how to get the base url from jsp request object?

Question

How to get the base url from the jsp request object? http://localhost:8080/SOMETHING/index.jsp, but I want the part till index.jsp, how is it possible in jsp?

Answer 1

So, you want the base URL? You can get it in a servlet as follows:

String url = request.getRequestURL().toString();
String baseURL = url.substring(0, url.length() - request.getRequestURI().length()) + request.getContextPath() + "/";
// ...

Or in a JSP, as , with little help of JSTL:

<%@taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
<%@taglib prefix="fn" uri="http://java.sun.com/jsp/jstl/functions" %>

${req.requestURL}

...

    

Note that this does not include the port number when it's already the default port number, such as 80. The java.net.URL doesn't take this into account.

See also:

• Browser can't access/find relative resources like CSS, images and links when calling a Servlet which forwards to a JSP