Question

我正在尝试使用Eclipselink将输入JSON解组为JAXB对象。但是，当我尝试这样做时，我发现嵌套对象最终被设置为null。我可以尝试自己解组嵌套对象，它将一直工作，直到它必须解组另一个嵌套对象，然后再设置为null。

例如，参加这个课程：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "event", propOrder = {
"objectBs"
})
public class ObjectA
implements Serializable
{

    private final static long serialVersionUID = 56347348765454329L;
    @XmlElement(required = true)
    protected ObjectA.ObjectBs objectBs;

    public ObjectA.ObjectBs getObjectBs() {
        return objectBs;
    }

    public void setObjectBs(ObjectA.ObjectBs value) {
        this.objectBs = value;
    }

    public boolean isSetObjectBs() {
        return (this.objectBs!= null);
    }

    @XmlAccessorType(XmlAccessType.FIELD)
    @XmlType(name = "", propOrder = {
        "objectB"
    })
    public static class ObjectBs
        implements Serializable
    {

        private final static long serialVersionUID = 56347348765454329L;
        @XmlElement(required = true)
        protected List<ObjectB> objectB;

        public List<ObjectB> getObjectB() {
            if (objectB == null) {
                objectB = new ArrayList<ObjectB>();
            }
            return this.objectB;
        }

        public boolean isSetObjectB() {
            return ((this.objectB!= null)&&(!this.objectB.isEmpty()));
        }

        public void unsetObjectB() {
            this.objectB = null;
        }

    }
}

其中ObjectA有一个名为ObjectBs的对象，其中包含ObjectB列表。当我尝试解组此类时，ObjectA具有的任何其他字段将被正确填充，并且将创建ObjectBs对象，但ObjectB的列表将为空。但是，如果我自己只解构一个ObjectB，它将在填充它的字段时创建，直到它自己的嵌套对象为止。

这是我用来解组JSON的代码：

JAXBContext jc = JAXBContextFactory.createContext(
    "com.package1"
    + ":com.package2"
    + ":com.package3"
    , null);
Unmarshaller um = jc.createUnmarshaller();
um.setProperty("eclipselink.media-type", "application/json"); 
um.setProperty(MarshallerProperties.JSON_INCLUDE_ROOT, false);
ObjectA objA = unmarshaller.unmarshal(new StreamSource(
    new StringReader(json)), ObjectA.class).getValue();

一些例子JSON：

        {
                "objectBs": [
                    {
                        "a": "blahblah",
                        "b": 123456,
                        "c": "abc blah 123 blah",
                        "nestedObject": {
                            "evenMoreNestedObjects": {
                                "d": "blah",
                                "e": "blahblahblah",
                                "anotherNestedObject": {
                                    "x": 25,
                                    "y": 50
                                }
                        }}}]
        }

Answer 1

你没有提供你试图解组的JSON，但我做了一些逆向工程，下面是一个使用你在问题中发布的模型的例子：

import javax.xml.bind.*;
import javax.xml.transform.stream.StreamSource;
import org.eclipse.persistence.jaxb.MarshallerProperties;
import org.eclipse.persistence.jaxb.UnmarshallerProperties;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(ObjectA.class);

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        unmarshaller.setProperty(UnmarshallerProperties.MEDIA_TYPE, "application/json");
        unmarshaller.setProperty(UnmarshallerProperties.JSON_INCLUDE_ROOT, false);
        StreamSource json = new StreamSource("src/forum17866155/input.json");
        ObjectA objectA = unmarshaller.unmarshal(json, ObjectA.class).getValue();

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(MarshallerProperties.MEDIA_TYPE, "application/json");
        marshaller.setProperty(MarshallerProperties.JSON_INCLUDE_ROOT, false);
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(objectA, System.out);
    }

}

<强> input.json /输出

{
   "objectBs" : {
      "objectB" : [ {
      }, {
      } ]
   }
}

使用JAXB从JSON解组嵌套对象

1 个答案: