APEX

Hi,

I am new to Apex and trying to develop an application. Need Help! Help with coding will be great!
My problem is to create apex page as popup window!

What i have done:
I have an image button(page item) which onclick apparently branches to another apex page - which displays several images and on submit returns the image information back to where it was called and setting the values in the respective text fields. I have implemented it.

What I need:
But I want to make this apex page with several images as a POPUP WINDOW or NEW WINDOW and upon clicking should close and return values to textfield in the parent apex page (like before). Is it possible ?

(2 is my parent page. 15 is my target page number)
What I have proceeded with:

Try 1: In the branch to the apex page (condition: button pressed-Go), i chose URL and in the URL target wrote :
javascript:popupURL('f?p=&APP_ID.:15:&SESSION.::::P15_PAGE_CALLED:2'')
But it branches to a blank page (not even pop up) which displays
"Location: f?p=145:2:4321906619921733::NO" instead of displaying page 15 as popup window.

Try2: Instead of Branching,
I pasted
onclick="popUp2('f?p=&APP_ID.:15:&SESSION.::::P15_PAGE_CALLED:2');" in Button attributes
I get a popup this time (yay!) but its an error display page which says:

Bad Request
Your browser sent a request that this server could not understand.
mod_plsql: /pls/apex/dev/f HTTP-400 Missing '=' in query string or post form

What Else Can I do to make it work? :(

*Need Help guys!
Popup page or New window page is required.*

Thanks in Advance.

Edited by: 865365 on Jun 12, 2011 8:50 PM