

ON t_score.sid = T3.sid AND T3.cid = ‘003‘

LEFT JOIN t_score AS T4

ON t_score.sid = T4.sid AND T4.cid = ‘004‘

WHERE student.sid= t_score.sid

and ISNULL(T1.grade ,0) + ISNULL(T2.grade ,0) + ISNULL(T3.grade ,0) + ISNULL(T4.grade ,0)

NOT IN (

SELECT DISTINCT TOP 15 WITH TIES ISNULL(T1.grade ,0) + ISNULL(T2.grade ,0) + ISNULL(T3.grade ,0) + ISNULL(T4.grade ,0)

FROM t_score

LEFT JOIN t_score AS T1

ON t_score.sid = T1.sid AND T1.cid = ‘k1‘

LEFT JOIN t_score AS T2

ON t_score.sid = T2.sid AND T2.cid = ‘k2‘

LEFT JOIN t_score AS T3

ON t_score.sid = T3.sid AND T3.cid = ‘k3‘

LEFT JOIN t_score AS T4

ON t_score.sid = T4.sid AND T4.cid = ‘k4‘

ORDER BY ISNULL(T1.grade ,0) + ISNULL(T2.grade ,0) + ISNULL(T3.grade ,0) + ISNULL(T4.grade ,0) DE t_score

);

23、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]

SELECT t1.cid as 课程ID, t2.cname as 课程名称,

SUM(CASE WHEN t1.grade BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85],

SUM(CASE WHEN t1.grade BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70],

SUM(CASE WHEN t1.grade BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60],

SUM(CASE WHEN t1.grade < 60 THEN 1 ELSE 0 END) AS [60 -]

FROM t_score t1,t_course t2 where t1.cid = t2.cid

GROUP BY t1.cid,t2.cname;

24、查询学生平均成绩及其名次

SELECT

1+(

SELECT COUNT( distinct avggrade )

FROM (

SELECT sid,AVG(grade ) AS avggrade FROM t_score GROUP BY sid

) AS T1 WHERE avggrade > T2.avggrade ) as 名次,

sid as 学生学号,

avggrade

FROM (

SELECT sid,AVG(grade )as avggrade FROM t_score GROUP BY sid

) AS T2

ORDER BY avggrade desc;

25、查询各科成绩前三名的记录:(不考虑成绩并列情况)

SELECT t1.sid as 学生ID,t1.cid as 课程ID,grade as 分数

FROM t_score t1 WHERE grade IN (

SELECT TOP 3 grade FROM t_score t2 WHERE t1.cid = t2.cid ORDER BY grade DESC

)

ORDER BY t1.cid;

26、查询每门课程被选修的学生数

select cid,count(sid) from t_score group by cid;

27、查询出只选修了一门课程的全部学生的学号和姓名

select t1.sid,t2.sname,count(t1.cid) AS 选课数 from t_score t1,t_student t2

where t1.sid=t2.sid

group by t1.sid ,t2.sname

having count(t1.cid)=1;

28、查询男生、女生人数

Select count(ssex) as 男生人数 from t_student group by ssex having ssex=‘男‘;

Select count(ssex) as 女生人数 from t_student group by ssex having ssex=‘女‘;

29、查询姓“张”的学生名单

SELECT sname FROM t_student WHERE sname like ‘张%‘;

30、查询同名同性学生名单，并统计同名人数

select sname,count(*) from t-student group by sname having count(*)>1;

31、1981年出生的学生名单(注：Student表中Sage列的类型是datetime)

select sname, CONVERT(char(11),DATEPART(year,sage)) as age

from t_tudent

where CONVERT(char(11),DATEPART(year,Sage))=‘1981‘;

32、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 cid,Avg(grade ) from t_score group by cid order by Avg(grade ),cid DESC

33、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 t1.sname, t2.sid ,avg(t2.grade ) from t_student t1, t_score t2

where t1.sid=t2.sid group by t2.sid,t1.sname

having avg(t2.grade )>85;

34、查询课程名称为“数据库”，且分数低于60的学生姓名和分数

Select t1.sname,isnull(t2.grade ,0) from t_student t1, t_score t2,t_course t3

where t2.sid = t1.sid and t2.cid = t3.cid

and t3.cname=‘数据库‘and t2.grade <60;

35、查询所有学生的选课情况;

SELECT t1.sid, t1.cid,t2.sname,t3.cname FROM t_score t1,t_student t2,t_course t3

where t1.sid = t2.sid and t1.cid = t3.cid ;

36、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;

SELECT distinct t1.sid,t1.sname,t2.cid, t2.grade FROM t_student t1, t_score t2

WHERE t2.grade >=70 AND t2.sid = t1.sid;

37、查询不及格的课程，并按课程号从大到小排列

select cid from t_score where grade < 60 order by cid ;

38、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 t1.sid,t2.sname from t_score t1,t_student t2

where t1.sid = t2.sid and t1.grade > 80 and t1.cid = ‘003‘;

39、求选了课程的学生人数

select count(*) from t_score;

40、查询选修“叶平”老师所授课程的学生中，成绩最高的学生姓名及其成绩

select t1.sname,t2.grade from t_student t1, t_score t2,t_course t3,t_teacher t4

where t1.sid = t2.sid

and t2.cid = t3.cid

and t3.tid = t4.tid

and t4.Tname = ‘叶平‘

and t2.grade = (select max(grade )from t_score where cid=t3.cid );

41、查询各个课程及相应的选修人数

select count(*) from t_score group by cid

42、查询不同课程成绩相同的学生的学号、课程号、学生成绩

select distinct t1.sid,t2.grade from t_score t1 , t_score t2

where t1.grade = t2.grade and t1.cid <>t2.cid ;

43、查询每门功成绩最好的前两名

SELECT t1.sid as 学生ID,t1.cid as 课程ID,grade as 分数 FROM t_score t1

WHERE grade IN (

SELECT TOP 2 grade FROM t_score

WHERE t1.cid = cid ORDER BY grade DESC

)

ORDER BY t1.cid;

44、统计每门课程的学生选修人数(超过10人的课程才统计)。要求输出课程号和选修人数，查询结果按人数降序排列，查询结果按人数降序排列，若人数相同，按课程号升序排列

select cid as 课程号,count(*) as 人数 from t_score

group by cid

order by count(*) desc,cid

45、检索至少选修两门课程的学生学号

select sid from t_score group by sid having count(*) > = 2

46、查询全部学生都选修的课程的课程号和课程名

select cid,cname from t_course where cid in (select cid from t_score group by cid)

47、查询没学过“叶平”老师讲授的任一门课程的学生姓名

select sname from t_student

where sid not in (

select sid from t_course t1,t_teacher t2, t_score t3

where t1.tid = t2.tid

and t3.cid =t 1.cid

and t2.tname = ‘叶平‘

);

48、查询两门以上不及格课程的同学的学号及其平均成绩

select sid,avg(isnull(grade ,0)) from t_score

where sid in (

select sid from t_score

where grade <60 group by sid having count(*)>2

)

group by sid;

49、检索“004”课程分数小于60，按分数降序排列的同学学号

select sid from t_score where cid=‘004‘and grade <60 order by grade desc;

50、删除“002”同学的“001”课程的成绩

delete from t_score where sid=‘001‘and cid = ‘001‘;

SQL高级查询