Hi Jérôme, >> You can of course use the "https://server/login"; url to call >> your CAS server to log in, but... Got it, I need to put service parameter, but I want a better form to do it.
In advance: I'm using Java web apps with JSF. I want to make a more userfrendly application. In general way, my users always will enter in an non-protected area of any apps of our intranet. And menu application will be filled with restricted actions available for that user. I managed to make the gateway mod (https://lists.wisc.edu/read/messages?id=18381813), so if the user is logged in CAS, he will always be logged in any app, thats good and working. Now, if the user is entering in an application but he is not logged in CAS - so menu is limited - then I want to present him a "login link". So far I created a link with "https://server/login"; an fill service parameter with request info. Thats work but no good. I want a better way to scape hardcoding, a way of getting cas login url programatically, or maybe, a way of request the authentication filter to make his magic. Tx in advance, Claudio Weiler -- You are currently subscribed to [email protected] as: [email protected] To unsubscribe, change settings or access archives, see http://www.ja-sig.org/wiki/display/JSG/cas-user
