JAXB：为内部元素创建属性

Question

我需要创建具有以下结构的XML文件：

<Persons>
   <Person ID="1">
      <Name></Name>
      ...
   </Person>
   <Person ID="2">
      <Name></Name>
      ...
   </Person>
   ...
</Persons>

我已经在Stack Overflow上尝试了所有解决方案，使用*@XmlPath("/Person/@ID")* ， *@XmlAttribute(name="ID")*并没有成功。 我的代码在根节点中创建属性，或者在“ 人”节点之前创建其他节点。 但是属性应该在Person节点内

Main.class中的方法：

public void savePersonDataToFile(File file) {
    try {
        JAXBContext context = JAXBContext
                .newInstance(PersonListXMLWrapper.class);
        Marshaller m = context.createMarshaller();
        m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

        // Wrapping our person data.
        PersonListXMLWrapper wrapper = new PersonListXMLWrapper();
        wrapper.setPersons(personData);

        // Marshalling and saving XML to the file.
        m.marshal(wrapper, file);

        // Save the file path to the registry.
        setPersonFilePath(file);
    } catch (Exception e) { // catches ANY exception
        Alert alert = new Alert(AlertType.ERROR);
        alert.setTitle("Error");
        alert.setHeaderText("Cant save to file:\n" + file.getPath());
        alert.setContentText("Ooops, there was an error!");
    }
}

Person.class具有变量声明，更改和获取它们的方法以及构造函数：

public class Person {
    private final IntegerProperty ID;
    private final StringProperty name;
    private final StringProperty surname;
    private final StringProperty prevSurname;
    private final StringProperty patronymic;
    ...

PersonListXMLWrapper.class：

@XmlRootElement(name = "Persons")
public class PersonListXMLWrapper {

    private List<Person> persons;

    @XmlElement(name = "Person")
    public List<Person> getPersons() {
        return persons;
    }

    public void setPersons(List<Person> persons) {
        this.persons = persons;
    }


    @XmlPath("/Person/@IDAtt")
    public String IDAtt ="123";

    @XmlAttribute(name="ID")
    public String getID() {
        return IDAtt;
    }

    public void setID(String ID) {
        this.IDAtt = ID;
    }
 }

Answer 1

如果可以使用另一个XML解析器，我建议使用Scilca XML Progression编写一个简单的XML文件（对其进行验证）。

 import org.scilca.sxp.*; // The SXP package

 Node rootNode = new XMLNode("Persons");

 // First Persion
 Node p1 = Node.constructNode("<Person Id=\"1\"?");
 p1.add("Name").setValue("Jacob"); // Add a child element and set its value
 p1.add("Habit").setValue("Eating"); 

 rootNode.add(p1); // Add it to root node

 // Second Person
 Node p2 = Node.constructNode("<Person Id=\"2\">");
 p1.add("Name").setValue("Jacks");

 rootNode.add(p2);

 Document XmlDocument = new Document(rootNode);
 XmlWriter xw = XmlDocument.getWriter();
 xw.saveXml("D:/file.txt"); // text file to see if the file is correct

Answer 2

解决方案非常简单。 我只需要向Person.class中的方法声明中添加*@XmlAttribute(name="ID")*而不是PersonListXMLWrapper.class：

public class Person {
    private final StringProperty ID;
    private final StringProperty name;
    private final StringProperty surname;
    private final StringProperty prevSurname;
    private final StringProperty patronymic;
    ...

    @XmlAttribute(name="ID")
    public String getID() {
        return ID.get();
    }
    public void setID(String ID) {
        this.ID.set(ID);
    }

    public StringProperty IDProperty() {
        return ID;
    }

所以我得到以下结构：

<Persons>
    <Person ID="1">
        <name>Name1</name>
        <surname></surname>
        <prevSurname></prevSurname>
        <patronymic></patronymic>
        ...
    </Person>
    <Person ID="2">
        <name>Name2</name>
        <surname></surname>
        <prevSurname></prevSurname>
        <patronymic></patronymic>
        ...
    </Person>
    ...
</Persons>