How can I get the request URL from a Java Filter?

Question

I am trying to write a filter that can retrieve the request URL, but I'm not sure how to do so.

Here is what I have so far:

import javax.servlet.*;
import javax.servlet.http.HttpServletRequest;
import java.io.IOException;

public class MyFilter implements Filter {
    public void init(FilterConfig config) throws ServletException { }

    public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws ServletException, IOException {
        chain.doFilter(request, response);

        String url = ((HttpServletRequest) request).getPathTranslated();
        System.out.println("Url: " + url);
    }

    public void destroy() { }
}

When I hit a page on my server, the only output I see is "Url: null".

What is the correct way to get the requested URL from a given ServletRequest object in a Filter?

Buhake Sindi · Accepted Answer · 2012-08-14 07:48:52Z

206

Is this what you're looking for?

if (request instanceof HttpServletRequest) {
 String url = ((HttpServletRequest)request).getRequestURL().toString();
 String queryString = ((HttpServletRequest)request).getQueryString();
}

To Reconstruct:

System.out.println(url + "?" + queryString);

Info on HttpServletRequest.getRequestURL() and HttpServletRequest.getQueryString().

edited Aug 14, 2012 at 7:48

answered Dec 8, 2010 at 16:11

Buhake Sindi

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4 Comments

BalusC Over a year ago

getRequestURL() returns StringBuffer, not String.

Kazuhiro Sera Over a year ago

It's better to consider the pattern of null queryString.

Alexander Taylor Over a year ago

if you want the "blabla:8080" part stripped away for you, getRequestURI() ('I' not 'l') returns a String starting with "/"

Polyana Fontes Over a year ago

If you want only the path to the servlet you can use request.getServletPath(), it's useful on JavaServer Faces to retrieve the equivalent of the view id (the path from the webapp root to the xhtml page, excluding domains, deployment prefix, etc)

Brad Parks · Accepted Answer · 2017-06-29 16:19:29Z

6

Building on another answer on this page,

public static String getCurrentUrlFromRequest(ServletRequest request)
{
   if (! (request instanceof HttpServletRequest))
       return null;

   return getCurrentUrlFromRequest((HttpServletRequest)request);
}

public static String getCurrentUrlFromRequest(HttpServletRequest request)
{
    StringBuffer requestURL = request.getRequestURL();
    String queryString = request.getQueryString();

    if (queryString == null)
        return requestURL.toString();

    return requestURL.append('?').append(queryString).toString();
}

answered Jun 29, 2017 at 16:19

Brad Parks

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Comments

Jess · Accepted Answer · 2021-03-31 06:04:05Z

3

If you use Spring, you can use OncePerRequestFilter or others.

import org.springframework.web.filter.OncePerRequestFilter;
import org.springframework.stereotype.Component;

@Component
public class MyFilter extends OncePerRequestFilter {
    @Override
    protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain) throws ServletException, IOException {
         String url = request.getRequestURL();
         filterChain.doFilter(request, response);
    }

}

answered Mar 31, 2021 at 6:04

Jess

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Comments

Öktem Erbay · Accepted Answer · 2023-07-06 11:28:58Z

0

If you want to acquire them separately you can use as the following

((HeaderWriterFilter.HeaderWriterRequest) request).getRequest().getAttribute("org.apache.catalina.AccessLog.ServerName") => host info
((HeaderWriterFilter.HeaderWriterRequest) request).getRequest().getAttribute("org.apache.catalina.AccessLog.ServerPort") => port info
((HttpServletRequest)request).getServletPath() or ((HttpServletRequest)request).getRequestURI() => requested source path
((HttpServletRequest)request).getQueryString() => query parameters

answered Jul 6, 2023 at 11:28

Öktem Erbay

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Comments

Jigar Joshi · Accepted Answer · 2010-12-08 16:10:39Z

-2

request.getRequestURL();

Documentation

answered Dec 8, 2010 at 16:10

Jigar Joshi

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2 Comments

Anthony Hayward Over a year ago

ServletRequest does not implement getRequestURL(). You need to cast it to HttpServletRequest if possible as per Buhake's answer

Stepan Yakovenko Over a year ago

its missing parameters

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