CODEWITHSUNDEEP
javaweb-servicestomcatjax-ws
tsukanomon
tsukanomon

Reputation: 1270

How to get the URL of my WSDL?

I have created a very simple web service using Netbeans, Java EE6, JAX-WS and TomCat. It just have one @WebMethod getWsdlURL(), which is supposed to return the URL of my wsdl, and it should be something similar to:

http://192.168.70.44:8088/SimpleWebService/WebService?wsdl

However I don't have any idea on how to do that.

Can anyone help me?

Upvotes: 6

Views: 67404

Answers (3)

VGR
VGR

Reputation: 44338

If you are generating classes from a WSDL when you build, your service class will extend javax.xml.ws.Service and therefore will inherit the getWSDLDocumentLocation method.

Upvotes: 1

David
David

Reputation: 20063

Don't add your own service to return the URL. Depend on the runtime, that already supplies it.

Just open the url in a browser. The url would be....

http://<hostname>:<port>/<webappname>/<servletEndpoint>?wsdl

Upvotes: 10

Ujjwal
Ujjwal

Reputation: 613

To add to answer of david99world, you can look for the endpoint url-pattern in sun-jaxws.xml configuration file present in WEB-INF dir.

The file may have a definition like:

<endpoints xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime" version="2.0">

    <endpoint name="ws/MyService" implementation="com.test.ws.services.MyService"
        url-pattern="/ws/MyService" enable-mtom="false" />

</endpoints>

Upvotes: 2

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