jaxb - 如何从多态类创建XML

Question

我刚刚开始使用JAXB从java对象生成XML输出.我的java类中存在一个多态性,它似乎在JAXB中不起作用.

下面是我尝试处理它的方式,但在输出中我没有预期字段:fieldA或fieldB.

@XmlRootElement(name = "root")
public class Root {
    @XmlElement(name = "fieldInRoot")
    private String fieldInRoot;
    @XmlElement(name = "child")
    private BodyResponse child;
    // + getters and setters
}

public abstract class BodyResponse {
}

@XmlRootElement(name = "ResponseA")
public class ResponseA extends BodyResponse {
    @XmlElement(name = "fieldA")
    String fieldB;
    // + getters and setters
}

@XmlRootElement(name = "ResponseB")
public class ResponseB extends BodyResponse {
    @XmlElement(name = "fieldB")
    String fieldB;  
    // + getters and setters  
}

在我开始发明一些错综复杂的遗产之前,有没有什么好方法可以做到这一点？

Answer 1

对于您可能想要利用的用例@XmlElementRefs,这对应于XML Schema中替换组的概念:

@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class Root {
    @XmlElement
    private String fieldInRoot;
    @XmlElementRef
    private BodyResponse child;
    // + getters and setters
}

• http://blog.bdoughan.com/2010/11/jaxb-and-inheritance-using-substitution.html

您还可以将该xsi:type属性用作继承指示符:

• http://blog.bdoughan.com/2010/11/jaxb-and-inheritance-using-xsitype.html

EclipseLink JAXB(MOXy)也有@XmlDescriminatorNode/ @XmlDescriminatorValueextension:

• http://bdoughan.blogspot.com/2010/11/jaxb-and-inheritance-moxy-extension.html