Question

我有一个SEI

@WebService
@SOAPBinding
(   
    style = javax.jws.soap.SOAPBinding.Style.DOCUMENT,
    use = javax.jws.soap.SOAPBinding.Use.LITERAL
)
@HandlerChain(file = "securityhandler.xml")
public interface UserService
{
    @WebMethod
    public AuthenticateResponse authenticateUser(AuthenticateRequest request,@WebParam(header = true, mode = Mode.IN) ApplicationCredentials credential);
}

我正在使用wsgen生成服务器端类文件和wsdl。我注意到wsgen在我的请求（AuthenticateRequest）上生成了包装器

package com.ecourt.ws.jaxws;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlType;
import com.ecourt.ws.impl.request.AuthenticateRequest;

@XmlRootElement(name = "authenticateUser", namespace = "http://ws.ecourt.com/")
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "authenticateUser", namespace = "http://ws.ecourt.com/")
public class AuthenticateUser {

    @XmlElement(name = "arg0", namespace = "")
    private AuthenticateRequest arg0;

    /**
     * 
     * @return
     *     returns AuthenticateRequest
     */
    public AuthenticateRequest getArg0() {
        return this.arg0;
    }

    /**
     * 
     * @param arg0
     *     the value for the arg0 property
     */
    public void setArg0(AuthenticateRequest arg0) {
        this.arg0 = arg0;
    }

}

现在我使用LogicalHandler拦截SOAP Message并尝试将有效负载解组为AuthenticateRequest，它会引发异常：

javax.xml.ws.WebServiceException：javax.xml.bind.UnmarshalException：unexpected element（uri：“http://ws.ecourt.com/”，local：“authenticateUser”）。预期元素为＆lt; {} authenticateRequest＆gt;，＆lt; {} baseRequest＆gt;

我的处理程序的句柄消息实现是

@Override
    public boolean handleMessage(LogicalMessageContext messageContext) {
        try
        {
            System.out.println("in handle messgae of AuthenticationHandler");
            Boolean outboundProperty = (Boolean)
            messageContext.get (MessageContext.MESSAGE_OUTBOUND_PROPERTY);
            if(outboundProperty == false)
            {
                JAXBContext jc = JAXBContext.newInstance(com.ecourt.ws.impl.request.AuthenticateRequest.class);
                Object obj = messageContext.getMessage().getPayload(jc);
}
}
catch(Exception ex)
        {
            ex.printStackTrace();
return false;
        }
        return true;
    }

这是xsd

的一部分

<xs:element name="authenticateUser" type="tns:authenticateUser"/>


  <xs:complexType name="authenticateUser">
    <xs:sequence>
      <xs:element name="arg0" type="tns:authenticateRequest" minOccurs="0"/>
    </xs:sequence>
  </xs:complexType>

我只想将UnMarshall payLoad直接发送给AuthenticateRequest而不是WSGen生成的AuthenticateUser。有什么办法吗？任何帮助都将受到高度赞赏

Answer 1

这解决了，我需要使用LogicalMessage Class中的另一个重载方法： Source source = messageContext.getMessage（）。getPayload（）; 通过源代码，我们可以处理完整的Soap地址。

由于

在Logic_SAPHandler inJax-WS中解组的问题

1 个答案: