Please do not mind I figured out How to do it. Its simple. Just use URLLoader, URLRequest, URLVariables Classes and emulate what HTML form does. here is the code var urlVar:URLVariables; var urlReq:URLRequest; urlVar = new URLVariables(); urlVar.j_username = "CEO"; urlVar.j_password = "raja1248"; urlVar._j_spring_security_remember_me = "false"; urlReq = new URLRequest("http://dell-15rse:8080/Tej-EIS/j_spring_security_check"); urlReq.method = "POST"; urlReq.data = urlVar;
var loader:URLLoader = new URLLoader(); loader.addEventListener(Event.COMPLETE, completeHandler); loader.addEventListener(Event.OPEN, openHandler); // loader.addEventListener(ProgressEvent.PROGRESS, progressHandler); loader.addEventListener(SecurityErrorEvent.SECURITY_ERROR, securityErrorHandler); loader.addEventListener(HTTPStatusEvent.HTTP_STATUS, httpStatusHandler); // loader.addEventListener(IOErrorEvent.IO_ERROR, ioErrorHandler); try { loader.load(urlReq); } catch(error:Error) { trace("Unable to load requested document."); } completeHandler receives whatever page mentioned for login success/login failure in form-login element of spring security config. On login success, subsequent requests to secured resources are possible by default nothing special is to be done. I am now trying to explore possibility of raising exception in case of login failure so that I can use HTTPService and its result and fault events by spring security config modifications. Hope this helps to someone who fumbles on AIR login issue. Thanks and warm regards Raja Patil