Apache CXF and tomcat

Question

I am fairly new to Apache CXF and tomcat. I am trying to build a simple web service and deploy it on tomcat. below is my web.xml However when I try to access the 'services' folder using my browser it says No services have been found. I tried creating java web service client but it is not able to locate the service either. What could be wrong in this?

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
    <display-name>Sample web service provider</display-name>
    <listener>
        <!-- For Metro, use this listener-class instead: 
             com.sun.xml.ws.transport.http.servlet.WSServletContextListener -->
        <listener-class>
              org.springframework.web.context.ContextLoaderListener
        </listener-class>
    </listener>
    <!-- Remove below context-param element if using Metro -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
              classpath:META-INF/cxf/cxf.xml
        </param-value>
    </context-param>
    <servlet>
        <servlet-name>WebServicePort</servlet-name>
        <!-- For Metro, use this servlet-class instead: 
             com.sun.xml.ws.transport.http.servlet.WSServlet  -->
        <servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>WebServicePort</servlet-name>
        <url-pattern>/services/*</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>60</session-timeout>
    </session-config>
</web-app>

Answer 1

This means that you don't have any services exposed in your application. Your web.xml seems to be correct but I've just missed one thing, your Spring configuration. Add your Spring config location in your web.xml, for e.g.:

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>WEB-INF/applicationContext.xml</param-value>
</context-param>

Also, you have to create a class which will implement your web service interface and expose it as the CXF endpoint in your Spring applicationContext.xml configuration file. For e.g.:

<bean id="candidateImpl" class="some.pckg.CandidateImpl"/>

<jaxws:endpoint id="candidateEndpoint"
                implementor="#candidateImpl"
                address="/Candidate"
        />

Your CandidateImpl class should have @WebService annotation. For e.g.:

@WebService(targetNamespace = "http://something.com/ws/candidate",
        portName = "CandidateService",
        serviceName = "Candidate",
        endpointInterface = "some.pckg.types.CandidateService",
        wsdlLocation = "WEB-INF/wsdl/CandidateService.wsdl")
public class CandidateImpl implements CandidateService {
     //Implementation of all methods from CandidateService.
}

If you've done everything correctly you should see that there is one service available under:

http(s)://whateverhost.com:<somePort>/SomeContextPath/services

And you should be able to get the WSDL file like this:

http(s)://whateverhost.com:<somePort>/SomeContextPath/services/Candidate?wsdl

See also:

• Writing a Web Service with Spring

Answer 2

You need to set the spring configuration file location to make this work. You can set it as follows.

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>WEB-INF/applicationContext.xml</param-value>
</context-param>