如何在Apache骆驼中保存肥皂播放文件？

Question

试图通过下面的代码将soap请求（jaxb有效载荷）保存为apache camel中的xml有效载荷，但失败了。

谁知道怎么做？

任何提示都将非常受欢迎！

码：

public class CamelRouteBuilder extends RouteBuilder {

    private static final String SOAP_ENDPOINT_URI = "cxf://http://localhost:{{soapEndpointPort}}/soap" +
        "?serviceClass=org.yw.springbootcamelesb.soap.CreateFileService";

    @Override
    public void configure() throws Exception {
        from(SOAP_ENDPOINT_URI).process(new CreateFileProcessor()).to("file:target/reports");;
    }
}

我得到的例外情况如下

pringbootcamelesb.soap.FileCreationStatus on：消息[ID-YandeMBP-1536989276152-2-2]。引发者：没有类型转换器可以从类型：org.yw.springbootcamelesb.soap.FileCreationStatus转换为所需类型：java.io.InputStream，其值为org.yw.springbootcamelesb.soap.FileCreationStatus@75ec37c5。交换[ID-YandeMBP-1536989276152-2-1]。引起：[org.apache.camel.NoTypeConversionAvailableException - 没有类型转换器可用于从类型：org.yw.springbootcamelesb.soap.FileCreationStatus转换为所需类型：java.io.InputStream，其值为org.yw.springbootcamelesb.soap。 FileCreationStatus @ 75ec37c5]在org.apache.camel.impl.MessageSupport.getMandatoryBody（MessageSupport.java:117）〜[camel-core-2.22.0.jar：2.22.0] org.apache.camel.component.file。 FileOperations.storeFile（FileOperations.java:333）〜[camel-core-2.22.0.jar：2.22.0] ...省略了40个常用帧

Answer 1

它适用于以下代码

        JaxbDataFormat xmlDataFormat = new JaxbDataFormat();
    JAXBContext con = JAXBContext.newInstance(FileCreationStatus.class);
    xmlDataFormat.setContext(con);
    from(SOAP_ENDPOINT_URI).process(new CreateFileProcessor())
            .marshal(xmlDataFormat).to("file:target/output").unmarshal(xmlDataFormat);