WebServiceTransportException: Not Found [404]

Question

Attualmente sto implementando un servizio Web Spring utilizzando jaxb. Ma quando provo a consumare il servizio web creato si incontra un errore WebServiceTransportException: Not Found [404]. Ho provato a cercare in rete ma non sono riuscito a trovare una possibile causa principale. Di seguito ho mostrato i miei codici sorgente.

applicazione context.xml

<bean 
    class="org.springframework.ws.server.endpoint.adapter.GenericMarshallingMethodEndpointAdapter"> 
    <constructor-arg ref="marshaller" /> 
</bean> 

<bean id="marshaller" class="org.springframework.oxm.jaxb.Jaxb2Marshaller"> 
    <property name="classesToBeBound"> 
     <list> 
      <value>com.ph.domain.EightBallRequest</value> 
      <value>com.ph.domain.EightBallResponse</value> 
     </list> 
    </property> 
</bean> 

<bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver"> 
    <property name="prefix"> 
     <value>/jsp/</value> 
    </property> 
    <property name="suffix"> 
     <value>.jsp</value> 
    </property> 
</bean> 

<bean id="simpleUrlHandlerMapping" 
    class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping" 
    lazy-init="true"> 
    <property name="mappings"> 
     <props> 
      <prop key="/test.asp">LandingController</prop> 
     </props> 
    </property> 
</bean>  

<bean name="LandingController" class="com.ph.controller.LandingController"> 
    <property name="stub" ref="eightBallClient"/> 
</bean> 

Client per webservice

public class EightBallClient extends WebServiceGatewaySupport { 

private Resource request; 

public void setRequest(Resource request) { 
    this.request = request; 
} 

public String AskQuestion(String question) throws IOException { 
    String responseString = null; 

    EightBallRequest request = new EightBallRequest(); 
    request.setQuestion(question); 

    EightBallResponse response = new EightBallResponse(); 

    response = (EightBallResponse) getWebServiceTemplate() 
      .marshalSendAndReceive(request); 
    responseString = response.getAnswer().toString(); 
    return responseString; 
} 
} 

definizione del mio servizio web

<bean id="schema" class="org.springframework.xml.xsd.SimpleXsdSchema"> 
    <property name="xsd" value="/WEB-INF/eightball.xsd" /> 
</bean> 

E sotto è la stack errori:

SEVERE: Servlet.service() for servlet dispatcher threw exception 
org.springframework.ws.client.WebServiceTransportException: Not Found [404] 
    at org.springframework.ws.client.core.WebServiceTemplate.handleError(WebServiceTemplate.java:626) 
    at org.springframework.ws.client.core.WebServiceTemplate.doSendAndReceive(WebServiceTemplate.java:550) 
    at org.springframework.ws.client.core.WebServiceTemplate.sendAndReceive(WebServiceTemplate.java:501) 
    at org.springframework.ws.client.core.WebServiceTemplate.marshalSendAndReceive(WebServiceTemplate.java:350) 
    at org.springframework.ws.client.core.WebServiceTemplate.marshalSendAndReceive(WebServiceTemplate.java:344) 
    at org.springframework.ws.client.core.WebServiceTemplate.marshalSendAndReceive(WebServiceTemplate.java:336) 

Answer 1

Forse il vostro URI:

<bean name="webserviceTemplate" 
class="org.springframework.ws.client.core.WebServiceTemplate"> 
    <property name="defaultUri" value="http://localhost:8080/mywebservice" /> 

Controllare questo valore:

"http://mylocal:8080/mywebservice"

Answer 2

Qui è come ho risolto questo errore:

1. Dichiarare un SoapActionCallback.

2. Utilizzare questa richiamata nel marshalSendAndReceive() come segue.

   final EightBallResponse response = new EightBallResponse(); 
   final SoapActionCallback soapActionCallback = new SoapActionCallback("<the operation name as defined in the WSDL>"); 
   response = (EightBallResponse) getWebServiceTemplate() 
       .marshalSendAndReceive(request, soapActionCallback); 
   responseString = response.getAnswer().toString(); 
   

Answer 3

Nel mio caso, la soluzione è stata di prestare attenzione al caso in l'URI. Ho avuto tutto in minuscolo, ma il webservice si aspettava un nome di azione CamelCase.