Root URl of the servlet

Question

I want to get the root url of my web application from one of the servlet.

If I deploy my application in \"www.mydomain.com\" I want to get the root url like \"http:/

Answer 1

You do realize that the URL client sees (and/or types into his browser) and the URL served by the container your servlet is deployed on can be very different?

In order to get the latter, though, you have a few methods available on HttpServletRequest:

• You can either call getScheme(), getServerName(), getServerPort() and getContextPath() and combine them using appropriate separators
• OR you can call getRequestURL() and remove getServletPath() and getPathInfo() from it.

Answer 2

Generally, you can't obtain the URL; but, there are workarounds for specific cases. See Finding your application’s URL with only a ServletContext

Answer 3

1. Write a scriptlet in the welcome file to capture the root path. I assume index.jsp is the default file. So put the following code in that

   <% RootContextUtil rootCtx = RootContextUtil.getInstance(); if( rootCtx.getRootURL()==null ){ String url = request.getRequestURL().toString(); String uri = request.getRequestURI(); String root = url.substring( 0, url.indexOf(uri) ); rootCtx.setRootURL( root ); } %>

2. Use this variable wherever needed within the application directly by calling the value as

String rootUrl = RootContextUtil.getInstance().getRootURL();

NOTE: No need to worry about protocols/ports/etc.. Hope this helps every one

Answer 4

public static String getBaseUrl(HttpServletRequest request) {
    String scheme = request.getScheme();
    String host = request.getServerName();
    int port = request.getServerPort();
    String contextPath = request.getContextPath();

    String baseUrl = scheme + "://" + host + ((("http".equals(scheme) && port == 80) || ("https".equals(scheme) && port == 443)) ? "" : ":" + port) + contextPath;
    return baseUrl;
}

Answer 5

This function helps you to get your base URL from a HttpServletRequest:

  public static String getBaseUrl(HttpServletRequest request) {
    String scheme = request.getScheme() + "://";
    String serverName = request.getServerName();
    String serverPort = (request.getServerPort() == 80) ? "" : ":" + request.getServerPort();
    String contextPath = request.getContextPath();
    return scheme + serverName + serverPort + contextPath;
  }