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Extracting parameters from a URL in Oracle ApEx

Del17Aug 31 2017 — edited Sep 1 2017

I am trying to pass a URL from one APEX app through to another APEX app as a parameter in the URL.  So that the destination app knows from where it was called.

I have to return from the destination app back to one of two sources because they are either side of the architecture that my employer uses under its SSO access control for different arms of my employer.

I understand I can place a link to a URL in the originating APEX app as a parameter using the URL structure
Protocol/HostName/Path?parameter&parameter.

The URL link would look like the follow with the parameter SEND_URL being what I am trying to retrieve:

http://hostname:port/ords/f?p=app_name&SEND_URL=http://hostname:port/ords/f?p=app_name

This works however in the receiving app I cannot find a way to get APEX to extract the parameter  SEND_URL from the URL.  All my research points to that it can be done with something called APEX Salesforce but not in standard APEX.

Salesforce suggests ApexPages.currentPage().getParameters().get(SEND_URL) but if I put this in as a URL in a vertical list I get the literal string returned and of course the web browser can't find it.

I tried writing some PLSQL in an application process using the above function to assign to a bind variable or APEX substitution string but PLSQL does not recognise getParameter().

I am using Oracle Application Express 5.1.1.00.08.  Any help would be very much appreciated.

Thank You,

Del17.

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Locked on Sep 29 2017
Added on Aug 31 2017
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