Question

我需要编写一个消息处理程序。我正在学习本教程https://www.mkyong.com/webservices/jax-ws/jax-ws-soap-handler-in-server-side/。

我创建了我的处理程序并将其用作我的服务实现类，如

@Component(value = "AccountValidationsoapImpl")
@WebService(endpointInterface = "com.abc.validation.api.IAccountValidationService")
@HandlerChain(file = "./handler/handler.xml")
public class AccountValidationSOAPImpl implements IAccountValidationService

我的handler.xml是

  <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
  <javaee:handler-chains 
       xmlns:javaee="http://java.sun.com/xml/ns/javaee" 
       xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <javaee:handler-chain>
      <javaee:handler>
        <javaee:handler-class>com.abc.validation.api.handler.WSHandler</javaee:handler-class>
      </javaee:handler>
    </javaee:handler-chain>
  </javaee:handler-chains>

我运行我的应用程序并发送SOAP请求，但我从未看到我的处理程序被调用。

我的handler.xml和WSHandler都存在于com.abc.validation.api.handler中而web服务类存在于com.abc.validation.api包中。我错过了什么？

我的WSHandler

    public class WSHandler implements SOAPHandler<SOAPMessageContext>{

    @Override
    public boolean handleMessage(SOAPMessageContext context) {

      System.out.println("Server : handleMessage()......");

      return true;
    }

    @Override
    public boolean handleFault(SOAPMessageContext context) {

      System.out.println("Server : handleFault()......");

      return true;
    }

    @Override
    public void close(MessageContext context) {
      System.out.println("Server : close()......");
    }

    @Override
    public Set<QName> getHeaders() {
      System.out.println("Server : getHeaders()......");
      return null;
    }

    private void generateSOAPErrMessage(SOAPMessage msg, String reason) {
      try {
        SOAPBody soapBody = msg.getSOAPPart().getEnvelope().getBody();
        SOAPFault soapFault = soapBody.addFault();
        soapFault.setFaultString(reason);
        throw new SOAPFaultException(soapFault);
      }
      catch(SOAPException e) { }
    }

  }

在服务器端实现SOAP消息处理程序

0 个答案: