使用 JAXB 从 XML 字符串创建对象

问题描述

如何使用以下代码解组 XML 字符串并将其映射到下面的 JAXB 对象?

How can I use the below code to unmarshal a XML string an map it to the JAXB object below?

JAXBContext jaxbContext = JAXBContext.newInstance(Person.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
Person person = (Person) unmarshaller.unmarshal("xml string here");

@XmlRootElement(name = "Person")
public class Person {
    @XmlElement(name = "First-Name")
    String firstName;
    @XmlElement(name = "Last-Name")
    String lastName;
    public String getFirstName() {
        return firstName;
    }
    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }
    public String getLastName() {
        return lastName;
    }
    public void setLastName(String lastName) {
        this.lastName = lastName;
    }
}

推荐答案

要传递 XML 内容，您需要将内容包装在 Reader 中，然后对其进行解组:

To pass XML content, you need to wrap the content in a Reader, and unmarshal that instead:

JAXBContext jaxbContext = JAXBContext.newInstance(Person.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();

StringReader reader = new StringReader("xml string here");
Person person = (Person) unmarshaller.unmarshal(reader);

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