Sql Server2005对t-sql的增强之排名函数

GO
if object_id('t_ntile','U') is not null
drop table t_ntile;
GO
create table t_ntile
(
id int unique not null,
categoryId int not null,
name nvarchar(20)
)
go
INSERT INTO t_ntile VALUES(1,1,'A')
INSERT INTO t_ntile VALUES(2,4,'B')
INSERT INTO t_ntile VALUES(3,2,'C')
INSERT INTO t_ntile VALUES(4,1,'D')
INSERT INTO t_ntile VALUES(5,3,'E')
INSERT INTO t_ntile VALUES(6,3,'F')
INSERT INTO t_ntile VALUES(7,2,'G')
INSERT INTO t_ntile VALUES(8,2,'H')
INSERT INTO t_ntile VALUES(9,2,'I')
Go

SELECT id,categoryId,name
,'ntile value' = NTILE(3) OVER(PARTITION BY categoryId ORDER BY categoryId) 
FROM t_ntile

if object_id('student_class_grade','U') is not null
drop table student_class_grade;
GO
create table student_class_grade
(
student_id int, --学生id
class_no int, --班级编号
grade int --成绩
);
GO
INSERT INTO student_class_grade VALUES(1,1,90);
INSERT INTO student_class_grade VALUES(2,1,85);
INSERT INTO student_class_grade VALUES(3,1,80);
INSERT INTO student_class_grade VALUES(4,1,80);
INSERT INTO student_class_grade VALUES(5,1,90);
INSERT INTO student_class_grade VALUES(6,1,75);
INSERT INTO student_class_grade VALUES(7,1,89);
INSERT INTO student_class_grade VALUES(11,2,90);
INSERT INTO student_class_grade VALUES(12,2,85);
INSERT INTO student_class_grade VALUES(13,2,80);
INSERT INTO student_class_grade VALUES(14,2,80);
INSERT INTO student_class_grade VALUES(15,2,90);
INSERT INTO student_class_grade VALUES(16,2,75);
INSERT INTO student_class_grade VALUES(17,2,89);
GO
--显示各个班级学生的成绩排名
SELECT student_id
,class_no,grade
,'名次' = RANK() OVER(PARTITION BY class_no ORDER BY grade desc)
FROM student_class_grade
GO
SELECT student_id
,class_no,grade
,'名次' = DENSE_RANK() OVER(PARTITION BY class_no ORDER BY grade desc)
FROM student_class_grade

WITH CTE_rn (student_id,class_no,grade,rn) AS(
SELECT student_id,class_no,grade,rn = ROW_NUMBER() OVER(ORDER BY student_id ASC)
FROM student_class_grade
WHERE 0=0 --可以在此处加一些过滤条件，这样下面的分页的sql中就都不需要加条件了
)
--获得第-10条的数据
SELECT student_id,class_no,grade FROM CTE_rn WHERE rn BETWEEN 6 AND 10;
SELECT totalCn = COUNT(*) FROM student_class_grade WHERE 0=0