Java Httpservletrequest Get Url in Browsers Url Bar

Java HttpServletRequest get URL in browsers URL bar

If your current request is coming from an "inside the app-server" forward or include, the app-server is expected to preserve request information as request attributes. The specific attributes, and what they contain, depends on whether you're doing a forward or an include.

For <jsp:include>, the original parent URL will be returned by request.getRequestURL(), and information about the included page will be found in the following request attributes:

     javax.servlet.include.request_uri
     javax.servlet.include.context_path
     javax.servlet.include.servlet_path
     javax.servlet.include.path_info
     javax.servlet.include.query_string

For <jsp:forward>, the new URL will be returned by request.getRequestURL(), and the original request's information will be found in the following request attributes:

     javax.servlet.forward.request_uri
     javax.servlet.forward.context_path
     javax.servlet.forward.servlet_path
     javax.servlet.forward.path_info
     javax.servlet.forward.query_string

These are set out in section 8.3 and 8.4 of the Servlet 2.4 specification.

However, be aware that this information is only preserved for internally-dispatched requests. If you have a front-end web-server, or dispatch outside of the current container, these values will be null. In other words, you may have no way to find the original request URL.

Get current URL in Webapplication

If you want a javascript solution, you can use window.document.location object and its properties:

console.log(window.document.location.protocol);
http:
console.log(window.document.location.host);
stackoverflow.com
console.log(window.document.location.port);

console.log(window.document.location.href);
http://stackoverflow.com/questions/10845606/get-current-url-in-webapplication
console.log(window.document.location.pathname);
/questions/10845606/get-current-url-in-webapplication

You can understand other parameters reading this article at MDN.

Get full URL and query string in Servlet for both HTTP and HTTPS requests

By design, getRequestURL() gives you the full URL, missing only the query string.

In HttpServletRequest, you can get individual parts of the URI using the methods below:

// Example: http://myhost:8080/people?lastname=Fox&age=30

String uri = request.getScheme() + "://" +   // "http" + "://
             request.getServerName() +       // "myhost"
             ":" +                           // ":"
             request.getServerPort() +       // "8080"
             request.getRequestURI() +       // "/people"
             "?" +                           // "?"
             request.getQueryString();       // "lastname=Fox&age=30"

• .getScheme() will give you "https" if it was a https://domain request.
• .getServerName() gives domain on http(s)://domain.
• .getServerPort() will give you the port.

Use the snippet below:

String uri = request.getScheme() + "://" +
             request.getServerName() + 
             ("http".equals(request.getScheme()) && request.getServerPort() == 80 || "https".equals(request.getScheme()) && request.getServerPort() == 443 ? "" : ":" + request.getServerPort() ) +
             request.getRequestURI() +
            (request.getQueryString() != null ? "?" + request.getQueryString() : "");

This snippet above will get the full URI, hiding the port if the default one was used, and not adding the "?" and the query string if the latter was not provided.

Proxied requests

Note, that if your request passes through a proxy, you need to look at the X-Forwarded-Proto header since the scheme might be altered:

request.getHeader("X-Forwarded-Proto")

Also, a common header is X-Forwarded-For, which show the original request IP instead of the proxys IP.

request.getHeader("X-Forwarded-For")

If you are responsible for the configuration of the proxy/load balancer yourself, you need to ensure that these headers are set upon forwarding.

getting a HttpServletRequest response address

There is an HTTP Header called "Referer" that contains the information from which page the request was triggered. For AJAX requests, this header normally contains the URL of the page in which the AJAX request was triggered. You can get this header using request.getHeader("Referer"), so you could try if that is what you need.

But it's up to the client to send that header, so you cannot be sure that it will always be there. And this approach may also lead to maintainability issues, e.g. if you move you web page to a different URL. So it may be the better design to parameterize your JavaScript and you servlet, i.e. the client adds a parameter to the URL that tells the servlet from which context the request is coming from or what the expected output format is.

Wicket: Get URL from browser

You need to query the underlying HTTPServletRequest:

public class DummyPage extends WebPage{

    private String getRequestUrl(){
        // this is a wicket-specific request interface
        final Request request = getRequest();
        if(request instanceof WebRequest){
            final WebRequest wr = (WebRequest) request;
            // but this is the real thing
            final HttpServletRequest hsr = wr.getHttpServletRequest();
            String reqUrl = hsr.getRequestURL().toString();
            final String queryString = hsr.getQueryString();
            if(queryString != null){
                reqUrl += "?" + queryString;
            }
            return reqUrl;
        }
        return null;

    }

}

Reference:

• (Wicket) Component.getRequest()
• (Wicket) Request
• (Wicket) WebRequest
• (Servlet API) HTTPServletRequest

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