如何使用下面的代码解码 XML 字符串并将其映射到下面的 JAXB 对象?
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;
}
}
最佳答案
要传递 XML 内容,您需要将内容包装在 Reader
中,然后将其解码:
JAXBContext jaxbContext = JAXBContext.newInstance(Person.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
StringReader reader = new StringReader("xml string here");
Person person = (Person) unmarshaller.unmarshal(reader);
关于java - 使用 JAXB 从 XML 字符串创建对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5458833/