How to change webservice url endpoint?

JavaWeb ServicesJbossJax Ws

Java Problem Overview


I generated a web-service client using JBoss utils (JAX-WS compatible) using Eclipse 'web service client from a wsdl'.

So, the only thing I provided was a url to a web-service WSDL.

Now, the web service provider tells me to change the "url of client endpoint application access" of the web-service.

What is it and how to change it?

Java Solutions


Solution 1 - Java

IMO, the provider is telling you to change the service endpoint (i.e. where to reach the web service), not the client endpoint (I don't understand what this could be). To change the service endpoint, you basically have two options.

Use the Binding Provider to set the endpoint URL

The first option is to change the BindingProvider.ENDPOINT_ADDRESS_PROPERTY property value of the BindingProvider (every proxy implements javax.xml.ws.BindingProvider interface):

...
EchoService service = new EchoService();
Echo port = service.getEchoPort();

/* Set NEW Endpoint Location */
String endpointURL = "http://NEW_ENDPOINT_URL";
BindingProvider bp = (BindingProvider)port;
bp.getRequestContext().put(BindingProvider.ENDPOINT_ADDRESS_PROPERTY, endpointURL);

System.out.println("Server said: " + echo.echo(args[0]));
...

The drawback is that this only works when the original WSDL is still accessible. Not recommended.

Use the WSDL to get the endpoint URL

The second option is to get the endpoint URL from the WSDL.

...
URL newEndpoint = new URL("NEW_ENDPOINT_URL");
QName qname = new QName("http://ws.mycompany.tld","EchoService"); 

EchoService service = new EchoService(newEndpoint, qname);
Echo port = service.getEchoPort();

System.out.println("Server said: " + echo.echo(args[0]));
...

Solution 2 - Java

To add some clarification here, when you create your service, the service class uses the default 'wsdlLocation', which was inserted into it when the class was built from the wsdl. So if you have a service class called SomeService, and you create an instance like this:

SomeService someService = new SomeService();

If you look inside SomeService, you will see that the constructor looks like this:

public SomeService() {
        super(__getWsdlLocation(), SOMESERVICE_QNAME);
}

So if you want it to point to another URL, you just use the constructor that takes a URL argument (there are 6 constructors for setting qname and features as well). For example, if you have set up a local TCP/IP monitor that is listening on port 9999, and you want to redirect to that URL:

URL newWsdlLocation = new URL("http://theServerName:9999/somePath");
SomeService someService = new SomeService(newWsdlLocation);

and that will call this constructor inside the service:

public SomeService(URL wsdlLocation) {
    super(wsdlLocation, SOMESERVICE_QNAME);
}

Solution 3 - Java

I wouldn't go so far as @Femi to change the existing address property. You can add new services to the definitions section easily.

<wsdl:service name="serviceMethodName_2">
  <wsdl:port binding="tns:serviceMethodNameSoapBinding" name="serviceMethodName">
    <soap:address location="http://new_end_point_adress"/>
  </wsdl:port>
</wsdl:service>

This doesn't require a recompile of the WSDL to Java and making updates isn't any more difficult than if you used the BindingProvider option (which didn't work for me btw).

Solution 4 - Java

To change the end address property edit your wsdl file

<wsdl:definitions.......
  <wsdl:service name="serviceMethodName">
    <wsdl:port binding="tns:serviceMethodNameSoapBinding" name="serviceMethodName">
      <soap:address location="http://service_end_point_adress"/>
    </wsdl:port>
  </wsdl:service>
</wsdl:definitions>

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionEugenePView Question on Stackoverflow
Solution 1 - JavaPascal ThiventView Answer on Stackoverflow
Solution 2 - JavaMattCView Answer on Stackoverflow
Solution 3 - JavaTastyWheatView Answer on Stackoverflow
Solution 4 - JavaFemiView Answer on Stackoverflow