查询数据库 中第31-40条记录，id自增长不是连续，SQL如何实现？

[程序一员]

查询数据库一张表 中第31-40条记录，id自增长不是连续，SQL如何实现？

[Mr_Nice]

row_number() 

[「已注销」]

1.    
select top m * from tablename where id not in (select top n id from tablename order by id asc/*|desc*/)    
  
2.    
select top m * into 临时表(或表变量) from tablename order by columnname -- 将top m笔插入到临时表    
set rowcount n   --只取n条结果   
select * from 表变量 order by columnname desc    
  
3.    
select top n * from     
(select top m * from tablename order by columnname) a    
order by columnname desc    
  
  
4.如果tablename里没有其他identity列，那么：    
先生成一个序列,存储在一临时表中.   
select identity(int) id0,* into #temp from tablename    
  
取n到m条的语句为：    
select * from #temp where id0 > =n and id0  <= m    
  
如果你在执行select identity(int) id0,* into #temp from tablename这条语句的时候报错,那是因为你的DB中间的select into/bulkcopy属性没有打开要先执行：    
exec sp_dboption 你的DB名字,'select into/bulkcopy',true    
  
  
5.如果表里有identity属性，那么简单：    
select * from tablename where identity_col between n and m     
  
6.SQL2005开始.可以使用row_number() over()生成行号   
;with cte as  
(   
 select id0=row_number() over(order by id),* from tablename   
)   
select * from cte where id0 between n to m

本文来自CSDN博客，转载请标明出处：http://blog.csdn.net/ws_hgo/archive/2009/02/18/3906309.aspx

[Mr_Nice]

[Quote=引用 1 楼 orchidcat 的回复:]

SQL code
row_number()
[/Quote]
参考！http://technet.microsoft.com/zh-cn/library/ms186734(SQL.90).aspx

[abuying]

--2000

select * from 
   (select * ,id1=(select count(1)+1 from tb a where a.id<tb.id)   from tb) t
    where id1 between n and m

[catherinelzl]

oracle的写法：
SELECT * FROM
(
SELECT A.*, ROWNUM RN
FROM (SELECT * FROM tb) A
WHERE ROWNUM <= 40
)
WHERE RN >= 31

[kevin87923]

select top 10 * from A where id not in (select top 30 id from A)

[catherinelzl]

还可以这样

SELECT * FROM
(
SELECT A.*, ROWNUM RN
FROM (SELECT * FROM tb) A
)
WHERE RN >= 31 and rn<=40

[程序一员]

with cte as(
select row_number() over(order by UserID asc) as row_number,* from Users
)
select * from cte where row_number between 31 and 40