Избиране на първите n елемента от група в Oracle

Това ли търсите?

Моята тестова настройка:

SQL> alter session set nls_date_format = 'DD-Mon-YYYY HH24:Mi:SS';

Session altered.

SQL> drop table so_test;

Table dropped.

SQL> create table so_test  (
  2    n varchar2(32)
  3  , v varchar2(32)
  4  , t date );

Table created.

SQL> 
SQL> insert into so_test values ( 'X' , 'Test1', to_date('01-Jan-2011 12:00:00','DD-Mon-YYYY HH24:Mi:SS') );

1 row created.

SQL> insert into so_test values ( 'X' , 'Test2', to_date('01-Jan-2011 13:00:00','DD-Mon-YYYY HH24:Mi:SS') );

1 row created.

SQL> insert into so_test values ( 'X' , 'Test3', to_date('01-Jan-2011 14:00:00','DD-Mon-YYYY HH24:Mi:SS') );

1 row created.

SQL> insert into so_test values ( 'Y' , 'Test5', to_date('02-Jan-2011 12:00:00','DD-Mon-YYYY HH24:Mi:SS') );

1 row created.

SQL> insert into so_test values ( 'Y' , 'Test6', to_date('03-Jan-2011 12:00:00','DD-Mon-YYYY HH24:Mi:SS') );

1 row created.

SQL> insert into so_test values ( 'Y' , 'Test7', to_date('04-Jan-2011 12:00:00','DD-Mon-YYYY HH24:Mi:SS') );

1 row created.

SQL> 

Ето заявката:

SQL> select n,v,t from (
  2  select n, v , t , rank() over ( partition by n order by t desc) r
  3  from so_test
  4  ) where r <= 2;

N                V                T
-------------------------------- -------------------------------- --------------------
X                Test3                01-Jan-2011 14:00:00
X                Test2                01-Jan-2011 13:00:00
Y                Test7                04-Jan-2011 12:00:00
Y                Test6                03-Jan-2011 12:00:00

SQL>