如何使用Jaxb从XML获取属性？

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Unmarshaller;
import java.io.File;

public class JAXBExample {
    public static void main(String[] args) {
        try {
            // 初始化JAXB上下文
            JAXBContext jaxbContext = JAXBContext.newInstance(YourClass.class);

            // 创建Unmarshaller对象
            Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();

            // 从XML文件中读取数据并转换为Java对象
            YourClass yourObject = (YourClass) unmarshaller.unmarshal(new File("your_xml_file.xml"));

            // 获取属性的值
            String attributeValue = yourObject.getAttribute();

            System.out.println("Attribute value: " + attributeValue);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}