--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