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大家好
上周写了匿名方法一文，很多读者，很高兴，相信我们已经从大伙的回复中，对.NET又有了更深刻的认识.
好，现在说主题，各类数据库都有相应更新本表top n的方案.现在我一一举例
首先看表结构如下：
数据库以及表创建命令初始化数据库语句
1 CREATE TABLE student( 2 id varchar(256) NOT NULL, 3 name varchar(256) NULL, 4 age int NULL)
1 insert into student values(‘001‘,‘wfg‘,20); 2 insert into student values(‘002‘,‘lxx‘,21); 3 insert into student values(‘003‘,‘wly‘,3); 4 insert into student values(‘004‘,‘jcj‘,60); 5 insert into student values(‘005‘,‘wss‘,60); 6 insert into student values(‘006‘,‘xsm‘,60); 7 insert into student values(‘007‘,‘lcf‘,60); 8 insert into student values(‘008‘,‘wjy‘,35); 9 insert into student values(‘009‘,‘hyf‘,35); 10 insert into student values(‘010‘,‘lwl‘,12);
表格结构如下：
| ID | Name | Age |
| 001 | wfg | 20 |
| 002 | lxx | 21 |
| 003 | wly | 3 |
| 004 | jcj | 60 |
| 005 | wss | 60 |
| 006 | xsm | 60 |
| 007 | lcf | 60 |
| 008 | wjy | 35 |
| 009 | hyf | 35 |
| 010 | hwl | 12 |
需求如下：按姓名顺序后，更新前５个的年龄为100岁，如何办到.
首先说下MSSQL
1 --方案1失败 2 update top(5) student set age = 100 order by name; 3 --方案2ok 4 update student set age = 100 where id in (select top 5 id from student order by name ); 5 select top(5) * from student order by name;
很成功，干得漂亮！
接下来我们看看MYSQL呢，拭目以待...
1 update student set age = 100 order by name limit 5; 2 select * from student order by name limit 5;
Good news, MySql干得不错，
接下来我们再看看oracle呢？
1 --方式１失败 2 update student set age = 100 where rownum <5 order by name ; 3 commit; 4 --方式２失败 5 update student set age = 100 where id in (select id from student where rownum<5 order by name); 6 commit; 7 select * from student where rownum<5 order by name;
目前为此，还没有在oracle里找到办法，请数据库专家指导，感谢为盼！
Oracle Update TOP N Issue, 请专家解答
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原文地址：http://www.cnblogs.com/xhu218/p/4285453.html
