oracle中如何取最新的数据

select * from

(select t.*,rownum rc from

(select *

from sys_log t where '"+today1+"' = to_char(operate_date,'yyyy-MM-dd')

order by operate_date desc

) t

and rownum<=5) a where a.rc>=1

1、通过rownum对数据库的记录进行编号

SELECT ROWNUM RN1,A.* FROM A//从表A中获取记录，并用rownum生成RN1

2、通过RN1的标记进行倒序排序

SELECT B.*,ROWNUM RN2 FROM (SELECT ROWNUM RN1,A.* FROM A) ORDER BY RN1 DESC

3、通过RN2取小于等于10的数据记录

SELECT * FROM (SELECT B.*,ROWNUM RN2 FROM (SELECT ROWNUM RN1,A.* FROM A) ORDER BY RN1 DESC) WHERE RN2<=10

表里面有时间字段吗？

select * from tablename t where t.updatedate >=(select max(updatedate) from tablename)

如果只考虑数据插入的早晚，可以用rowid最大的。

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