Get Url of endpoint using on a handler (JAX-WS)

Question

I am implementing many JAX-WS web services with a common Handler class to validate the correct structure of incoming SOAP messages.

Is there some way to obtain the URL to which is directed the current message so i could get the schema from this url automatically and get the message validated?

Spitz

make different methods and put first element as method name so that you can direct varisous soap message – Sorn

Answer 1

Found!!!

public class HeaderHandler implements SOAPHandler<SOAPMessageContext> {

public boolean handleMessage(SOAPMessageContext smc) {
    System.out.println("URL of Endpoint" +smc.get(JAXWSProperties.HTTP_REQUEST_URL));
 }
}

Spitz

com.sun.xml.ws.developer for JAXWSProperties seems not to be available in Java 7. Any alternatives? – Sharanshard

Answer 2

In my application smc.get(JAXWSProperties.HTTP_REQUEST_URL) returns null. I've fount another way, I hope this helps:

public class HeaderHandler implements SOAPHandler<SOAPMessageContext> {

    public boolean handleMessage(SOAPMessageContext smc) {
        String endpointAddress = (String) smc.get(BindingProvider.ENDPOINT_ADDRESS_PROPERTY);
        log.debug("endpointAddress " +endpointAddress);
    }
}

Adamo

Yes, you are right. I found that to be filled when the handled message was the response from the server. In this case, the HTTP method used is not GET, but POST, so the HTTP_REQUEST_URL is empty. So summarizing, if you need to get the Url as a server on an incoming message should use HTTP_REQUEST_URL while if you are doing it like a client ENDPOINT_ADDRESS_PROPERTY. At least in tomcat 7!!! – Spitz

Answer 3

Neither JAXWSProperties.HTTP_REQUEST_URL nor BindingProvider.ENDPOINT_ADDRESS_PROPERTY worked for me.

This is how i got url in an outbound message on WildFly 10 (Java 7):

public boolean handleMessage(SOAPMessageContext context) {
    String url = (String) context.get(MessageContext.PATH_INFO);
    ...
}

Ianiana