url-pattern

Useing Spring for Security, I can get the program running using the following code. <intercept-url pattern="/web/admin**/**" access="ROLE_ADMIN" requires-channel="https"/> <intercept-url pattern="/web/**/" access="ROLE_USER,ROLE_ADMIN" requires-channel="https"/> I am trying to do this within a web.xml currently. Using JBOSS to deploy a .war file. Below is what I have, The url-pattern is what is causing me the problems in the first security-constraint. The pages are located at, and named /web/adminarchive /web/adminsettings /web/adminstuff etc... The code above within Spring handled it the way

When I use http.HandleFunc("/", serveRest) //serveRest is the method to handle request http.ListenAndServe("localhost:4000", nil) It will accept all request starting with "/" . How do I restrict it to serve only "localhost:4000" instead of every address like "localhost:4000/*" ? And can you guys suggest me a good tutorial for Go? icza The URL pattern to which you register your handlers is documented in the http.ServeMux type: Patterns name fixed, rooted paths, like "/favicon.ico", or rooted subtrees, like "/images/" (note the trailing slash). Longer patterns take precedence over shorter ones,

When I query this url http://mywebsite.com/foos/ Django give me : Page not found (404) Request Method: GET Request URL: http://mywebsite.com/foos// The current URL, , didn't match any of these. the errors : - in Request URL it add me '/' at the end, - in current URL is empty. my spec : I run my django website with nginx as reverseproxy to the fast_cgi. Here my website conf for nginx : server { listen 80; server_name mywebsite.com; location / { fastcgi_pass unix:/tmp/_var_wwwdjango_mywebsite.socket; include /etc/nginx/fastcgi_params; } } here is my fastcgi_params file : fastcgi_param QUERY

I'm developing a JSF 2 web application. For prestige purpouses I would like that every URL ends with .jsf extension. Now it ends with .xhtml . If I change it directly to .jsf in web browser address bar, then a HTTP 500 error is shown. How can I set it to .jsf ? The URL pattern of JSF pages is specified by <servlet-mapping> of the FacesServlet in web.xml . As you mentioned that .xhtml works fine, you have apparently configured it as follows: <servlet> <servlet-name>Faces Servlet</servlet-name> <servlet-class>javax.faces.webapp.FacesServlet</servlet-class> </servlet> <servlet-mapping> <servlet

I use a filter to check URL patterns for the logged in user. But I have many URL patterns I need to filter. { "/table/*", "/user/*", "/contact/*", "/run/*", "/conf/*", ..., ..., ...} It's becoming unmaintainable. It will be simpler just to exclude: { "/", "/login", "/logout", "/register" } How can I achieve this? @WebFilter(urlPatterns = { "/table/*","/user/*", "/contact/*","/run/*","/conf/*"}) public class SessionTimeoutRedirect implements Filter { protected final Logger logger = LoggerFactory.getLogger("SessionFilter"); @Override public void doFilter(ServletRequest req, ServletResponse res,

I was reading Head First JSP and Servlets book. I was going through the mapping of servlet. And my doubt here is <servlet> <servlet-name>test</servlet-name> <servlet-class>com.avinash.TestServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>test</servlet-name> <url-pattern>/ServletBeer.do</url-pattern> </servlet-mapping> Why should the url-pattern start with forward slash( / )? What does the forward slash( / ) represent? Does it represent our webapp name? What happens if the url-pattern do not start with forward slash( / )? Is it a specification to start with forward slash( / )?

I'm following a tutorial where my urlpatterns are: urlpatterns = patterns('', url(r'^passwords/$', PasswordListView.as_view(), name='passwords_api_root'), url(r'^passwords/(?P<id>[0-9]+)$', PasswordInstanceView.as_view(), name='passwords_api_instance'), ...other urls here..., ) The PasswordListView and PasswordInstanceView are supposed to be class based views. I could not figure out the meaning of the name parameter. Is it a default parameter passed to the view? No. It is just that django gives you the option to name your views in case you need to refer to them from your code, or your

Having an @WebServlet(urlPatterns = "/myServlet/") . If the user goes to myapp/myServlet/other , I still want my servlet to catch. So to say, wildcard anything on after the servlet path. How could I do this? BalusC You can use * as prefix or suffix wildcard. In your case, you can use /myServlet/* for a folder mapping. @WebServlet("/myServlet/*") The path info (the part after the mapping in the URL) is in the servlet by the way available as: String pathInfo = request.getPathInfo(); This would in case of myapp/myServlet/other return /other . See also: Servlet and path parameters like /xyz/{value