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Home/ Questions/Q 8981087
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T20:14:47+00:00 2026-06-15T20:14:47+00:00

I have created one CXF webservice sample: @WebService public interface InterfaceWebService { boolean doLogin(@WebParam(name=username)String

  • 0

I have created one CXF webservice sample:

@WebService
public interface InterfaceWebService {

    boolean doLogin(@WebParam(name="username")String username,@WebParam(name="password")String password);

}

Server code:

public class WebServer {

    protected WebServer() throws Exception {
        // START SNIPPET: publish
        System.out.println("Starting Server");
        WebServiceImpl implementor = new WebServiceImpl();
        String address = "http://192.168.0.76:9000/sample";
        Endpoint.publish(address, implementor);
        // END SNIPPET: publish
    }

    public static void main(String args[]) throws Exception {
        new WebServer();
        System.out.println("Server ready...");

        Thread.sleep(5 * 60 * 5000);
        System.out.println("Server exiting");
        System.exit(0);
    }
}

WebServiceImpl class

@WebService(endpointInterface = "com.nextenders.services.InterfaceWebService",
serviceName = "sample")

public class WebServiceImpl  implements InterfaceWebService{    


    @Override
    public boolean doLogin(String username, String password) {
          //Here some business logic call
        return true;
    }


}

Now I am trying to call this webservice through below url:
http://192.168.0.76:9000/sample/services/doLogin?username=abc&password=abc

But I am getting wsdl xml structure. but I required only particular method result!!.
am I doing any wrong here? how can I pass parameter in CXF webservice?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T20:14:48+00:00Added an answer on June 15, 2026 at 8:14 pm

    The problem is solved. I forgot to put service name and try to access direct method.

    http://192.168.0.76:9000/sample/services/login_service/doLogin?username=abc&password=abc

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