求SQL代码：学分排名次（请看清，要求有点儿特)
名词解释 bj:班级 zf: 总分 zmc: 总名次 bmc: 班名次由cj表：
bj zf
1 98
2 97
1 96
2 96
1 95
2 94
1 94
2 94
1 93
得出mc表：
bj zmc bmc zf
1 1 1 98
2 2 1 97
1 3 2 96
2 3 2 96
1 5 3 95
2 6 3 94
1 6 4 94
2 6 3 94
1 9 5 93
请注意：
1. zf相同的学生，名次要相同，请注意观察规律。
2. zmc是所有班学生参加排名，bmc是各班内部。
3. 班级个数是可变的，请代码有一定适应性。
名词解释 bj:班级 zf: 总分 zmc: 总名次 bmc: 班名次由cj表：
bj zf
1 98
2 97
1 96
2 96
1 95
2 94
1 94
2 94
1 93
得出mc表：
bj zmc bmc zf
1 1 1 98
2 2 1 97
1 3 2 96
2 3 2 96
1 5 3 95
2 6 3 94
1 6 4 94
2 6 3 94
1 9 5 93
请注意：
1. zf相同的学生，名次要相同，请注意观察规律。
2. zmc是所有班学生参加排名，bmc是各班内部。
3. 班级个数是可变的，请代码有一定适应性。
zmc=(select count(distinct zf) from tb where zj>a.zj),
bmc=(select count(distinct zf) from tb where zj>a.zj and bj=a.bj),
zf
from tb a
CREATE TABLE tb(
ID1 char(2) NOT NULL,
ID2 char(4) NOT NULL,
col int,
PRIMARY KEY(ID1,ID2))
INSERT tb SELECT 'aa','0001',1
UNION ALL SELECT 'aa','0003',2
UNION ALL SELECT 'aa','0004',3
UNION ALL SELECT 'bb','0005',4
UNION ALL SELECT 'bb','0006',5
UNION ALL SELECT 'cc','0007',6
UNION ALL SELECT 'cc','0009',7
GO--重排编号处理
DECLARE @ID1 char(2),@ID2 int
UPDATE a SET ID2=RIGHT(10000
+(SELECT COUNT(*) FROM tb WHERE ID1=a.ID1 AND ID2<=a.ID2)
,4)
FROM tb a
SELECT * FROM tb
/*--结果
ID1 ID2 col
---- ---- -----------
aa 0001 1
aa 0002 2
aa 0003 3
bb 0001 4
bb 0002 5
cc 0001 6
cc 0002 7
--*/
ROW_NUMBER、RANK、DENSE_RANK的用法
(爱新觉罗.毓华 2007-12-16 广东深圳)SQL Server 2005 引入几个新的排序(排名)函数,如ROW_NUMBER、RANK、DENSE_RANK等。
这些新函数使您可以有效地分析数据以及向查询的结果行提供排序值。 --------------------------------------------------------------------------
ROW_NUMBER()说明：返回结果集分区内行的序列号，每个分区的第一行从 1 开始。
语法：ROW_NUMBER () OVER ( [ <partition_by_clause> ] <order_by_clause> ) 。
备注：ORDER BY 子句可确定在特定分区中为行分配唯一 ROW_NUMBER 的顺序。
参数：<partition_by_clause> ：将 FROM 子句生成的结果集划入应用了 ROW_NUMBER 函数的分区。
<order_by_clause>：确定将 ROW_NUMBER 值分配给分区中的行的顺序。
返回类型：bigint 。示例：
/*以下示例将根据年初至今的销售额，返回 AdventureWorks 中销售人员的 ROW_NUMBER。*/USE AdventureWorks
GO
SELECT c.FirstName, c.LastName, ROW_NUMBER() OVER(ORDER BY SalesYTD DESC) AS 'Row Number', s.SalesYTD, a.PostalCode
FROM Sales.SalesPerson s JOIN Person.Contact c on s.SalesPersonID = c.ContactID
JOIN Person.Address a ON a.AddressID = c.ContactID
WHERE TerritoryID IS NOT NULL AND SalesYTD <> 0
/*
FirstName LastName Row Number SalesYTD PostalCode
--------- ---------- ---------- ------------ ----------------------------
Shelley Dyck 1 5200475.2313 98027
Gail Erickson 2 5015682.3752 98055
Maciej Dusza 3 4557045.0459 98027
Linda Ecoffey 4 3857163.6332 98027
Mark Erickson 5 3827950.238 98055
Terry Eminhizer 6 3587378.4257 98055
Michael Emanuel 7 3189356.2465 98055
Jauna Elson 8 3018725.4858 98055
Carol Elliott 9 2811012.7151 98027
Janeth Esteves 10 2241204.0424 98055
Martha Espinoza 11 1931620.1835 98055
Carla Eldridge 12 1764938.9859 98027
Twanna Evans 13 1758385.926 98055
(13 行受影响)
*/
/*以下示例将返回行号为 50 到 60（含）的行，并以 OrderDate 排序。*/
USE AdventureWorks;
GO
WITH OrderedOrders AS
(SELECT SalesOrderID, OrderDate,
ROW_NUMBER() OVER (order by OrderDate)as RowNumber
FROM Sales.SalesOrderHeader )
SELECT *
FROM OrderedOrders
WHERE RowNumber between 50 and 60;
/*
SalesOrderID OrderDate RowNumber
------------ ----------------------- --------------------
43708 2001-07-03 00:00:00.000 50
43709 2001-07-03 00:00:00.000 51
43710 2001-07-03 00:00:00.000 52
43711 2001-07-04 00:00:00.000 53
43712 2001-07-04 00:00:00.000 54
43713 2001-07-05 00:00:00.000 55
43714 2001-07-05 00:00:00.000 56
43715 2001-07-05 00:00:00.000 57
43716 2001-07-05 00:00:00.000 58
43717 2001-07-05 00:00:00.000 59
43718 2001-07-06 00:00:00.000 60
(11 行受影响)
*/--------------------------------------------------------------
RANK()说明：返回结果集的分区内每行的排名。行的排名是相关行之前的排名数加一。
语法：RANK () OVER ( [ < partition_by_clause > ] < order_by_clause > )
备注：如果两个或多个行与一个排名关联，则每个关联行将得到相同的排名。
例如，如果两位顶尖销售员具有同样的 SalesYTD 值，他们将并列第一。
由于已有两行排名在前，所以具有下一个最大 SalesYTD 的销售人员将排名第三。
因此，RANK 函数并不总返回连续整数。
用于整个查询的排序顺序决定了行在结果集中的显示顺序。这也隐含了行在每个分区中的排名。
参数：< partition_by_clause > ：将 FROM 子句生成的结果集划分为要应用 RANK 函数的分区。
< order_by_clause >：确定将 RANK 值应用于分区中的行时所基于的顺序。
返回类型：bigint示例：
/*以下示例按照数量对清单中的产品进行了排名。行集按 LocationID 分区，按 Quantity 排序。
USE AdventureWorks;
GO
SELECT i.ProductID, p.Name, i.LocationID, i.Quantity, RANK() OVER (PARTITION BY i.LocationID order by i.Quantity) as RANK
FROM Production.ProductInventory i JOIN Production.Product p
ON i.ProductID = p.ProductID
ORDER BY p.Name
GO
/*
ProductID Name LocationID Quantity RANK
----------- -------------------------------------------------- ---------- -------- --------------------
1 Adjustable Race 6 324 71
1 Adjustable Race 1 408 78
1 Adjustable Race 50 353 117
2 Bearing Ball 6 318 67
2 Bearing Ball 1 427 85
2 Bearing Ball 50 364 122
3 BB Ball Bearing 50 324 106
3 BB Ball Bearing 1 585 110
3 BB Ball Bearing 6 443 115
4 Headset Ball Bearings 1 512 99
4 Headset Ball Bearings 6 422 108
4 Headset Ball Bearings 50 388 140
316 Blade 10 388 33
......
(1069 行受影响)
*/ --接上.
-------------------------------------------------------------------------------------
DENSE_RANK()说明：返回结果集分区中行的排名，在排名中没有任何间断。行的排名等于所讨论行之前的所有排名数加一。
语法：DENSE_RANK () OVER ( [ < partition_by_clause > ] < order_by_clause > )
备注：如果有两个或多个行受同一个分区中排名的约束，则每个约束行将接收相同的排名。
例如，如果两位顶尖销售员具有相同的 SalesYTD 值，则他们将并列第一。
接下来 SalesYTD 最高的销售人员排名第二。该排名等于该行之前的所有行数加一。
因此，DENSE_RANK 函数返回的数字没有间断，并且始终具有连续的排名。
整个查询所用的排序顺序确定了各行在结果中的显示顺序。这说明排名第一的行可以不是分区中的第一行。
参数：< partition_by_clause > ：将 FROM 子句所生成的结果集划分为数个将应用 DENSE_RANK 函数的分区。
< order_by_clause >：确定将 DENSE_RANK 值应用于分区中各行的顺序。
返回类型：bigint示例：
/*以下示例返回各位置上产品数量的 DENSE_RANK。 */
USE AdventureWorks;
GO
SELECT i.ProductID, p.Name, i.LocationID, i.Quantity, DENSE_RANK() OVER (PARTITION BY i.LocationID order by i.Quantity) as DENSE_RANK
FROM Production.ProductInventory i JOIN Production.Product p ON i.ProductID = p.ProductID
ORDER BY Name;
GO
/*
ProductID Name LocationID Quantity DENSE_RANK
----------- -------------------------------------------------- ---------- -------- --------------------
1 Adjustable Race 1 408 57
1 Adjustable Race 6 324 52
1 Adjustable Race 50 353 82
879 All-Purpose Bike Stand 7 144 34
712 AWC Logo Cap 7 288 38
3 BB Ball Bearing 50 324 74
3 BB Ball Bearing 6 443 81
3 BB Ball Bearing 1 585 82
*/
将上面三个函数放在一起计算，更能明显看出各个函数的功能。CREATE TABLE rankorder(orderid INT,qty INT)
INSERT rankorder VALUES(30001,10)
INSERT rankorder VALUES(10001,10)
INSERT rankorder VALUES(10006,10)
INSERT rankorder VALUES(40005,10)
INSERT rankorder VALUES(30003,15)
INSERT rankorder VALUES(30004,20)
INSERT rankorder VALUES(20002,20)
INSERT rankorder VALUES(20001,20)
INSERT rankorder VALUES(10005,30)
INSERT rankorder VALUES(30007,30)
INSERT rankorder VALUES(40001,40)
INSERT rankorder VALUES(30007,30)
GO
--对一个列qty进行的排序
SELECT orderid,qty,
ROW_NUMBER() OVER(ORDER BY qty) AS rownumber,
RANK() OVER(ORDER BY qty) AS rank,
DENSE_RANK() OVER(ORDER BY qty) AS denserank
FROM rankorder
ORDER BY qty
/*
orderid qty rownumber rank denserank
----------- ----------- -------------------- -------------------- --------------------
30001 10 1 1 1
10001 10 2 1 1
10006 10 3 1 1
40005 10 4 1 1
30003 15 5 5 2
30004 20 6 6 3
20002 20 7 6 3
20001 20 8 6 3
10005 30 9 9 4
30007 30 10 9 4
30007 30 11 9 4
40001 40 12 12 5
(12 行受影响)
*/ --对两个列qty,orderid进行的排序
SELECT orderid,qty,
ROW_NUMBER() OVER(ORDER BY qty,orderid) AS rownumber,
RANK() OVER(ORDER BY qty,orderid) AS rank,
DENSE_RANK() OVER(ORDER BY qty,orderid) AS denserank
FROM rankorder
ORDER BY qty,orderid
drop table rankorder
/*
orderid qty rownumber rank denserank
----------- ----------- -------------------- -------------------- --------------------
10001 10 1 1 1
10006 10 2 2 2
30001 10 3 3 3
40005 10 4 4 4
30003 15 5 5 5
20001 20 6 6 6
20002 20 7 7 7
30004 20 8 8 8
10005 30 9 9 9
30007 30 10 10 10
30007 30 11 10 10
40001 40 12 12 11
(12 行受影响)
*/
CREATE TABLE tb(Name varchar(10),Score decimal(10,2))
INSERT tb SELECT 'aa',99
UNION ALL SELECT 'bb',56
UNION ALL SELECT 'cc',56
UNION ALL SELECT 'dd',77
UNION ALL SELECT 'ee',78
UNION ALL SELECT 'ff',76
UNION ALL SELECT 'gg',78
UNION ALL SELECT 'ff',50
GO--1. 名次生成方式1,Score重复时合并名次
SELECT *,Place=(SELECT COUNT(DISTINCT Score) FROM tb WHERE Score>=a.Score)
FROM tb a
ORDER BY Place
/*--结果
Name Score Place
---------------- ----------------- -----------
aa 99.00 1
ee 78.00 2
gg 78.00 2
dd 77.00 3
ff 76.00 4
bb 56.00 5
cc 56.00 5
ff 50.00 6
--*/--2. 名次生成方式2,Score重复时保留名次空缺
SELECT *,Place=(SELECT COUNT(Score) FROM tb WHERE Score>a.Score)+1
FROM tb a
ORDER BY Place
/*--结果
Name Score Place
--------------- ----------------- -----------
aa 99.00 1
ee 78.00 2
gg 78.00 2
dd 77.00 4
ff 76.00 5
bb 56.00 6
cc 56.00 6
ff 50.00 8
--*/
create table cj(bj int,zf int)
insert into cj select 1, 98
insert into cj select 2, 97
insert into cj select 1, 96
insert into cj select 2, 96
insert into cj select 1, 95
insert into cj select 2, 94
insert into cj select 1, 94
insert into cj select 2, 94
insert into cj select 1, 93 select bj,
zmc=(select count(distinct zf) from cj where zf>a.zf)+1,
bmc=(select count(distinct zf) from cj where zf>a.zf and bj=a.bj)+1,
zf
from cj ago
drop table cj/*
bj zmc bmc zf
----------- ----------- ----------- -----------
1 1 1 98
2 2 1 97
1 3 2 96
2 3 2 96
1 4 3 95
2 5 3 94
1 5 4 94
2 5 3 94
1 6 5 93（所影响的行数为 9 行）
*/
insert into cj select 1,98
insert into cj select 2,97
insert into cj select 1,96
insert into cj select 2,96
insert into cj select 1,95
insert into cj select 2,94
insert into cj select 1,94
insert into cj select 2,94
insert into cj select 1,93select a.bj,zmc=(select count(1) from cj where a.zf<=zf),bmc=(select count(1) from cj where a.bj=bj and a.zf<=zf),zf from cj abj zmc bmc zf
----------- ----------- ----------- -----------
1 1 1 98
2 2 1 97
1 4 2 96
2 4 2 96
1 5 3 95
2 8 4 94
1 8 4 94
2 8 4 94
1 9 5 93(9 行受影响)
create table cj(bj int,zf int)
insert into cj select 1, 98
insert into cj select 2, 97
insert into cj select 1, 96
insert into cj select 2, 96
insert into cj select 1, 95
insert into cj select 2, 94
insert into cj select 1, 94
insert into cj select 2, 94
insert into cj select 1, 93 select bj,
zmc=(select count( zf) from cj where zf>a.zf)+1,
bmc=(select count(distinct zf) from cj where zf>a.zf and bj=a.bj)+1,
zf
from cj ago
drop table cj/*
bj zmc bmc zf
----------- ----------- ----------- -----------
1 1 1 98
2 2 1 97
1 3 2 96
2 3 2 96
1 5 3 95
2 6 3 94
1 6 4 94
2 6 3 94
1 9 5 93（所影响的行数为 9 行）*/
insert into #tab values(2,97)
insert into #tab values(1,96)
insert into #tab values(2,96)
insert into #tab values(1,95)
insert into #tab values(2,94)
insert into #tab values(1,94)
insert into #tab values(2,94)
insert into #tab values(1,93)
select bj,zmc=(select count(1) from (select distinct zf from #tab) b where zf>=a.zf),bmc=(select count(1) from (select distinct zf from #tab where bj=a.bj) b where zf>=a.zf),zf from #tab abj zmc bmc zf
----------- ----------- ----------- -----------
1 1 1 98
2 2 1 97
1 3 2 96
2 3 2 96
1 4 3 95
2 5 3 94
1 5 4 94
2 5 3 94
1 6 5 93(9 行受影响)
--上面写错了，修正一下
create table cj( bj int,zf int)
insert into cj select 1,98
insert into cj select 2,97
insert into cj select 1,96
insert into cj select 2,96
insert into cj select 1,95
insert into cj select 2,94
insert into cj select 1,94
insert into cj select 2,94
insert into cj select 1,93select a.bj,zmc=(select count(1) from cj where a.zf<zf)+1,bmc=(select count(1) from cj where a.bj=bj and a.zf<zf)+1,zf from cj abj zmc bmc zf
----------- ----------- ----------- -----------
1 1 1 98
2 2 1 97
1 3 2 96
2 3 2 96
1 5 3 95
2 6 3 94
1 6 4 94
2 6 3 94
1 9 5 93(9 行受影响)
select bj,zmc=(select count(1) from #tab where zf>a.zf)+1,bmc=(select count(1) from (select distinct zf from #tab where bj=a.bj) b where zf>=a.zf),zf from #tab abj zmc bmc zf
----------- ----------- ----------- -----------
1 1 1 98
2 2 1 97
1 3 2 96
2 3 2 96
1 5 3 95
2 6 3 94
1 6 4 94
2 6 3 94
1 9 5 93(9 行受影响)
create table #cj( bj int,zf int)
insert into #cj select 1,98
insert into #cj select 2,97
insert into #cj select 1,96
insert into #cj select 2,96
insert into #cj select 1,95
insert into #cj select 2,94
insert into #cj select 1,94
insert into #cj select 2,94
insert into #cj select 1,93select bj,
rank() over (ORDER BY zf desc) zmc,
rank() over (PARTITION BY bj ORDER BY zf desc) bmc,
zf
from #cj
order by zf descmssql2005 代码
谢谢侠义相助
也谢谢其它热心的哥们儿