--查询每门课程的前2名成绩
CREATE TABLE StudentGrade(
stuId CHAR(4), --学号
subId INT, --课程号
grade INT, --成绩
PRIMARY KEY (stuId,subId)
)
GO
--表中数据如下
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('001',1,97);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('001',2,50);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('001',3,70);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('002',1,92);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('002',2,80);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('002',3,30);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('003',1,93);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('003',2,95);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('003',3,85);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('004',1,73);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('004',2,78);
INSERT INTO StudentGrade(stuId,subId,grade) VALUES('004',3,87);
GO
/*
要查询每门课程的前2名成绩
001 1 97
003 1 93
003 2 95
002 2 80
004 3 87
003 3 85
如何实现？
*/
--查看数据
select * from StudentGrade
--假如出现并列时，也只取两个同学的话。
--方法一：
select distinct *
from studentgrade as t1
where stuid in
(select top 2 stuid
from studentgrade as t2
where t1.subid=t2.subid
order by t2.grade desc)
order by subid, grade desc
--方法二：
select * from StudentGrade a where (select count(1) from studentGrade where subId=a.subId and grade>=a.grade)<=2
--方法三：
select * from StudentGrade t
where (select count(1) from StudentGrade where subid=t.subid and grade>t.grade)<=1
order by subId,grade desc
--结果
/*
stuId subId grade
----- ----------- -----------
001 1 97
003 1 93
003 2 95
002 2 80
004 3 87
003 3 85
(6 row(s) affected)
*/
共有三种方案，从难易程度上讲我倾向于后两种，从查询逻辑思想上来讲后两种是一样的
select * from StudentGrade t
where (select count(1) from StudentGrade where subid=t.subid and grade>t.grade)<=1
order by subId,grade desc
我是这样理解的，看成两张表A和B，条件为A表的学科=B表的学科，select count(1) from StudentGrade where subid=t.subid and grade>t.grade，返回A表的学科=B表的学科并且A表的成绩小于B表的成绩的影响行数，如果所影响的行数为零说明它的成绩是最高的，如果等于1的话就是最高的两个成绩。这就是查询条件，再按 subId,grade 排序。这种查询思想很值得我学习
转自：http://blog.csdn.net/cnkiminzhuhu/article/details/2048879