我使用JAXB创建了这样的XML文件:
-<persons>
-<person>
<active>Active</active>
<amountOwed>500 Galleons</amountOwed>
<email>harrypotter@hogwarts.edu</email>
<firstName>harry</firstName>
<lastName>potter</lastName>
<memberNum>1234</memberNum>
<school>Hogwarts</school>
<state>some state</state>
<yearJoined>1991</yearJoined>
</person>
</persons>
我想使用JAXB附加到这个文件:
-<persons>
-<person>
<active>Active</active>
<amountOwed>500 Galleons</amountOwed>
<email>harrypotter@hogwarts.edu</email>
<firstName>harry</firstName>
<lastName>potter</lastName>
<memberNum>1234</memberNum>
<school>Hogwarts</school>
<state>some state</state>
<yearJoined>1991</yearJoined>
</person>
<person>
<active>Inactive</active>
<amountOwed>123412362 Galleons</amountOwed>
<email>ronweasley@hogwarts.edu</email>
<firstName>ron</firstName>
<lastName>weasley</lastName>
<memberNum>2342</memberNum>
<school>hogwarts</school>
<state>some state</state>
<yearJoined>1991</yearJoined>
</person>
</persons>
我知道XML不适合记录数据,但我必须将它用于我的项目。我怎么能这样做?
答案 0 :(得分:0)
假设存在以下类
@XmlRootElement
public class Persons{
List<Person> person;
...getters and setters
}
public class Person{
...fields, getters and setters...
}
首先解组原始xml
JAXBContext context = JAXBContext.newInstance(Persons.class);
Unmarshaller unmarshaller = context.createUnmarshaller();
File f = new File("the original xml");
JAXBElement<Persons> personsElement = (JAXBElement<Persons>) unmarshaller.unmarshal(f);
然后你得到人物对象
Persons persons = personsElement.getValue();
然后。把你要追加的对象放在
Person newPerson = new Person();
... put values into newPerson
persons.getPerson().add(newPerson);
最后,你整理了它
Marshaller marshaller = context.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
marshaller.marshal(personsElement, the output Stream to your file);
你也可以在oracle的教程中找到很多例子