java - 通过使用 JAXB 解码将 XML 转换为对象

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我想通过使用 JAXB 解码将 XML 转换为 java。

这是我的employee.xml。我尝试转换，但显示错误 javax.xml.bind.UnmarshalException

<?xml version="1.0" encoding="UTF-8"?>
<deptname name="Research">
 <employee>
    <eid>r-001</eid>
    <ename>Dinesh R</ename>
    <age>35</age>
    <deptcode>d1</deptcode>
    <deptname>Research</deptname>
    <salary>20000</salary>
</employee>
</deptname>
<deptname name="Sales">
<employee>
    <eid>s-001</eid>
    <ename>Kanmani S</ename>
    <age>35</age>
    <deptcode>d2</deptcode>
    <deptname>Sales</deptname>
    <salary>30000</salary>
   </employee>
  </deptname>
<deptname name="Delivery">
  <employee>
    <eid>d-001</eid>
    <ename>Kanchana M</ename>
    <age>35</age>
    <deptcode>d3</deptcode>
    <deptname>Delivery</deptname>
    <salary>20000</salary>
</employee>
  </deptname>

Answer 1

根据XML documentation ，您只能有一个根元素，因此您应该将 xml 更改为:

<?xml version="1.0" encoding="UTF-8"?>
<departments>
    <deptname name="Research">
        <employee>
            <eid>r-001</eid>
            <ename>Dinesh R</ename>
            <age>35</age>
            <deptcode>d1</deptcode>
            <deptname>Research</deptname>
            <salary>20000</salary>
        </employee>
    </deptname>
    <deptname name="Sales">
        <employee>
            <eid>s-001</eid>
            <ename>Kanmani S</ename>
            <age>35</age>
            <deptcode>d2</deptcode>
            <deptname>Sales</deptname>
            <salary>30000</salary>
        </employee>
    </deptname>
    <deptname name="Delivery">
        <employee>
            <eid>d-001</eid>
            <ename>Kanchana M</ename>
            <age>35</age>
            <deptcode>d3</deptcode>
            <deptname>Delivery</deptname>
            <salary>20000</salary>
        </employee>
    </deptname>
</departments>

那么你需要一个 Departments 类来阅读这篇文章:

@XmlRootElement(name="departments")
public class Departments {
  List<Department> deptname;

  // Getters and setters
}

测试用例是:

@Test
  public void testXML() {
    try {
      File file = new File("./src/main/resources/employee.xml");
      JAXBContext jaxbContext = JAXBContext.newInstance(Departments.class);

      Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
      Departments departments = (Departments) jaxbUnmarshaller.unmarshal(file);
      System.out.println(departments);
    } catch (JAXBException e) {
      e.printStackTrace();
    }
  }

其次，您没有“研究”或“销售”或“交付”属性，您有一个可以采用任何字符串值的属性“名称”，因此您需要更改实现:

public class Department {
  @XmlAttribute(name = "name")
  private String name;

  @XmlElement(name = "employee")
  private List<Employee> employee = new ArrayList<>();

  // Getters and setters
}