Hi, My web Application URL where I want to re-direct CAS after successful log in: http://<machinename>:28080/<applicationName>/Login.do
My CAS server URL which I am accessing from browser and then enter my username/password on CAS login page: https://localhost:8443/cas/login?service=http://<machinename>:28080/test/Login.do Below in the entry in my web.xml file of my application. <filter> <filter-name>CAS Filter</filter-name> <filter-class>edu.yale.its.tp.cas.client.filter.CASFilter</filter-class> <init-param> <param- name>edu.yale.its.tp.cas.client.filter.loginUrl</param-name> <param-value>https://localhost:8443/cas/login</param-value> </init-param> <init-param> <param-name>edu.yale.its.tp.cas.client.filter.validateUrl</param-name> <param-value>https://localhost:8443/cas/serviceValidate</param-value> </init-param> <init-param> <param-name>edu.yale.its.tp.cas.client.filter.serverName</param-name> <param-value>http://ps2078:28080</param-value> </init-param> </filter> <filter-mapping> <filter-name>CAS Filter</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> I have doubt about the "edu.yale.its.tp.cas.client.filter.serverName" param value. Whether it should be "http://ps2078:28080"; or "http://<machinename>:28080/test/Login.do". Also, in SecureURL.java class, I get an "An exception occurred: javax.net.ssl.SSLHandshakeException" at line 84 "r = new BufferedReader(new InputStreamReader(uc.getInputStream()));" where we read the input stream. Please help me in correcting the web.xml entry, if any and also the probable reason for the exception which I am encountering. Thanks, Sagar -- 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
