3

I have a JAX-WS 2.2 WebService and I must take the IP Address of each client that communicate with it. I write a SOAP protocol handler but I can't see the addresses because the handlers doesn't contain this information and using the mimeheaders I can't also see this information. The code of my handler is the follow:

public class AddressHandler implements SOAPHandler<SOAPMessageContext> {

private void takeIPAddress(SOAPMessageContext context) {

    try {
        SOAPMessage original = context.getMessage();
        MimeHeaders mimeheaders = original.getMimeHeaders();
        MimeHeader mimeheader = null;

        Iterator<?> iter = mimeheaders.getAllHeaders();

        for (; iter.hasNext();) {
            mimeheader = (MimeHeader) iter.next();

            System.out.println("name=" + mimeheader.getName() + ", value="
                    + mimeheader.getValue());
        }


     } catch (Exception e) {
         e.printStackTrace();
     }


}

@Override
public void close(MessageContext arg0) {
    // TODO Auto-generated method stub

}

@Override
public boolean handleFault(SOAPMessageContext arg0) {
    // TODO Auto-generated method stub
    return false;
}

@Override
public boolean handleMessage(SOAPMessageContext context) {
    takeIPAddress(context);
    return true;
}

@Override
public Set<QName> getHeaders() {
    // TODO Auto-generated method stub
    return null;
}
}

Now I'm seeing that would be possible to see the addresses using the following code:

SOAPMessageContext jaxwsContext = (SOAPMessageContext)wsContext.getMessageContext();
HttpServletRequest request = HttpServletRequest)jaxwsContext.get(SOAPMessageContext.SERVLET_REQUEST);
String ipAddress = request.getRemoteAddr();

But I can't import correctly the HttpServletRequest class. Do you have any ideas?

UPDATE

Thanks to A Nyar Thar, I've seen that exists another method to take address and I've implemented this in my code, that now is:

private void takeIPAddress(SOAPMessageContext context) {

    HttpExchange exchange = (HttpExchange)context.get("com.sun.xml.ws.http.exchange");
    InetSocketAddress remoteAddress = exchange.getRemoteAddress();
    String remoteHost = remoteAddress.getHostName();

    System.out.println(remoteHost);     

}

But the code execution create this error (where row 39 is where I do exchange.getRemoteAddress()):

java.lang.NullPointerException
at server.AddressHandler.takeIPAddress(AddressHandler.java:39)
at server.AddressHandler.handleMessage(AddressHandler.java:80)
at server.AddressHandler.handleMessage(AddressHandler.java:1)
at com.sun.xml.internal.ws.handler.HandlerProcessor.callHandleMessage(HandlerProcessor.java:282)
at com.sun.xml.internal.ws.handler.HandlerProcessor.callHandlersRequest(HandlerProcessor.java:125)
at com.sun.xml.internal.ws.handler.ServerSOAPHandlerTube.callHandlersOnRequest(ServerSOAPHandlerTube.java:123)
at com.sun.xml.internal.ws.handler.HandlerTube.processRequest(HandlerTube.java:105)
at com.sun.xml.internal.ws.api.pipe.Fiber.__doRun(Fiber.java:626)
at com.sun.xml.internal.ws.api.pipe.Fiber._doRun(Fiber.java:585)
at com.sun.xml.internal.ws.api.pipe.Fiber.doRun(Fiber.java:570)
at com.sun.xml.internal.ws.api.pipe.Fiber.runSync(Fiber.java:467)
at com.sun.xml.internal.ws.server.WSEndpointImpl$2.process(WSEndpointImpl.java:299)
at com.sun.xml.internal.ws.transport.http.HttpAdapter$HttpToolkit.handle(HttpAdapter.java:593)
at com.sun.xml.internal.ws.transport.http.HttpAdapter.handle(HttpAdapter.java:244)
at com.sun.xml.internal.ws.transport.http.server.WSHttpHandler.handleExchange(WSHttpHandler.java:95)
at com.sun.xml.internal.ws.transport.http.server.WSHttpHandler.handle(WSHttpHandler.java:80)
at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:77)
at sun.net.httpserver.AuthFilter.doFilter(AuthFilter.java:83)
at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:80)
at sun.net.httpserver.ServerImpl$Exchange$LinkHandler.handle(ServerImpl.java:677)
at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:77)
at sun.net.httpserver.ServerImpl$Exchange.run(ServerImpl.java:649)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
at java.lang.Thread.run(Thread.java:744)

I think that the real problem is that I don't know how take WebServiceContext from my class AddressHandler. Do you have ideas?

hasmet
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  • Look into possible duplicate of http://stackoverflow.com/questions/12727989/jax-ws-getting-client-ip – Wundwin Born Jul 30 '14 at 09:55
  • I've seen the other question, but the difference is that in my situation I've not a MessageContext but a SOAPMessageContext. Do you now any way to obtain this object? – hasmet Jul 30 '14 at 10:04
  • I think `SOAPMessageContext extends MessageContext`. In the mentioned link, you should get `HttpExchange` way for JAX-WS based ws then get remote address. – Wundwin Born Jul 30 '14 at 10:07
  • Ok I've understood the mentioned link, but I can't find a method to take MessageContext in my case. Do you have any idea? – hasmet Jul 30 '14 at 11:36
  • Ok I will post as an answer soon. – Wundwin Born Jul 30 '14 at 11:38

1 Answers1

4

For JAX-WS based webservice, you can access remote host info from javax.xml.ws.spi.http.HttpExchange, that can be accessed based on JAX-WS version,

  • JAX-WS 2.1

    SOAPMessageContext soapContext = (SOAPMessageContext)wsContext.getMessageContext();
HttpExchange exchange = (HttpExchange)soapContext.get(JAXWSProperties.HTTP_EXCHANGE);

  • JAX-WS 2.2

    SOAPMessageContext soapContext = (SOAPMessageContext)wsContext.getMessageContext();
HttpExchange exchange = (HttpExchange)soapContext.get("com.sun.xml.ws.http.exchange");

Note that wsContext.getMessageContext() will return MessageContext. If you want it, don't cast to SOAPMessageContext, like that,

MessageContext msgContext = wsContext.getMessageContext();

Finally you can access remote address info,

InetSocketAddress remoteAddress = exchange.getRemoteAddress();
String remoteHost = remoteAddress.getHostName();
Wundwin Born
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