CODEWITHSUNDEEP
javasoapjax-ws
abishkar bhattarai
abishkar bhattarai

Reputation: 7651

Is not a valid service exception in JAX-WS

I am taking reference from http://www.mkyong.com/webservices/jax-ws/jax-ws-hello-world-example/

This is my HelloWorldClient class 

package WebService;


import java.net.URL;
import javax.xml.namespace.QName;
import javax.xml.ws.Service;



public class HelloWorldClient{

    public static void main(String[] args) throws Exception {

    URL url = new URL("http://localhost:8099/dummy1/dummy2?wsdl");

        //1st argument service URI, refer to wsdl document above
    //2nd argument is service name, refer to wsdl document above
        QName qname = new QName("http://localhost:8099/dummy1/dummy2?wsdl", "HelloWorldImplService");


        Service service = Service.create(url, qname);

        HelloWorld hello = service.getPort(HelloWorld.class);

        System.out.println(hello.getHelloWorldAsString("mkyong"));

    }

}

When running this class i am getting error from below line of code

Service service = Service.create(url, qname);

The error is 

Exception in thread "main" javax.xml.ws.WebServiceException: {http://localhost:8099/dummy1/dummy2?wsdl}HelloWorldImplService is not a valid service. Valid services are: {http://WebService/}HelloWorldImplService
    at com.sun.xml.internal.ws.client.WSServiceDelegate.<init>(WSServiceDelegate.java:220)
    at com.sun.xml.internal.ws.client.WSServiceDelegate.<init>(WSServiceDelegate.java:165)
    at com.sun.xml.internal.ws.spi.ProviderImpl.createServiceDelegate(ProviderImpl.java:93)
    at javax.xml.ws.Service.<init>(Service.java:56)
    at javax.xml.ws.Service.create(Service.java:680)
    at WebService.HelloWorldClient.main(HelloWorldClient.java:19)

In the reference example in HelloWorldClient class it has 

    QName qname = new QName("http://ws.mkyong.com/", "HelloWorldImplService");

In my case i have replaced it with 

    QName qname = new QName("http://localhost:8099/dummy1/dummy2?wsdl", "HelloWorldImplService");

I could not figure out where i have made mistake .When i run http://localhost:8099/dummy1/dummy2?wsdl it is working fine.But ,when i access from client i am getting above mentioned exception.Any help please ?

Upvotes: 4

Views: 15340

Answers (5)

Bobojonov Farruh
Bobojonov Farruh

Reputation: 1

I solved this problem. I created WebServiceClient and WebServices projects. And same files: WebServiceClient :: webservices.HelloWorld.java webservices.HelloWorldClient.java

WebServices :: 
webservices.HelloWorld.java
webservices.HelloWorldImpl.java
webservices.HelloWorldPublisher.java

I used NetBeans 8. In both project must have same name of package and 

QName qname = new QName("http://webservices/", "HelloWorldImplService");
in webservices.HelloWorldClient.java.
The end. It runs ! Sorry My english. (Bobojonov Farruh)

Upvotes: 0

Alex Henao
Alex Henao

Reputation: 31

The error message tells you what to fix :

Valid services are: {http://WebService/}HelloWorldImplService

for me the following was necessary: 

QName qname = new QName("http://WebService/" , "HelloWorldImplService");

Upvotes: 1

Alexander Belenov
Alexander Belenov

Reputation: 412

Here is my recipe to solve this issue:

1. run the publisher class written by Mkyong;

2. open the url (ex: http://localhost:8099/dummy1/dummy2?wsdl) in browser;

3. check if "targetNamespace" property in WSDL equals to the 1st argument in QName constructor. If it doesn't, set it from WSDL;

4. check if "name" property in WSDL equals to the 2nd argument in QName constructor. If it doesn't, set it from WSDL;

5. stop both the client and the publisher;

6. run the publisher;

7. run the client;

8. enjoy the result =)

Upvotes: 3

Paolo
Paolo

Reputation: 1691

Try to replace 

QName qname = new QName("http://localhost:8099/dummy1/dummy2?wsdl", "HelloWorldImplService");

with

QName qname = new QName("http://WebService/", "HelloWorldImplService");

Upvotes: 6

roxolid
roxolid

Reputation: 135

I haven't tried it, but I do believe that first argument in QName instantiation should be without that ?wsdl. You are asked for providing namespace, not the URI of WSDL document.

Upvotes: 0

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