如何在使用Mule Payload将数据发布到Web Service后在客户端设置响应

Question

我有2个Mule流…第一个流公开了执行DB CRUD操作的SOAP Web服务…我的第一个流程是:-

<flow name="ServiceFlow" doc:name="ServiceFlow">
<http:inbound-endpoint exchange-pattern="request-response" host="localhost" port="8082" path="mainData" doc:name="HTTP"/>
<cxf:jaxws-service  serviceClass="com.test.services.schema.maindata.v1.MainData"  doc:name="SOAP"/>
<component class="com.test.services.schema.maindata.v1.Impl.MainDataImpl" doc:name="JavaMain_ServiceImpl"/>
</flow>

web服务的SOAP请求:-

<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:v1="http://services.test.com/schema/MainData/V1">
   <soapenv:Header/>
   <soapenv:Body>
      <v1:insertDataRequest>
         <v1:Id>477</v1:Id>
         <v1:Name>ttttt</v1:Name>
         <v1:Age>56</v1:Age>
         <v1:Designation>aaaaaa </v1:Designation>
      </v1:insertDataRequest>
   </soapenv:Body>
</soapenv:Envelope>

， SOAP响应是:-

<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
   <soap:Body>
      <insertDataResponse xmlns="http://services.test.com/schema/MainData/V1">
         <Response>Data inserted Successfully</Response>
         <Id>0</Id>
         <Age>0</Age>
      </insertDataResponse>
   </soap:Body>
</soap:Envelope> 

现在我有另一个流，这是一个客户端流到这个web服务....客户端流程为:-

    <flow name="ClientFlow" doc:name="ClientFlow">
            <http:inbound-endpoint exchange-pattern="request-response" host="localhost" port="8888" path="clientpath" doc:name="HTTP"/>
    <set-payload doc:name="Set Payload" value="#[import com.test.services.schema.maindata.v1.*; dRequest = new DataRequest();dRequest.id = 7;dRequest.age = 55;dRequest.name = 'aValue';dRequest.designation = 'hhhhh'; dRequest]"/>
     <async doc:name="Async">
    <mulexml:object-to-xml-transformer doc:name="Object to XML"/>
    <logger message="payload :- #[message.payload]" level="INFO" doc:name="Logger"/>
    </async>
    <cxf:jaxws-client doc:name="SOAP" serviceClass="com.test.services.schema.maindata.v1.MainData" operation="insertDataOperation" port="MainDataPort" />  
 <http:outbound-endpoint exchange-pattern="request-response" host="localhost" port="8082" path="mainData" doc:name="HTTP" method="POST"/>
</flow>

现在问题是……在这个客户端流程中，我正在设置来自Mule <set-payload的请求…服务工作正常，来自客户端的请求使用Http出站发布到主服务…但我从回应中得到了例外…我不知道如何设置这里的响应…我需要在http出站后再次使用设置有效载荷吗?如果是……那我怎么设置它??请帮忙……下面是我得到的例外:-

Exception stack is:
1. unable to marshal type "com.test.services.schema.maindata.v1.DataResponse" as an element because it is missing an @XmlRootElement annotation (com.sun.istack.SAXException2)
  com.sun.xml.bind.v2.runtime.XMLSerializer:244 (null)
2. null (javax.xml.bind.MarshalException)
  com.sun.xml.bind.v2.runtime.MarshallerImpl:328 (http://java.sun.com/j2ee/sdk_1.3/techdocs/api/javax/xml/bind/MarshalException.html)
3. failed to mashal objec tto XML (java.io.IOException)
  org.mule.module.xml.transformer.jaxb.JAXBMarshallerTransformer$1:110 (null)
4. failed to mashal objec tto XML (java.io.IOException). Message payload is of type: HttpResponse (org.mule.execution.ResponseDispatchException)
  org.mule.transport.http.HttpMessageProcessTemplate:141 (http://www.mulesoft.org/docs/site/current3/apidocs/org/mule/execution/ResponseDispatchException.html)
--------------------------------------------------------------------------------
Root Exception stack trace:
com.sun.istack.SAXException2: unable to marshal type "com.test.services.schema.maindata.v1.DataResponse" as an element because it is missing an @XmlRootElement annotation
    at com.sun.xml.bind.v2.runtime.XMLSerializer.reportError(XMLSerializer.java:244)
    at com.sun.xml.bind.v2.runtime.ClassBeanInfoImpl.serializeRoot(ClassBeanInfoImpl.java:303)
    at com.sun.xml.bind.v2.runtime.XMLSerializer.childAsRoot(XMLSerializer.java:490)
    + 3 more (set debug level logging or '-Dmule.verbose.exceptions=true' for everything)

---

UPDATED FLOW:-我使用Java组件为客户端流设置响应，它工作正常，没有任何问题…:-

<flow name="ClientFlow" doc:name="ClientFlow">
<http:inbound-endpoint exchange-pattern="request-response" host="localhost" port="8888" path="aa" doc:name="HTTP"/>
<set-payload doc:name="Set Payload" value="#[import com.test.services.schema.maindata.v1.*; dRequest = new DataRequest();dRequest.id = 7;dRequest.age = 55;dRequest.name = 'aValue';dRequest.designation = 'hhhhh'; dRequest]"/>
<http:outbound-endpoint exchange-pattern="request-response" host="localhost" port="8082" path="mainData" doc:name="HTTP">
<cxf:jaxws-client doc:name="SOAP" serviceClass="com.test.services.schema.maindata.v1.MainData" operation="insertDataOperation" port="MainDataPort" />
</http:outbound-endpoint>
<custom-transformer class="com.test.request.ResponseTransformer" doc:name="JavaTransformerForResponse"/> <!-- Response Java Class -->
<mulexml:object-to-xml-transformer doc:name="Object to XML"/>
<logger message="#[message.payload]" level="INFO" doc:name="JSON Logging"/>
</flow>

现在工作正常，日志中的响应是:-

Now Entering Method:: com.test.services.schema.maindata.v1.Impl.MainDataImpl.insertDataOperation() *****
Data inserted Successfully
[2014-07-31 13:06:23,150] [INFO ] [[SOAPHeaderInterceptor].connector.http.mule.default.receiver.04] <com.test.services.schema.maindata.v1.DataResponse>
  <response>Data inserted Successfully</response>
  <id>0</id>
  <age>0</age>
</com.test.services.schema.maindata.v1.DataResponse>

但是我不想使用Java组件的响应…请建议如何使用setpayload或Expression在Mule…

Answer 1

您可以删除custom-transformer，因为它确实没有为您做任何有用的事情。

我相信原文:

无法封送类型"com.test.services.schema.maindata.v1 "。DataResponse作为元素，因为它缺少@XmlRootElement注释

问题是由于Mule需要一种方法将DataResponse对象从cxf:jaxws-client封送到可以通过HTTP返回给调用者的形式。

mulexml:object-to-xml-transformer的添加通过提示Mule如何封送解决了这个问题(即不要使用JAXB，因为它根本无法工作，而是使用XStream)。

Answer 2

因此根据David的建议，最终解决方案是使用object-to-json-transformer或object-to-xml-transformer或object-to-string-transformer任何一个都可以在这里用于封送响应对象