--查询学生成绩--成绩表（包含了若干个学生的若干次考试）
create table exam
(
eid int identity(1,1), --考试编号 可看做自增
sid INT, --学生号
yuwen INT, --语文得分
shuxue INT--数学得分
)
insert into exam values(101,60,70);
insert into exam values(101,80,90);
insert into exam values(101,70,100);
insert into exam values(101,10,20);
insert into exam values(101,30,40);
insert into exam values(101,10,90);
insert into exam values(101,70,20);insert into exam values(102,80,90);
insert into exam values(102,60,70);
insert into exam values(102,70,100);
insert into exam values(102,20,10);
insert into exam values(102,30,40);
insert into exam values(102,0,0);
insert into exam values(102,10,90);
insert into exam values(102,70,20);insert into exam values(103,80,90);
insert into exam values(103,10,90); --学生成绩--查询条件表
create table con
(
cid int identity(1,1),
minyuwen int,
maxyuwen int,
minshuxue int,
maxshuxue int,
counts int
)insert into con values(60,100,60,100,2);
insert into con values(0,59,0,59,2); --意思是 查询条件为 查询两门都及格或者都不及格的 各两条 （每个人）--学生表
create table stu
(
sid int
)insert into stu values(101);
insert into stu values(102);
insert into stu values(103); ---三个学生--统计表
create table examlist
(
elid int identity(1,1),
eid INT, --学生号
yuwen INT, --语文得分
shuxue INT--数学得分
)/*
目标：
****
1.根据查询条件表con中的每个条件在考试表exam中查询出 每个学生相应条数的记录
意思是 查询两门都及格或者都不及格的 各两条 （每个人都满足）
可结合stu表的学号 唯一的
结果集如：（查询完后插入到examlist表中）(此eid随便写的)eid sid yuwen shuxue
1 101 60 70
2 101 80 90
3 101 10 20
4 101 30 405 102 80 90
6 102 60 70
7 102 20 10
8 102 30 409 103 80 90
****
2.此时学生103的新考试成绩出来了 还多了104
insert into stu values(103,80,70);
insert into stu values(103,60,0);insert into stu values(104,0,100);
insert into stu values(104,20,10);此时再按照前面条件查询出结果 并插入到examlist表中
此时注意 数据不能重复 也不能多了
最终结果如下(此eid随便写的)
eid sid yuwen shuxue
1 101 60 70
2 101 80 90
3 101 10 20
4 101 30 405 102 80 90
6 102 60 70
7 102 20 10
8 102 30 409 103 80 90
10 103 80 7011 104 20 10入门学习中。。请各位前辈指教，感激万分！
*/
create table exam
(
eid int identity(1,1), --考试编号 可看做自增
sid INT, --学生号
yuwen INT, --语文得分
shuxue INT--数学得分
)
insert into exam values(101,60,70);
insert into exam values(101,80,90);
insert into exam values(101,70,100);
insert into exam values(101,10,20);
insert into exam values(101,30,40);
insert into exam values(101,10,90);
insert into exam values(101,70,20);insert into exam values(102,80,90);
insert into exam values(102,60,70);
insert into exam values(102,70,100);
insert into exam values(102,20,10);
insert into exam values(102,30,40);
insert into exam values(102,0,0);
insert into exam values(102,10,90);
insert into exam values(102,70,20);insert into exam values(103,80,90);
insert into exam values(103,10,90); --学生成绩--查询条件表
create table con
(
cid int identity(1,1),
minyuwen int,
maxyuwen int,
minshuxue int,
maxshuxue int,
counts int
)insert into con values(60,100,60,100,2);
insert into con values(0,59,0,59,2); --意思是 查询条件为 查询两门都及格或者都不及格的 各两条 （每个人）--学生表
create table stu
(
sid int
)insert into stu values(101);
insert into stu values(102);
insert into stu values(103); ---三个学生--统计表
create table examlist
(
elid int identity(1,1),
eid INT, --学生号
yuwen INT, --语文得分
shuxue INT--数学得分
)/*
目标：
****
1.根据查询条件表con中的每个条件在考试表exam中查询出 每个学生相应条数的记录
意思是 查询两门都及格或者都不及格的 各两条 （每个人都满足）
可结合stu表的学号 唯一的
结果集如：（查询完后插入到examlist表中）(此eid随便写的)eid sid yuwen shuxue
1 101 60 70
2 101 80 90
3 101 10 20
4 101 30 405 102 80 90
6 102 60 70
7 102 20 10
8 102 30 409 103 80 90
****
2.此时学生103的新考试成绩出来了 还多了104
insert into stu values(103,80,70);
insert into stu values(103,60,0);insert into stu values(104,0,100);
insert into stu values(104,20,10);此时再按照前面条件查询出结果 并插入到examlist表中
此时注意 数据不能重复 也不能多了
最终结果如下(此eid随便写的)
eid sid yuwen shuxue
1 101 60 70
2 101 80 90
3 101 10 20
4 101 30 405 102 80 90
6 102 60 70
7 102 20 10
8 102 30 409 103 80 90
10 103 80 7011 104 20 10入门学习中。。请各位前辈指教，感激万分！
*/
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,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 a.S# from (select s#,score from SC where C#='001') a,(select s#,score
from SC where C#='002') b
where a.score>b.score and a.s#=b.s#;
2、查询平均成绩大于60分的同学的学号和平均成绩；
select S#,avg(score)
from sc
group by S# having avg(score) >60;
3、查询所有同学的学号、姓名、选课数、总成绩；
select Student.S#,Student.Sname,count(SC.C#),sum(score)
from Student left Outer join SC on Student.S#=SC.S#
group by Student.S#,Sname
4、查询姓“李”的老师的个数；
select count(distinct(Tname))
from 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 Student.S#,Student.Sname
from Student
where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.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 Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='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 S#,Sname
from Student
where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平'));
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；
Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2
from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 <score;
9、查询所有课程成绩小于60分的同学的学号、姓名；
select S#,Sname
from Student
where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60);
10、查询没有学全所有课的同学的学号、姓名；
select Student.S#,Student.Sname
from Student,SC
where Student.S#=SC.S# group by Student.S#,Student.Sname having count(C#) <(select count(C#) from 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 S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# from SC where S#='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 SC.S#,Sname
from Student,SC
where Student.S#=SC.S# and C# in (select C# from SC where S#='001');
13、把“SC”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 SC set score=(select avg(SC_2.score)
from SC SC_2
where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.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 S# from SC where C# in (select C# from SC where S#='1002')
group by S# having count(*)=(select count(*) from SC where S#='1002');
15、删除学习“叶平”老师课的SC表记录；
Delect SC
from course ,Teacher
where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平';
16、向SC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 SC select S#,'002',(Select avg(score)
from SC where C#='002') from Student where S# not in (Select S# from SC where C#='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 S# as 学生ID
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语
,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩
FROM SC AS t
GROUP BY S#
ORDER BY avg(t.score)
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 L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分
FROM SC L ,SC AS R
WHERE L.C# = R.C# and
L.score = (SELECT MAX(IL.score)
FROM SC AS IL,Student AS IM
WHERE L.C# = IL.C# and IM.S#=IL.S#
GROUP BY IL.C#)
AND
R.Score = (SELECT MIN(IR.score)
FROM SC AS IR
WHERE R.C# = IR.C#
GROUP BY IR.C#
);
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.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩
,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数
FROM SC T,Course
where t.C#=course.C#
GROUP BY t.C#
ORDER BY 100 * SUM(CASE WHEN isnull(score,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 C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分
,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数
,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分
,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数
,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分
,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数
,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分
,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数
FROM SC
SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ＩＤ,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
FROM SC AS T,Course AS C ,Teacher AS Z
where T.C#=C.C# and C.T#=Z.T#
GROUP BY C.C#
ORDER BY AVG(Score) 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
SC.S# As 学生学号,
Student.Sname AS 学生姓名 ,
T1.score AS 企业管理,
T2.score AS 马克思,
T3.score AS UML,
T4.score AS 数据库,
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分
FROM Student,SC LEFT JOIN SC AS T1
ON SC.S# = T1.S# AND T1.C# = '001'
LEFT JOIN SC AS T2
ON SC.S# = T2.S# AND T2.C# = '002'
LEFT JOIN SC AS T3
ON SC.S# = T3.S# AND T3.C# = '003'
LEFT JOIN SC AS T4
ON SC.S# = T4.S# AND T4.C# = '004'
WHERE student.S#=SC.S# and
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
NOT IN
(SELECT
DISTINCT
TOP 15 WITH TIES
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
FROM sc
LEFT JOIN sc AS T1
ON sc.S# = T1.S# AND T1.C# = 'k1'
LEFT JOIN sc AS T2
ON sc.S# = T2.S# AND T2.C# = 'k2'
LEFT JOIN sc AS T3
ON sc.S# = T3.S# AND T3.C# = 'k3'
LEFT JOIN sc AS T4
ON sc.S# = T4.S# AND T4.C# = 'k4'
ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC); 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 SC.C# as 课程ID, Cname as 课程名称
,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85]
,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70]
,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60]
,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]
FROM SC,Course
where SC.C#=Course.C#
GROUP BY SC.C#,Cname; 24、查询学生平均成绩及其名次
SELECT 1+(SELECT COUNT( distinct 平均成绩)
FROM (SELECT S#,AVG(score) AS 平均成绩
FROM SC
GROUP BY S#
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次,
S# as 学生学号,平均成绩
FROM (SELECT S#,AVG(score) 平均成绩
FROM SC
GROUP BY S#
) AS T2
ORDER BY 平均成绩 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.S# as 学生ID,t1.C# as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 3 score
FROM SC
WHERE t1.C#= C#
ORDER BY score DESC
)
ORDER BY t1.C#;
26、查询每门课程被选修的学生数
select c#,count(S#) from sc group by C#;
27、查询出只选修了一门课程的全部学生的学号和姓名
select SC.S#,Student.Sname,count(C#) AS 选课数
from SC ,Student
where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1;
28、查询男生、女生人数
Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男';
Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女'；
29、查询姓“张”的学生名单
SELECT Sname FROM 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 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 student
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 C#,Avg(score) from SC group by C# order by Avg(score),C# 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 Sname,SC.S# ,avg(score)
from Student,SC
where Student.S#=SC.S# group by SC.S#,Sname having avg(score)>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 Sname,isnull(score,0)
from Student,SC,Course
where SC.S#=Student.S# and SC.C#=Course.C# and Course.Cname='数据库'and score <60;
35、查询所有学生的选课情况；
SELECT SC.S#,SC.C#,Sname,Cname
FROM SC,Student,Course
where SC.S#=Student.S# and SC.C#=Course.C# ;
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 student.S#,student.Sname,SC.C#,SC.score
FROM student,Sc
WHERE SC.score>=70 AND SC.S#=student.S#;
37、查询不及格的课程，并按课程号从大到小排列
select c# from sc where scor e <60 order by C# ;
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 SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003';
39、求选了课程的学生人数
select count(*) from sc;
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 Student.Sname,score
from Student,SC,Course C,Teacher
where Student.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# );
41、查询各个课程及相应的选修人数
select count(*) from sc group by C#;
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 A.S#,B.score from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ;
43、查询每门功成绩最好的前两名
SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 2 score
FROM SC
WHERE t1.C#= C#
ORDER BY score DESC
)
ORDER BY t1.C#;
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 C# as 课程号,count(*) as 人数
from sc
group by C#
order by count(*) desc,c#
45、检索至少选修两门课程的学生学号
select S#
from sc
group by s#
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 C#,Cname
from Course
where C# in (select c# from sc group by c#)
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 Student where S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# and 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 S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#;
49、检索“004”课程分数小于60，按分数降序排列的同学学号
select S# from SC where C#='004'and score <60 order by score desc;
50、删除“002”同学的“001”课程的成绩
delete from Sc where S#='001'and C#='001';
tony辛苦了，这么晚，宝宝还没睡啊？
。。理解
这个题 我不答。。
留给 TONY哥。。
;with t as(
select * ,id=(select count(1) from exam where sid=a.sid )from exam as a where yuwen>=60 and shuxue>=60
union all
select * ,id=(select count(1)-2 from exam where sid=a.sid )from exam as a where yuwen<60 and shuxue<60
)SELECT eid , sid ,yuwen ,shuxue FROM T AS b WHERE eid IN(SELECT TOP 2 eid FROM T WHERE ID=b.ID ORDER BY yuwen DESC) /*
eid sid yuwen shuxue id
----------- ----------- ----------- ----------- -----------
2 101 80 90 7
3 101 70 100 7
8 102 80 90 8
10 102 70 100 8
16 103 80 90 2
4 101 10 20 5
5 101 30 40 5
11 102 20 10 6
12 102 30 40 6(9 行受影响)
*/
;with t as(
select * ,id=(select count(1) from exam where sid=a.sid )from exam as a where yuwen>=60 and shuxue>=60
union all
select * ,id=(select count(1)+2 from exam where sid=a.sid )from exam as a where yuwen<60 and shuxue<60
)SELECT eid , sid ,yuwen ,shuxue FROM T AS b WHERE eid IN(SELECT TOP 2 eid FROM T WHERE ID=b.ID ORDER BY yuwen DESC)
/*eid sid yuwen shuxue
----------- ----------- ----------- -----------
2 101 80 90
3 101 70 100
8 102 80 90
10 102 70 100
16 103 80 90
4 101 10 20
5 101 30 40
11 102 20 10
12 102 30 40(9 行受影响)*/
select * ,DENSE_RANK() over (partition by [sid] order by eid) as id from exam as a where yuwen>=60 and shuxue>=60
union all
select * ,DENSE_RANK() over (partition by [sid] order by eid) as id from exam as a where yuwen<60 and shuxue<60
)
SELECT eid , [sid] ,yuwen ,shuxue FROM T WHERE [id]<=2 ORDER BY [sid]
/*
eid sid yuwen shuxue id
----------- ----------- ----------- ----------- --------------------
1 101 60 70 1
2 101 80 90 2
4 101 10 20 1
5 101 30 40 2
11 102 20 10 1
12 102 30 40 2
8 102 80 90 1
9 102 60 70 2
16 103 80 90 1
18 103 80 70 2
21 104 20 10 1(11 行受影响)
*/
select
a.*,
b.*
into #
from
exam a,con b
where
a.yuwen between b.minyuwen and b.maxyuwen
and
a.shuxue between b.minshuxue and b.maxshuxueinsert into examlist(eid,yuwen,shuxue)
select
sid,
yuwen,
shuxue
from
(select *,px=(select count(1) from # where [counts]=t.[counts] and cid=t.cid and sid=t.sid and eid<t.eid) from # t) tt
where
px/[counts]=0
order by
sidselect * from examlist
/**
elid eid yuwen shuxue
----------- ----------- ----------- -----------
1 101 60 70
2 101 80 90
3 101 10 20
4 101 30 40
5 102 20 10
6 102 30 40
7 102 80 90
8 102 60 70
9 103 80 90（所影响的行数为 9 行）
**/
insert into exam values(103,80,70); --楼主怎么这么不小心，把表名也写错
insert into exam values(103,60,0);
insert into exam values(104,0,100);
insert into exam values(104,20,10);
delete examlist where eid in(select eid from examlist group by eid having count(1)<4)select
a.*,
b.*
into #2
from
(select * from exam where not exists(select 1 from examlist where eid=exam.sid)) a,
con b
where
a.yuwen between b.minyuwen and b.maxyuwen
and
a.shuxue between b.minshuxue and b.maxshuxueinsert into examlist(eid,yuwen,shuxue)
select
sid,
yuwen,
shuxue
from
(select *,px=(select count(1) from #2 where [counts]=t.[counts] and cid=t.cid and sid=t.sid and eid<t.eid) from #2 t) tt
where
px/[counts]=0
order by
sidselect * from examlist
/**
elid eid yuwen shuxue
----------- ----------- ----------- -----------
1 101 60 70
2 101 80 90
3 101 10 20
4 101 30 40
5 102 20 10
6 102 30 40
7 102 80 90
8 102 60 70
10 103 80 90
11 103 80 70
12 104 20 10（所影响的行数为 11 行）
**/
弄了老半天也弄不出来一样的效果，呵呵
go
create table [exam]([eid] int,[sid] int,[yuwen] int,[shuxue] int)
insert [exam]
select 1,101,60,70 union all
select 2,101,80,90 union all
select 3,101,70,100 union all
select 4,101,10,20 union all
select 5,101,30,40 union all
select 6,101,10,90 union all
select 7,101,70,20 union all
select 8,102,80,90 union all
select 9,102,60,70 union all
select 10,102,70,100 union all
select 11,102,20,10 union all
select 12,102,30,40 union all
select 13,102,0,0 union all
select 14,102,10,90 union all
select 15,102,70,20 union all
select 16,103,80,90 union all
select 17,103,10,90
--> 测试数据：[con]
if object_id('[con]') is not null drop table [con]
go
create table [con]([cid] int,[minyuwen] int,[maxyuwen] int,[minshuxue] int,[maxshuxue] int,[counts] int)
insert [con]
select 1,60,100,60,100,2 union all
select 2,0,59,0,59,2
--> 测试数据：[stu]
if object_id('[stu]') is not null drop table [stu]
go
create table [stu]([sid] int)
insert [stu]
select 101 union all
select 102 union all
select 103
--MS SQL SERVER 2005/2008方法
--------------开始查询--------------------------
SELECT y.sid as 学号,y.yuwen 语文成绩,y.shuxue as 数学成绩
from
(select row_number() over(partition by a.sid order by a.eid) as sn,1 as sn1,a.sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen)
and b.cid=2
union all
select row_number() over(partition by a.sid order by a.sid) as sn,2 as sn2,sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen) and b.cid=1
)y
where y.sn<=2
order by y.sid,y.sn1 desc
--MS SQL SERVER 2000方法
--------------开始查询--------------------------
SELECT y.sid as 学号,y.yuwen 语文成绩,y.shuxue as 数学成绩
from --由于第一次出现<=60的记录在第四行，固count需要-2
(select (select count(1)-2 from exam where sid=a.sid and eid<a.eid) as sn,1 as sn1,a.sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen)
and b.cid=2
union all
select (select count(1)+1 from exam where sid=a.sid and eid<a.eid) as sn,2 as sn2,sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen) and b.cid=1
)y
where y.sn<=2
order by y.sid,y.sn1 desc
/*
学号 语文成绩 数学成绩
101 60 70
101 80 90
101 10 20
101 30 40
102 80 90
102 60 70
102 20 10
102 30 40
103 80 90
*/
if object_id('[exam]') is not null drop table [exam]
go
create table [exam]([eid] int,[sid] int,[yuwen] int,[shuxue] int)
insert [exam]
select 1,101,60,70 union all
select 2,101,80,90 union all
select 3,101,70,100 union all
select 4,101,10,20 union all
select 5,101,30,40 union all
select 6,101,10,90 union all
select 7,101,70,20 union all
select 8,102,80,90 union all
select 9,102,60,70 union all
select 10,102,70,100 union all
select 11,102,20,10 union all
select 12,102,30,40 union all
select 13,102,0,0 union all
select 14,102,10,90 union all
select 15,102,70,20 union all
select 16,103,80,90 union all
select 17,103,10,90
insert into [exam] values(19,103,80,70);
insert into [exam] values(20,103,60,0); insert into [exam] values(21,104,0,100);
insert into [exam] values(22,104,20,10);
--> 测试数据：[con]
if object_id('[con]') is not null drop table [con]
go
create table [con]([cid] int,[minyuwen] int,[maxyuwen] int,[minshuxue] int,[maxshuxue] int,[counts] int)
insert [con]
select 1,60,100,60,100,2 union all
select 2,0,59,0,59,2
--> 测试数据：[stu]
if object_id('[stu]') is not null drop table [stu]
go
create table [stu]([sid] int)
insert [stu]
select 101 union all
select 102 union all
select 103
--MS SQL SERVER 2005/2008方法
--------------开始查询--------------------------
SELECT y.sid as 学号,y.yuwen 语文成绩,y.shuxue as 数学成绩
from
(select row_number() over(partition by a.sid order by a.eid) as sn,1 as sn1,a.sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen)
and b.cid=2
union all
select row_number() over(partition by a.sid order by a.sid) as sn,2 as sn2,sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen) and b.cid=1
)y
where y.sn<=2
order by y.sid,y.sn1 desc
--MS SQL SERVER 2000方法
--------------开始查询--------------------------
create table examlist
(
elid int identity(1,1),
eid INT, --学生号
yuwen INT, --语文得分
shuxue INT--数学得分
)
insert examlist(eid,yuwen,shuxue)
SELECT y.sid as 学号,y.yuwen 语文成绩,y.shuxue as 数学成绩
from --由于第一次出现<=60的记录在第四行，固count需要-2
(select (select count(1)-2 from exam where sid=a.sid and eid<a.eid) as sn,1 as sn1,a.sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen)
and b.cid=2
union all
select (select count(1)+1 from exam where sid=a.sid and eid<a.eid) as sn,2 as sn2,sid,a.yuwen,a.shuxue
from exam a,con b
where (a.shuxue between B.minshuxue and b.[maxshuxue]
and a.yuwen between b.minyuwen and b.maxyuwen) and b.cid=1
)y
where y.sn<=2
order by y.sid,y.sn1 desc
select * from examlist
--最终结果如下
/*
elid eid yuwen shuxue
1 101 60 70
2 101 80 90
3 101 10 20
4 101 30 40
5 102 80 90
6 102 60 70
7 102 20 10
8 102 30 40
9 103 80 90
10 104 20 10
*/
额..有个问题
可能exam有1000万条 符合大范围条件的也很多 比如500万
那么插入到临时表可能会花费上1，20分钟
那么第一步插入到临时表#的时候能不能尽量缩小范围呢？
比如指定学生的范围 和 指定一个通用的每人每条件的总数量 比如取所有counts中最大的
这样 插入的数据可能最大只有 人数*条件数*counts