Question

我有不同的Response类用于不同的Rest调用。但他们都处理相同的Entity。最初我是因为没有告诉JaxB来自Response类的实体类。它是使用@XmlSeeAlso({Entity.class})解决的。现在我正在尝试修复Response中的根元素名称。类
的 Issue1：

@XmlRootElement(name="?")
@XmlSeeAlso({Entity.class})
public class FindAllResponse 
    extends ListResponse<Entity> 
{
    /**
     * serial version uid
     */
    private static final long serialVersionUID = xxx;
}

预期回应是 <Response> <Entities> <Entity></Entity> </Entities> <Response>
如何将<Response><Entities> root添加到上面的响应类中。
的 Issue2：

@XmlRootElement(name="?")
@XmlSeeAlso({Entity.class})
public class FindEntityResponse 
    extends Response<JobType> 
{


    @JsonUnwrapped
    public Entity getValue() {
        return super.getValue();
    }

}

在这种情况下，预期的响应没有任何根元素。预期响应xml只是从<Entity></Entity>开始使用@XmlTransient和问题1的@XmlRootElement尝试但没有运气。

Answer 1

您需要将Entity类包装在分别定义Response和Entities的根元素的类中。

像基本的东西：

// this class will represent your Response element and all attributes
@XmlRootElement(name="Response")
@XmlAccessorType(XmlAccessType.FIELD)
public class Response {
    @XmlElement(name="Entities")
    private Entities myEntities;

    public Response() {
        myEntities = new Entities();
    }

    public Entities getMyEntities() { return myEntities; }
    public void setMyEntities(Entities entities) { this.myEntities = entities; }
}


// this class will represent the collection of Entities
@XmlRootElement(name="Entities")
@XmlAccessorType(XmlAccessType.FIELD)
public class Entities {
    @XmlElement(name="Entity")
    private List<Entity> entities;

    public Entities() {
        entities = new ArrayList<Entity>();
    }

    public List<Entity> getEntities() { return this.entities; }
    public void setEntities(List<Entity> entities) { this.entities = entities; }
}

// a simple Entity class (I don't know what yours is)
@XmlRootElement(name="Entity")
@XmlAccessorType(XmlAccessType.FIELD)
public class Entity {
    @XmlAttribute(name="id")
    private String id;

    public Entity() {}
    public Entity(String id) { setId(id); }

    public String getId() { return id; }
    public void setId(String id) { this.id = id; }
}

最后，一个简单的测试：

public class Test {
    public static void main(String[] args) throws Exception {
        Response r = new Response();
        Entities e = new Entities();
        e.getEntities().add(new Entity("1"));
        e.getEntities().add(new Entity("2"));
        e.getEntities().add(new Entity("3"));
        r.setMyEntities(e);

        JAXBContext context = JAXBContext.newInstance(Response.class);
        Marshaller m = context.createMarshaller();
        m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

        m.marshal(r, System.out);
    }
}

产生以下输出：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Response>
    <Entities>
        <Entity id="1"/>
        <Entity id="2"/>
        <Entity id="3"/>
    </Entities>
</Response>

我认为这可以让你更接近你所追求的目标。

JAXB：使用根元素处理

1 个答案: