Re: What's wrong with this cursor???

You need to open the cursor outside your loop.

I always open a cursor like this...just to be sure and avoid this error

if (this_cursor%isopen) then

    close this_cursor;
end if;
open this_cursor;
--miscellaneous code here.
close this_cursor;

Steve McClure
----- Original Message -----
From: Peter Hazelton <peterhazelton_at_hotmail.com> To: Multiple recipients of list ORACLE-L <ORACLE-L_at_fatcity.com> Sent: Friday, September 08, 2000 8:46 AM Subject: What's wrong with this cursor???

> Hi all
>
> I have created the following in Oracle Reports:
>
> function CF_25_percentFormula return Number is
> total_rec number(9);
> v_counter number(9) :=0;
> v_icons inpatient.icons%type;
> twenty_five_perc_rec number(9);
>
> cursor this_cursor is
> select icons
> from inpatient
> where icons < 100;
>
> begin
> select count(icons)
> into total_rec
> from inpatient
> where icons <=100;
> twenty_five_perc_rec := total_rec*.25 ;
> loop
> open this_cursor;
> fetch this_cursor into v_icons;
> v_counter := v_counter +1;
>
> if v_counter = twenty_five_perc_rec then return(v_icons);
> end if;
> exit when this_cursor%NOTFOUND;
> end loop;
> close this_cursor;
> END;
>
>
> The function compiles fine, but when I go to run it, I get the following
> error:
> ORA_06511: PL/SQL: cursor already open.
>
> Am I close with this function? I am looking to return the ICONS number for
> the 25th percentile record.
>
> Thanks in advance for your help.
>
> Peter
>
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> Author: Peter Hazelton
> INET: peterhazelton_at_hotmail.com
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Received on Fri Sep 08 2000 - 17:02:30 CDT