原文： https://howtodoinjava.com/jaxb/unmarshal-without-xmlrootelement/

很多时候，您将需要解组不具有诸如@XmlRootElement之类的 JAXB 注解的 Java 对象，并且不允许您对源代码进行任何更改。 当您使用旧代码或某些您没有源代码的客户端 jar 时，可能会发生这种情况。

1.不带 JAXB 注解的解组问题

在这种情况下，如果您尝试直接将 Java 对象解组为 XML ，则将出现类似的错误。

  1. javax.xml.bind.UnmarshalException: unexpected element (uri:"", local:"employee"). Expected elements are (none)
  2.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext.handleEvent(UnmarshallingContext.java:726)
  3.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportError(Loader.java:247)
  4.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportError(Loader.java:242)
  5.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportUnexpectedChildElement(Loader.java:109)
  6.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext$DefaultRootLoader.childElement(UnmarshallingContext.java:1131)
  7.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext._startElement(UnmarshallingContext.java:556)
  8.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext.startElement(UnmarshallingContext.java:538)
  9.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.SAXConnector.startElement(SAXConnector.java:153)
  10.     at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.startElement(AbstractSAXParser.java:509)
  11.     at com.sun.org.apache.xerces.internal.impl.XMLNSDocumentScannerImpl.scanStartElement(XMLNSDocumentScannerImpl.java:379)
  12.     at com.sun.org.apache.xerces.internal.impl.XMLNSDocumentScannerImpl$NSContentDriver.scanRootElementHook(XMLNSDocumentScannerImpl.java:605)
  13.     at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl$FragmentContentDriver.next(XMLDocumentFragmentScannerImpl.java:3138)
  14.     at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl$PrologDriver.next(XMLDocumentScannerImpl.java:880)
  15.     at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next(XMLDocumentScannerImpl.java:606)
  16.     at com.sun.org.apache.xerces.internal.impl.XMLNSDocumentScannerImpl.next(XMLNSDocumentScannerImpl.java:117)
  17.     at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(XMLDocumentFragmentScannerImpl.java:510)
  18.     at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:848)
  19.     at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:777)
  20.     at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:141)
  21.     at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1213)
  22.     at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(SAXParserImpl.java:649)
  23.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(UnmarshallerImpl.java:243)
  24.     at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal(UnmarshallerImpl.java:214)
  25.     at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:157)
  26.     at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:162)
  27.     at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:171)
  28.     at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:189)
  29.     at com.howtodoinjava.demo.JaxbExample.jaxbXmlFileToObject(JaxbExample.java:63)
  30.     at com.howtodoinjava.demo.JaxbExample.main(JaxbExample.java:21)

其中Employee.java类如下。 它没有任何@XmlRootElement这样的 JAXB 注解。

  1. package com.howtodoinjava.demo.model;
  3. import java.io.Serializable;
  5. public class Employee implements Serializable {
  7.     private static final long serialVersionUID = 1L;
  9.     private Integer id;
  10.     private String firstName;
  11.     private String lastName;
  12.     private Department department;
  14.     public Employee() {
  15.         super();
  16.     }
  18.     public Employee(int id, String fName, String lName, Department department) {
  19.         super();
  20.         this.id = id;
  21.         this.firstName = fName;
  22.         this.lastName = lName;
  23.         this.department = department;
  24.     }
  26.     //Getters and Setters
  28.     @Override
  29.     public String toString() {
  30.         return "Employee [id=" + id + ", firstName=" + firstName + ", lastName=" + lastName + ", department="
  31.                 + department + "]";
  32.     }
  33. }

2.不带 JAXB 注解的解组解决方案

缺少@XmlRootElement注解时，JAXB 无法为Employee对象构建JAXBElement实例。 因此，您必须在此帮助 JAXB 手动构建它。

2.1 语法

  1. /**
  2.   * Unmarshal XML data from the specified XML Source by <tt>declaredType</tt> and return the
  3.   * resulting content tree.
  4.   * @param  source       source the XML Source to unmarshal XML data from (providers are
  5.   *                        only required to support SAXSource, DOMSource, and StreamSource)
  6.   *
  7.   * @param  declaredType  appropriate JAXB mapped class to hold <tt>source</tt>'s xml root element
  8.   * 
  9.   * @return value           Java content rooted by JAXB Element
  10.   */
  11. public <T> JAXBElement<T> unmarshal( javax.xml.transform.Source source, Class<T> declaredType )
  12.         throws JAXBException;

例如：

  1. JAXBElement<Employee> jaxbElement = (JAXBElement<Employee>) jaxbUnmarshaller
  2.                     .unmarshal(new StreamSource(xmlFile), Employee.class);

2.2 解组

现在，让我们看看该解组代码的工作方式。

  1. <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
  2. <employee>
  3.     <department>
  4.         <id>101</id>
  5.         <name>IT</name>
  6.     </department>
  7.     <firstName>Lokesh</firstName>
  8.     <id>1</id>
  9.     <lastName>Gupta</lastName>
  10. </employee>
  1. package com.howtodoinjava.demo;
  3. import java.io.File;
  5. import javax.xml.bind.JAXBContext;
  6. import javax.xml.bind.JAXBElement;
  7. import javax.xml.bind.JAXBException;
  8. import javax.xml.bind.Unmarshaller;
  9. import javax.xml.transform.stream.StreamSource;
  11. import com.howtodoinjava.demo.model.Employee;
  13. public class JaxbExample 
  14. {
  15.     public static void main(String[] args) 
  16.     {
  17.         String fileName = "employee.xml";
  19.         jaxbXmlFileToObject(fileName);
  20.     }
  22.     private static void jaxbXmlFileToObject(String fileName) 
  23.     {
  24.         File xmlFile = new File(fileName);
  25.         JAXBContext jaxbContext;
  26.         try 
  27.         {
  28.             jaxbContext = JAXBContext.newInstance(Employee.class);
  29.             Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
  31.             JAXBElement<Employee> jaxbElement = (JAXBElement<Employee>) jaxbUnmarshaller
  32.                     .unmarshal(new StreamSource(xmlFile), Employee.class);
  34.             Employee employee = jaxbElement.getValue();
  36.             System.out.println(employee);
  37.         }
  38.         catch (JAXBException e) 
  39.         {
  40.             e.printStackTrace();
  41.         }
  42.     }
  43. }

程序输出：

  1. Employee [id=1, firstName=Lokesh, lastName=Gupta, department=Department [id=101, name=IT]]

将我的问题放在评论部分。

学习愉快！

参考： XmlRootElement Java 文档