Trying to deploy JAX WS webservice on Tomcat 7.0.32

[Paulo Carvalho]

Hello

I wrote a simple JaxWS webservice.

@WebService
public class WSReport {

	
	@WebMethod(operationName = "buildReport")
	public boolean buildReport(@WebParam(name = "xml") final String xml) {
		return true;
	}
}

I created the sun-jaxws.xml:

<?xml version="1.0" encoding="UTF-8"?>
<endpoints
  xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime"
  version="2.0">
  <endpoint
      name="Generate"
      implementation="com.tst.WSReport"
      url-pattern="/report"/>
</endpoints>

And my web.xml file is:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, 
Inc.//DTD Web Application 2.3//EN"
"http://java.sun.com/j2ee/dtds/web-app_2_3.dtd">

<web-app>
    <listener>
        <listener-class>
        	com.sun.xml.ws.transport.http.servlet.WSServletContextListener
        </listener-class>
    </listener>
    <servlet>
        <servlet-name>report</servlet-name>
        <servlet-class>
        	com.sun.xml.ws.transport.http.servlet.WSServlet
        </servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>report</servlet-name>
        <url-pattern>/report</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>120</session-timeout>
    </session-config>
</web-app>

I have also put all the jar files of JAX WS RI on the lib folders of the web service.

I've build the war and it has the expected structure:

META-INF/
WEB-INF/
WEB-INF/classes/...
WEB-INF/sun-jaxws.xml
WEB-INF/web.xml

Now, I've put the war file on the webapps tomcat folder. Finally, I started tomcat. No errors happened.

However, when I try to access the application using the URL : http://localhost:8080/Generate/report, I get the following error on the browser:

HTTP Status 404 - /HelloWorld/hello
type Status report
message /Generate/report
description The requested resource (/Generate/report) is not available.

Anyone knows the cause of the problem?


Thank you

[Meghana Reddy]

Paulo Carvalho wrote: URL : http://localhost:8080/Generate/report

What is Generate in the above URL? That should be the name of the war file(in other words, the context root of the webapp) that you've deployed in tomcat.

Thanks

[Ulf Dittmer]

Well, the servlet listens to "/report", so any URL needs to start with that. If the endpoint also listens to "/report", the URL would be something like "/report/report". And then you need to prepend the name of the web app, unless this is part of the ROOT web app.