Question

我刚开始使用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。

以下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

对于您的用例，您可能希望利用@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

您还可以利用xsi:type属性作为继承指标：

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

EclipseLink JAXB（MOXy）也有@XmlDescriminatorNode / @XmlDescriminatorValue扩展名：

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

Answer 2

@XmlRootElement(name = "root")
public class Root {
    ....

    @XmlElements({
        @XmlElement(type = ResponseA.class, name = "ResponseA"),
        @XmlElement(type = ResponseB.class, name = "ResponseB")})
    private BodyResponse child;

}

也许您的@XmlType(name = "ResponseX")课程需要Response。

jaxb - 如何从多态类创建XML

2 个答案: