Re: Getting server-url ?
On Tue, 2009-11-17 at 14:34 +, Peter Arnulf Lustig wrote: Hi, how can I get the server-url on which the wicket application is running? like http://www.serverurl.com/WicketApp/ it should return http://www.serverurl.com ((WebRequest) getRequest()).getHttpServletRequest().getServerHost() or something like that Thank you __ Do You Yahoo!? Sie sind Spam leid? Yahoo! Mail verfügt über einen herausragenden Schutz gegen Massenmails. http://mail.yahoo.com - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: Getting server-url ?
Martin Grigorov schrieb: it should return http://www.serverurl.com ((WebRequest) getRequest()).getHttpServletRequest().getServerHost() or something like that pass the httpServletRequest to public static StringBuffer getContextUrl(final HttpServletRequest req) { String protocol = req.isSecure() ? https://; : http://;; String hostname = req.getServerName(); int port = req.getServerPort(); StringBuffer url = new StringBuffer(128); url.append(protocol); url.append(hostname); if ((port != 80) (port != 443)) { url.append(:); url.append(port); } String ctx = req.getSession().getServletContext().getContextPath(); if (!ctx.startsWith(/)) { url.append('/'); } url.append(ctx); if (!ctx.endsWith(/)) { url.append('/'); } return url; } cu uwe - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org