Need help on a continuation of the "how to get the first 10 records with order by" problem

From: L Lee <linda.lee_at_ci.seattle.wa.us>
Date: Tue, 19 Oct 1999 20:59:48 GMT
Message-ID: <01bf1a74$e08b0940$a40c4a9c@MD13457>

I have the following code, using just one customer as an example but needing to get 13 "reads" back from each customer in order of most recent date first. There will not always be 13 reads available and there will definitely be more reads than 13 in the database.

select prem_code, billed_chg, billed_consump, to_char(action_date,'MM/DD/YYYY')
  from urrshis us,

       ubbchst ub,
	   uabopen ua
  where 13 >= (select count( distinct to_char(action_date,'MM/DD/YYYY'))
	   		   from urrshis ur
			   where ur.action_date <= us.action_date
			   and prem_code = '1110'
			   and ur.scat_code = us.scat_code)  

  and ua.prem_code = '1110'
  and scat_code in ('KW', 'KWH')

  and ub.scat_code = ua.scat_code
  and ub.prem_code = ua.prem_code
  and ur.prem_code = ub.prem_code

  order by ua.prem_code, us.action_date desc

For this particular customer there are exactly 10 records. This gets back what I need when I use a count of 10 or more, however if I use "where 9 >=" then it cuts out the most recent date and so forth on down. What the user wants is to look at the most recent records for all customers. There may be more than 13 read dates but they want only the most recent ones. The number of records coming back will vary for each customer. Is there a way to do this in a select statement without calling functions?

Thanks for any help you can give, I hope I haven't left out anything pertinent...    Received on Tue Oct 19 1999 - 15:59:48 CDT