Question

我有一个像这样的xml文件：

<passwords>
<use>
    <title>LoL</title>
    <username>hallo</username>
    <password>asdasd</password>
    <note>asdasdasdasd</note>
</use>
<E-Mail>
    <use>
        <title>GMail</title>
        <username>hallo</username>
        <password>asdasd</password>
        <note>asdasdasdasd</note>
    </use>
    <Webmail>
        <use>
            <title>Yahoo</title>
            <username>hallo</username>
            <password>asdasd</password>
            <note>asdasdasdasd</note>
        </use>
    </Webmail>
</E-Mail>

类别的类

public class Category {
    private ArrayList<Usage> usages;
    private ArrayList<Category> categories; 
    private String name;

    public Category(String name){
        this.name = name;
        usages = new ArrayList<>();
        categories = new ArrayList<>();
    }
    public void addUsage(Usage usage){
        usages.add(usage);
    }
    public void addCategory(Category category){
       categories.add(category);
    }
}

和使用

public class Usage {
private String title;
private String username;
private String password;
private String notes;


public void setTitle(String title) {
    this.title = title;
}
public void setUsername(String username) {
    this.username = username;
}
public void setPassword(String password) {
    this.password = password;
}
public void setNotes(String notes) {
    this.notes = notes;
}


public String getUsername() {
    return username;
}
public String getPassword() {
    return password;
}
public String getNotes() {
    return notes;
}
public String getTitle(){
    return title;
}

public String toString(){
    return "[[" + this.title + "][" + this.username + "][" + this.password + "][" + this.notes + "]]";
}
}

我想在arraylist中使用其他类别和用法获得1个类别对象，但我无法弄清楚如何使用标准解析器进行分析。

我知道如何只使用一个类别而不使用子类别。

Answer 1

如上所述，您可以使用JAXB，这是一个非常基本的示例，只是为了让您了解使用这个有用的工具是多么容易：

首先，如果将XML转换为模式会更好，您可以使用：

http://www.freeformatter.com/xsd-generator.html

但是如果你想和你的班级一起工作，那么你可以使用注释，例如，这是一个基本类：

@XmlRootElement(name = "programmer")
public class ProgrammerBean implements Serializable {
 //Here attributes and getters and setters.
}

我将声明一个用于管理我的类的单例类：

public class XMLFileHandler {

    private static XMLFileHandler instance;

    private XMLFileHandler() {
    }

    public static XMLFileHandler getInstance() {
        if (instance == null) {
            instance = new XMLFileHandler();
        }
        return instance;
    }

    public void writeXml(String filePath, Object targetObject)
            throws FileNotFoundException, JAXBException {
        File xmlFile = new File(filePath);
        OutputStream xmlOutput = new FileOutputStream(xmlFile);
        // Define a JAXBContext
        JAXBContext context = null;
        // When I write I use a marshaller
        Marshaller xmlWriter = null;

        // I set the class of the objet to write
        context = JAXBContext.newInstance(targetObject.getClass());
        // I create the marshaller
        xmlWriter = context.createMarshaller();
        // I set the properties for the output
        xmlWriter.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        // Finally write
        xmlWriter.marshal(targetObject, xmlOutput);

    }

    public Object readXml(String filePath, Class objType) throws JAXBException {
        Object read = null;
        JAXBContext context = null;
        Unmarshaller xmlReader = null;

        context = JAXBContext.newInstance(objType);
        xmlReader = context.createUnmarshaller();
        read = xmlReader.unmarshal(new File(filePath));
        return read;
    }
}

然后，您可以使用：

XmlFileHandler xmlHandler = XmlFileHandler.getInstance();
ProgrammerBean myObject = (ProgrammerBean) xmlHandler.readXml("/home/developer/programmer.xml", ProgrammerBean.class);

希望它能帮助您了解JaxB的基础知识，如上所述，如果您使用模式并从中生成jaxb类会更容易，那么它只是set和get方法的问题。最好的问候。

java xml到Object

1 个答案: