Ranking rows in Oracle

Oracle has a very nifty function to rank the usefullness of a row, it is sort of comparable to an aggregate but returns an rank value for the input instead of just the result row.
Perhaps you want to find the 5 most sold items for every month of the year or you need to figure out who worked the longest hours from a timesheet system. The rank query returns a sequence over the "top ranked" rows as ordered by the rank queried. If you just want the top 1 you just select the whole rank statement as an identifier and add "where identifier=1" in your outer where statement.

Example
I wrote this mockup query to show how it can be used.
We have a nice web based application where we, for audit purposes, store all applications logins to a logins table in the database. Lets say that you for some unknown reason need to figure out who was the first user to login each month (yes I know, terrible use case).

select
  user_id, 
  to_char(time,'YYYY-MM-DD HH24:MI:SS') LOGON_TIME
from 
  (
  select 
    user_id, 
    time, 
    rank() over(partition by user_id 
                order by time asc) rowrank
  from
    logins
  )
where
  rowrank=1
/

Returns

   USER_ID LOGIN_TIME
---------- -------------------
        68 2006-02-01 00:07:04
        68 2006-03-01 00:07:17
        68 2006-04-01 03:21:17
       108 2006-05-01 00:00:13

More reading:
http://www.adp-gmbh.ch/ora/sql/analytical/rank.html
http://download-uk.oracle.com/docs/cd/B19306_01/server.102/b14200/functions123.htm

Im working on the second part of the XML stuff.