Sure, I can explain it.

This is the SQL query on which the interactive report is based:

select f.FPID, f.TITLE,f.DATE,f.STATE 
FROM FORM f
left outer join MANAGER m on m.MID=f.MID
left outer join USERS u on u.UUID=m.UUID
where lower(u.username)=lower(:session_user_name);

With this query, the report is displayed normally.
However, when I try to retrieve another column from the Form table for the report, I get the error described above.

select f.FPID, f.TITLE,f.DATE,f.STATE, f.SPID
FROM FORM f
left outer join MANAGER m on m.MID=f.MID
left outer join USERS u on u.UUID=m.UUID
where lower(u.username)=lower(:session_user_name);

Hello Vinish,

this is how I set the P8_ID.

I created a pre-rendering process with the following PL/SQL Code:

declare 
l_id number;
begin

select f.id into l_id from formtable f
left outer join users u on u.UUID=f.UUID
where lower(u.username)=lower(:session_user_name);
apex_util.set_session_state('P8_ID',l_id);
end;

I checked it by viewing the session in the debug tool bar.

And the value is set.

By using this function:

DECLARE
l_branch_page number;
l_id number;

begin
select id into l_finalpaper_id from formtable where id=:P8_ID;
l_branch_page:= FORWARDING_TO_PAGE(:P8_ID);
:P8_URL :=apex_page.get_url(p_page => l_branch_page,p_items=>l_id);

END;

I get the error:

ERR-1002 Unable to find item ID for item "5" in application "100".

The number 5 is the ID of the form.