Question

我需要创建具有以下结构的XML文件：

*@XmlPath("/Person/@ID")*

我已经在Stack Overflow上尝试了所有解决方案，使用*@XmlAttribute(name="ID")*，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!"); } }以不同方式使用，但没有成功。我的代码在根节点中创建属性或在 Person 节点之前创建其他节点。但属性应该在 Person 节点

中

Main.class中的方法：

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

Person.class具有变量声明，更改和获取它们的方法以及构造函数：

@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;
    }
 }

PersonListXMLWrapper.class：

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Answer 1

如果您可以使用其他XML解析器，我建议使用Scilca XML Progression编写一个简单的XML文件（验证它）。

 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。我只需要在Person.class中添加*@XmlAttribute(name="ID")*方法声明，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>

JAXB：为内部元素创建属性

2 个答案: