如何从 SOAP 逻辑处理程序中获取有效负载对象答案

【问题标题】：How to get payload object from SOAP Logical Handler如何从 SOAP 逻辑处理程序中获取有效负载对象
【发布时间】：2016-04-25 21:58:18
【问题描述】：

我正在学习如何在 JAX-WS SOAP Web 服务中创建逻辑处理程序。在这里，我正在尝试获取有效负载数据并希望将其打印以进行测试。但我遇到了问题。

这是我的LogincalHandler 代码

public class HelloMessengerLogicalHandler implements LogicalHandler<LogicalMessageContext> {

    public void close(MessageContext ctx) {
    }

    public boolean handleFault(LogicalMessageContext ctx) {
        return false;
    }

    public boolean handleMessage(LogicalMessageContext ctx) {
        Boolean outbound = (Boolean) ctx.get(MessageContext.MESSAGE_OUTBOUND_PROPERTY);
        if(outbound) {
            LogicalMessage message = ctx.getMessage();
            //Source payload = message.getPayload();
            JAXBContext jaxbContext;
            try {
                jaxbContext = JAXBContext.newInstance(Person.class);
                Object jaxbPayload = message.getPayload(jaxbContext);
                System.out.println(jaxbPayload);
            } catch (JAXBException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

        }

        return true;
    }
}

这是我的网络服务代码：

@WebService
@HandlerChain(file = "handler-chain.xml")
public class HelloMessenger {
    public Person getPerson(String name) {
        return new Person(name);
    }
}

这是我的Person 班级：

public class Person {

    // Default Constructor & Getters, Setters
    private String name;

    public Person(String name) {
        this.name = name;
    }
    @Override
    public String toString() {
        return "Person [name=" + name + "]";
    }
}

当我发布我的代码并访问网络服务时，我遇到了以下异常：

javax.xml.ws.WebServiceException: javax.xml.bind.MarshalException
 - with linked exception:
[com.sun.istack.internal.SAXParseException2; unexpected element (uri:"http://simple/", local:"getPersonResponse"). Expected elements are (none)]
    at com.sun.xml.internal.ws.handler.LogicalMessageImpl.getPayload(LogicalMessageImpl.java:121)
    at simple.HelloMessengerLogicalHandler.handleMessage(HelloMessengerLogicalHandler.java:31)
    at simple.HelloMessengerLogicalHandler.handleMessage(HelloMessengerLogicalHandler.java:1)
    at com.sun.xml.internal.ws.handler.HandlerProcessor.callHandleMessageReverse(HandlerProcessor.java:326)
    at com.sun.xml.internal.ws.handler.HandlerProcessor.callHandlersResponse(HandlerProcessor.java:197)
    at com.sun.xml.internal.ws.handler.ServerLogicalHandlerTube.callHandlersOnResponse(ServerLogicalHandlerTube.java:158)
    at com.sun.xml.internal.ws.handler.HandlerTube.processResponse(HandlerTube.java:149)
    at com.sun.xml.internal.ws.api.pipe.Fiber.__doRun(Fiber.java:636)
    at com.sun.xml.internal.ws.api.pipe.Fiber._doRun(Fiber.java:585)
    at com.sun.xml.internal.ws.api.pipe.Fiber.doRun(Fiber.java:570)
    at com.sun.xml.internal.ws.api.pipe.Fiber.runSync(Fiber.java:467)
    at com.sun.xml.internal.ws.server.WSEndpointImpl$2.process(WSEndpointImpl.java:299)
    at com.sun.xml.internal.ws.transport.http.HttpAdapter$HttpToolkit.handle(HttpAdapter.java:593)
    at com.sun.xml.internal.ws.transport.http.HttpAdapter.handle(HttpAdapter.java:244)
    at com.sun.xml.internal.ws.transport.http.server.WSHttpHandler.handleExchange(WSHttpHandler.java:95)
    at com.sun.xml.internal.ws.transport.http.server.WSHttpHandler.handle(WSHttpHandler.java:80)
    at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:77)
    at sun.net.httpserver.AuthFilter.doFilter(AuthFilter.java:83)
    at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:80)
    at sun.net.httpserver.ServerImpl$Exchange$LinkHandler.handle(ServerImpl.java:677)
    at com.sun.net.httpserver.Filter$Chain.doFilter(Filter.java:77)
    at sun.net.httpserver.ServerImpl$Exchange.run(ServerImpl.java:649)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
    at java.lang.Thread.run(Thread.java:745)

我是新手，我的理解是payload对象是Person，所以它应该在日志中显示person对象，但是我看到了这个错误。

请解释此错误表示什么以及如何在逻辑处理程序中获取有效负载数据？

【问题讨论】：

标签： java web-services soap jax-ws

【解决方案1】：

源负载 = message.getPayload();

然后使用 xmltransformation 或其他方式将 DomSource 对象转换为字符串，它将为您提供有效负载。

【讨论】：

【解决方案2】：

正如@curious所说，使用

Source payload = message.getPayload();

使用 xmltransformation 将 DomSource 对象转换为字符串的示例代码如下：

StringWriter sw = new StringWriter();
Transformer transformer = TransformerFactory.newInstance().newTransformer();
Result output = new StreamResult( sw );
transformer.transform( payload, output );
System.out.println("Response : "+sw.toString());

【讨论】：