HttpServletRequest, csrf check for referrer header

Question

I need to add a check to see if the domain matches the referrer and completely new to csrf concepts and servlets. I would like to know if there is a way for me to validate if the referrer exists

If the referrer header is not https://[samedomain]/abc/sso?module=console, then fail. Note that the check should be very strict here. Cannot just compare using endswith “/abc/sso?module=console” since it could be bypass with https://attacker.com/abc/sso?module=console or https://[samedomain].attacker.com/abc/sso?module=console If not fail, proceed

I am looking for the right validation with regards to code like may be need to compare the port and the server name too. I am not looking for something overly complicated i have come this far ,

    String refererHeader = request.getHeader("referer");
    final String PATH = '/abc/sso?module=console',
    String host = request.getServerName();
    int port = request.getServerPort();

    String portstr="";
    if(port!=80 || port!= 443){
      portstr=":"+port;
    }

    if (refererHeader == null) {
        response.setStatus(HttpServletResponse.SC_BAD_REQUEST);
        return;
    }

    if (refererHeader != null &&  host!== null) {
     //check if port is not the default ports, in that case construct the url 

     //append with PATH and compare if the referrer header and this matches 

    }

Any help would be appreciated

Let us know where you're stuck - you'll likely want to start with a [servlet filter](https://docs.oracle.com/javaee/7/api/javax/servlet/Filter.html) and [take a look](https://stackoverflow.com/questions/37726903/how-to-prevent-my-web-app-from-csrfcross-site-request-forgery-in-java) at some examples to get you started. — stdunbar, Jul 10 '18 at 23:09
@stdunbar Thanks , i am looking for something very simple to compare if the servername and port match and then teh request goes through, please see above for what i have so far , may be this needs a regex too ? — looneytunes, Jul 11 '18 at 19:41

score 3 · Accepted Answer · answered Jul 12 '18 at 02:32

This was actually a bit harder than I thought so I thought I'd share what I came up with. The code could be optimized - there are too many if statements but it looks like you are coming from a different language so I tried to make it fairly straight forward. Additionally, there are probably some error conditions I missed but it should be close.

import java.io.IOException;
import java.net.URL;
import java.util.Arrays;
import java.util.List;
import java.util.regex.Pattern;

import javax.servlet.Filter;
import javax.servlet.FilterChain;
import javax.servlet.FilterConfig;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import javax.servlet.annotation.WebFilter;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

@WebFilter
public class RefererFilter implements Filter {
    private static final String PATH = "/abc/sso?module=console";
    // the domains that you will accept a referrer from
    private static final List<String> acceptableDomains = Arrays.asList("google.com", "mydomain.com");

    @Override
    public void init(FilterConfig filterConfig) throws ServletException {
        // unused in this application
    }

    @Override
    public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException {
        HttpServletRequest request = (HttpServletRequest) servletRequest;
        HttpServletResponse response = (HttpServletResponse) servletResponse;

        String refererHeader = request.getHeader("referer");
        // no need to continue if the header is missing
        if (refererHeader == null) {
            response.setStatus(HttpServletResponse.SC_BAD_REQUEST);
            return;
        }

        // parse the given referrer
        URL refererURL = new URL(refererHeader);
        // split the host name by the '.' character (but quote that as it is a regex special char)
        String[] hostParts = refererURL.getHost().split(Pattern.quote("."));

        if (hostParts.length == 1) { // then we have something like "localhost"
            if (!acceptableDomains.contains(hostParts[0])) {
                response.setStatus(HttpServletResponse.SC_BAD_REQUEST);
                return;
            }
        } else if (hostParts.length >= 2) { // handle domain.tld, www.domain.tld, and net1.net2.domain.tld
            if (!acceptableDomains.contains(hostParts[hostParts.length - 2] + "." + hostParts[hostParts.length - 1])) {
                response.setStatus(HttpServletResponse.SC_BAD_REQUEST);
                return;
            }
        }
        // if we've gotten this far then the domain is ok, how about the path and query?
        if( !(refererURL.getPath() + "?" + refererURL.getQuery()).equals(PATH) ) {
            response.setStatus(HttpServletResponse.SC_BAD_REQUEST);
            return;
        }
        // all tests pass - continue filter chain
        filterChain.doFilter(request, response);
    }

    @Override
    public void destroy() {
        // unused in this implementation
    }
}

HttpServletRequest, csrf check for referrer header

1 Answers1