我正在尝试编组实现公共接口的对象列表。 有3个类和1个接口:
社区类(有一种方法:列表< Person> getPeople(); )
人界面(有一种方法: String getName(); )
女孩类(实现人)
男孩类(实现人)
见下面的代码。
我想要一个看起来像这样的XML:
<community>
<people>
<girl>
<name>Jane</name>
</girl>
<boy>
<name>John</name>
</boy>
<girl>
<name>Jane</name>
</girl>
<boy>
<name>John</name>
</boy>
</people>
</community>
或可能:
<community>
<people>
<person>
<girl>
<name>Jane</name>
</girl>
</person>
<person>
<boy>
<name>John</name>
</boy>
</person>
</people>
</community>
到目前为止,我得到的是:
<community>
<people>
<person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="girl">
<name>Jane</name>
</person>
<person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="boy">
<name>John</name>
</person>
<person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="girl">
<name>Jane</name>
</person>
<person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="boy">
<name>John</name>
</person>
</people>
</community>
我意识到我可以将元素更改为其他内容,但我希望元素名称是Girl或Boy类中的名称。
可以这样做吗?感谢。
@XmlRootElement(name = "community")
public class Community {
private List<Person> people;
@XmlElementWrapper
@XmlElement(name="person")
public List<Person> getPeople() {
return people;
}
public Community() {
people = new ArrayList<Person>();
people.add(new Girl());
people.add(new Boy());
people.add(new Girl());
people.add(new Boy());
}
}
@XmlRootElement(name = "girl")
public class Girl implements Person {
@XmlElement
public String getName() {
return "Jane";
}
}
@XmlRootElement(name = "boy")
public class Boy implements Person {
@XmlElement
public String getName() {
return "John";
}
}
@XmlJavaTypeAdapter(AnyTypeAdapter.class)
public interface Person {
public String getName();
}
public class AnyTypeAdapter extends XmlAdapter<Object, Object> {
@Override
public Object marshal(Object v) throws Exception {
return v;
}
@Override
public Object unmarshal(Object v) throws Exception {
return v;
}
}
答案 0 :(得分:45)
对于这种情况,我建议使用@XmlElements。 @XmlElements用于表示选择的XML模式概念:
以下是您查找示例的方式:
@XmlElements({
@XmlElement(name="girl", type=Girl.class),
@XmlElement(name="boy", type=Boy.class)
})
@XmlElementWrapper
public List<Person> getPeople() {
return people;
}
@XmlElementRef对应于XML模式中替换组的概念。这就是为什么之前的答案要求将Person从接口更改为类。
答案 1 :(得分:12)
好的,如果您准备将Person从界面更改为抽象基类,那么您就是金色的。这是代码:
public class Main {
public static void main(String[] args) throws Exception {
Community community = new Community();
JAXBContext context = JAXBContext.newInstance(Community.class);
Marshaller marshaller = context.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.marshal(community, System.out);
}
}
@XmlRootElement(name = "community")
@XmlSeeAlso({Person.class})
public class Community {
private List<Person> people;
@XmlElementWrapper(name="people")
@XmlElementRef()
public List<Person> getPeople() {
return people;
}
public Community() {
people = new ArrayList<Person>();
people.add(new Girl());
people.add(new Boy());
people.add(new Girl());
people.add(new Boy());
}
}
@XmlRootElement(name="boy")
public class Boy extends Person {
public String getName() {
return "John";
}
}
@XmlRootElement(name="girl")
public class Girl extends Person {
public String getName() {
return "Jane";
}
}
@XmlRootElement(name = "person")
@XmlSeeAlso({Girl.class,Boy.class})
public abstract class Person {
@XmlElement(name="name")
public abstract String getName();
}
主要技巧是在社区列表中使用@XmlElementRef。这通过它的@XmlRootElement标识类的类型。您可能还对@XmlSeeAlso感兴趣,这有助于组织上下文声明。
输出
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<community>
<people>
<girl>
<name>Jane</name>
</girl>
<boy>
<name>John</name>
</boy>
<girl>
<name>Jane</name>
</girl>
<boy>
<name>John</name>
</boy>
</people>
</community>