Navigation menu - current page = HOW??

Question

I've been pulling my hair out for days over such a simple thing as having a navigation page with page links, and wanting the current page to stand out in the navigation menu (a different style from the other links).
I created a html content layout template to achieve this. In my template (which my page link items use) I have tried to figure out what page I am on. Nothing works at all. I've googled the hell out of this forum and the internet at large, and no go. How do I get the value (url) of the current page? as in http://myserver.com/myportal/mypage/mysubpage sot hat I may compare it to the items destination url?
In my template, I have an if else construct, but since I don't know what to compare my #item.pageurl# (or some such) to, I'm getting absolutely nowhere. owa_util.get_cgi_env is not helping, nothing useful can be grabbed from those headers (just urls for individual portlets or items etc).
I do not administer this server myself, nor do I have access to it other than this one pagegroup through the navigator/builder GUI. My permissions are pretty restricted too I guess.
So is there a simple way to get the url of the page I'm on? Or has someone already done this and I've been at it for naught? Help! :(
Portal version is OracleAS Portal 10g Release 2 (10.1.4)

Answer 1

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by saying html content layout template, do you mean that on Navigator, you using html template (in the pageGroups tab/any-pageGroup/html templates) or a User Interface Template (within the providers tab options)?Yes, my page group -> html templates -> HTML Content Layout
I'm using this template to style the page link items in my navigation page (you know, some divs, css and so on), my template looks like this (ignore the if construct owa_util.bla_blah('BLAH'), it obviously doesn't work but it's there to illustrate exactly what I need:
<oracle>
declare
luokka varchar2(300);
begin
htp.p('page_url: #PAGE.EDITPAGE.URL#');
if  owa_util.get_cgi_env('BLAH')='#ITEM.PAGENAME#' then
class:= 'active';
else
class:='nonactive';
end if;
htp.p('<div style="padding-left:15px; font-weight:bold;background-image: url(https://myportal.com/portal/pls/portal/docs/1/1911205.PNG);background-repeat: no-repeat;background-position:left;" class='|| class||'>#ITEM.CONTENT#</div>');
end;
</oracle>
if you are using the first one of them, then try the following to get page_url;
- in the body section of the template, use <ORACLE> </ORACLE> tags-pair to write a plsql code;
<oracle>
declare
v_pageUrl varchar2(50):= '#PAGE.EDITPAGE.URL#';This doesn't give me anything, and #PAGE.EDITPAGE.URL# is not available in the substitution string list. I only have substittion strings that begin with #ITEM
begin
htp.p('page url = '||substr(v_page_url,1,instr(v_page_url,'?',1)-1 )||'<br>');
end;
</oracle>
this should give you the url of the page you are on.
hope that helps!Edited by: Baguette on 25-Feb-2009 09:02
Edited by: Baguette on 25-Feb-2009 09:05