我想将对象编组为XML。
但是，它失败，并带有以下异常：
javax.xml.bind.MarshalException
- with linked exception:
[com.sun.istack.SAXException2: unable to marshal type "FreightOfferDetail" as an element because it is missing an @XmlRootElement annotation]
at com.sun.xml.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:331)
at com.sun.xml.bind.v2.runtime.MarshallerImpl.marshal(MarshallerImpl.java:257)
at javax.xml.bind.helpers.AbstractMarshallerImpl.marshal(AbstractMarshallerImpl.java:96)
at com.wktransportservices.fx.test.util.jaxb.xmltransformer.ObjectTransformer.toXML(ObjectTransformer.java:27)
at com.wktransportservices.fx.test.sampler.webservice.connect.FreightOfferToConnectFreight.runTest(FreightOfferToConnectFreight.java:59)
at org.apache.jmeter.protocol.java.sampler.JavaSampler.sample(JavaSampler.java:191)
at org.apache.jmeter.threads.JMeterThread.process_sampler(JMeterThread.java:429)
at org.apache.jmeter.threads.JMeterThread.run(JMeterThread.java:257)
at java.lang.Thread.run(Thread.java:662)
Caused by: com.sun.istack.SAXException2: unable to marshal type "FreightOfferDetail" 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)
at com.sun.xml.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:328)
实际上，存在此注释（对于父类和交付的类）：
@XmlRootElement(name = "Freight_Offer")
@XmlAccessorType(XmlAccessType.FIELD)
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonInclude(JsonInclude.Include.NON_EMPTY)
public class FreightOffer {
@JsonIgnore
@XmlTransient
private String freightId;
private String id;
private String externalSystemId;
private AddressLocation pickUp;
private AddressLocation delivery;
private FreightDescription freightDescription;
private ListContacts contacts;
private Customer customer;
private ListSla slas;
private String pushId;
private CompanyProfile company;
private Route route;
private String href;
private Lifecycle lifecycle;
private Visibility visibility;
private Boolean unfoldedVXMatching;
// getters / setters
子班：
@XmlAccessorType(XmlAccessType.PROPERTY)
public class FreightOfferDetail extends FreightOffer {
private List<Contact> contact;
@XmlElement(name = "contacts")
@JsonProperty("contacts")
public List<Contact> getContact() {
return contact;
}
public void setContact(List<Contact> contact) {
this.contact = contact;
}
正是在这种方法下失败了toXML()：
public class ObjectTransformer<T> implements Transformer<T> {
protected final JAXBContext context;
protected final Marshaller marshaller;
protected final int okStatusCode = 200;
protected final String okSubErrorCode = "OK";
public ObjectTransformer(JAXBContext context) throws JAXBException {
this.context = context;
marshaller = context.createMarshaller();
marshaller.setProperty("jaxb.encoding", "UTF-8");
marshaller.setProperty("jaxb.formatted.output", Boolean.TRUE);
}
public String toXML(T object) throws JAXBException {
StringWriter writer = new StringWriter();
marshaller.marshal(object, writer);
String xmlOffer = writer.toString();
return xmlOffer;
}
它应该工作，但是不应该。
我在这里找不到遗漏或错误的内容。
更新：
这是测试的片段：
public SampleResult runTest(JavaSamplerContext context) {
AbstractSamplerResults results = new XMLSamplerResults(new SampleResult());
results.startAndPauseSampler();
if (failureCause != null) {
results.setExceptionFailure("FAILED TO INSTANTIATE connectTransformer", failureCause);
} else {
FreightOfferDTO offer = null;
FreightOffer freightOffer = null;
try {
results.resumeSampler();
RouteInfo routeDTO = SamplerUtils.getRandomRouteFromRepo(context.getIntParameter(ROUTES_TOUSE_KEY));
offer = FreightProvider.createRandomFreight(routeDTO, createUserWithLoginOnly(context));
freightOffer = connectTransformer.fromDTO(offer);
String xmlOfferString = connectTransformer.toXML(freightOffer); // <- it fails here.
我从CSV文件获取日期并将其转换为DTO对象。这种方法回到我身边FreightOfferDetail.
这是此方法的摘录：
public FreightOfferDetail freightFromDTO(FreightOfferDTO freightDTO, boolean fullFormat){
FreightOfferDetail freight = new FreightOfferDetail();
freight.setFreightId(freightDTO.getIds().getAtosId());
freight.setId(freightDTO.getIds().getFxId());
// ...
在这种情况下，如何将对象编组为XML文件？
public String toXML(T object) throws JAXBException {
StringWriter stringWriter = new StringWriter();
JAXBContext jaxbContext = JAXBContext.newInstance(T.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
// format the XML output
jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
QName qName = new QName("com.yourModel.t", "object");
JAXBElement<T> root = new JAXBElement<Bbb>(qName, T.class, object);
jaxbMarshaller.marshal(root, stringWriter);
String result = stringWriter.toString();
LOGGER.info(result);
return result;
}
这是我必须在没有@XmlRootElement的情况下进行编组/解组时使用的文章：http : //www.source4code.info/2013/07/jaxb-marshal-unmarshal-with-missing.html
最好的希望是它会有所帮助:)
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我来说两句