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Oracle Apex : How to get the URL requested?

YounMay 21 2021

Hello Community,
I have an Apex 19.2 application.
I'm trying to log in the database all the pages requested of my application.
I created an Application Process that runs On load Before the header on all the pages that saves the page requested in a table.
So far, I can get the AppId, the PageId and the Session that I store in a table :

insert into PAGESVISITED 
(Application, PAGE, IPAdress, UserAgent, Session) 
values 
(:APP_ID, :APP_PAGE_ID, owa_util.get_cgi_env('X-FORWARDED-FOR'), owa_util.get_cgi_env('user-agent'), :APP_SESSION));

I would like to know please if there is a way to get the entire URL requested. Means including the pages items set as well their values.
When we know the Page Item set in a URL, It's easy to get its value. But is it possible to get any Page Item name and its value for a generic page ? or any way to get the entire url requested for example get :

f?p=App:Page:Session:Request:Debug:ClearCache:itemNames:itemValues:PrinterFriendly

I'm not asking how to use the function APEX_UTIL.PREPARE_URL to generate a URL. My question is more : assuming a URL requested, how to get all its details ?
Thank you in advance. Cheers,

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Added on May 21 2021
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