【发布时间】:2015-02-11 06:16:48
【问题描述】:
我在 XStream 上乱搞以习惯使用它。我可以将我的 Person 变量转换为 XML 来给我这个
<list>
<Person>
<name>Mitch</name>
<age>17</age>
<adress>Yehaaa</adress>
<fav-hobbie>Programming</fav-hobbie>
</Person>
<Person>
<name>Ant</name>
<age>18</age>
<adress>Mitch's House</adress>
<fav-hobbie>Football</fav-hobbie>
</Person>
</list>
我想知道如何读取 XML 文件并从 xml 文件中创建一个包含姓名、地址、年龄和爱好的新人员变量?
这是我的代码
public class base {
static XStream xstream = new XStream(new DomDriver());
static NewPerson person1 = new NewPerson();
static NewPerson person2 = new NewPerson();
static List<NewPerson> persons = new ArrayList();
public base(){
}
public static void main(String[] args) throws FileNotFoundException{
persons.add(person1);
persons.add(person2);
person1.name = "Mitch";
person1.adress = "52 Hope Street";
person1.age = 17;
person1.hobbie = "Programming";
person2.name = "Ant";
person2.adress = "Mitch's House";
person2.age = 18;
person2.hobbie = "Football";
String str = "res/file.xml";
xstream.processAnnotations(NewPerson.class);
xstream.toXML(persons, System.out);
}
}
@XStreamAlias("Person")
class NewPerson {
@XStreamAlias("name")
String name;
@XStreamAlias("age")
int age;
@XStreamAlias("adress")
String adress;
@XStreamAlias("fav-hobbie")
String hobbie;
}
谁能提供任何示例代码来演示如何从 xml 文件创建一个新的 Person 变量
【问题讨论】:
-
阅读2 min tutorial怎么样?
-
我已经阅读了它,但无法让它工作