question: public URL createServerURL ( URL server , String path ) throws MalformedURLException { return new URL ( server . getProtocol ( ) , server . getHost ( ) , server . getPort ( ) , localServerDeploymentPath + "/" + path ) ; }
target:  Creates a new URL by adding the provided path to the end of the provided server's URL. This assumes that the provided server URL does not include the deployment path of OpenAM since it can be assumed to be the same as this server's deployment path. e.g. Given the server URL "https://remote.example.com:8080" and the path "GetHttpSession?op=create", if this OpenAM server is deployed to "/openam" then the resulting URL will be https://remote.example.com:8080/openam/GetHttpSession?op=create