7

这是我的课程：

@XmlRootElement(name="Zoo")
class Zoo {
    //@XmlElementRef
    public Collection<? extends Animal> animals;
}

@XmlAccessorType(XmlAccessType.FIELD)
@XmlSeeAlso({Bird.class, Cat.class, Dog.class})
@XmlDiscriminatorNode("@type")
abstract class Animal {
    @XmlElement
    public String name; 
}

@XmlDiscriminatorValue("Bird")
@XmlRootElement(name="Bird")
class Bird extends Animal {
    @XmlElement
    public String wingSpan;
    @XmlElement
    public String preferredFood;
}

@XmlDiscriminatorValue("Cat")
@XmlRootElement(name="Cat")
class Cat extends Animal {
    @XmlElement
    public String favoriteToy;
}

@XmlDiscriminatorValue("Dog")
@XmlRootElement(name="Dog")
class Dog extends Animal {
    @XmlElement
    public String breed;
    @XmlElement
    public String leashColor;
}

这是序列化的 JSON：

   {
        "animals": [
            {
                "type": "Bird",
                "name": "bird-1",
                "wingSpan": "6 feets",
                "preferredFood": "food-1"
            },
            {
                "type": "Cat",
                "name": "cat-1",
                "favoriteToy": "toy-1"
            },
            {
                "type": "Dog",
                "name": "dog-1",
                "breed": "bread-1",
                "leashColor": "black"
            }
        ]
    }

这是反序列化程序代码：

public static <T> T Deserialize_Moxy(String jsonStr, Class<?>[] cl) throws JAXBException {
    InputStream is = new ByteArrayInputStream(jsonStr.getBytes());
    JAXBContext jc = JAXBContext.newInstance(cl);         
    Unmarshaller unmarshaller = jc.createUnmarshaller();

    // Marshal to JSON
    unmarshaller.setProperty(MarshallerProperties.MEDIA_TYPE, "application/json");
    unmarshaller.setProperty(MarshallerProperties.JSON_INCLUDE_ROOT, false);
    @SuppressWarnings("unchecked")
    T obj = (T)unmarshaller.unmarshal(is);
    return obj;
}

这是一个例外：

Exception in thread "main" javax.xml.bind.UnmarshalException
 - with linked exception:
[Exception [EclipseLink-25008] (Eclipse Persistence Services - 2.4.1.v20121003-ad44345): org.eclipse.persistence.exceptions.XMLMarshalException
Exception Description: A descriptor with default root element  was not found in the project]
    at org.eclipse.persistence.jaxb.JAXBUnmarshaller.handleXMLMarshalException(JAXBUnmarshaller.java:1014)
    at org.eclipse.persistence.jaxb.JAXBUnmarshaller.unmarshal(JAXBUnmarshaller.java:147)
    at com.bp.samples.json.generics.Foo.Deserialize_Moxy(Foo.java:271)
    at com.bp.samples.json.generics.Foo.main(Foo.java:111)
Caused by: Exception [EclipseLink-25008] (Eclipse Persistence Services - 2.4.1.v20121003-ad44345): org.eclipse.persistence.exceptions.XMLMarshalException
Exception Description: A descriptor with default root element  was not found in the project
    at org.eclipse.persistence.exceptions.XMLMarshalException.noDescriptorWithMatchingRootElement(XMLMarshalException.java:143)
    at org.eclipse.persistence.internal.oxm.record.SAXUnmarshallerHandler.startElement(SAXUnmarshallerHandler.java:222)
    at org.eclipse.persistence.internal.oxm.record.json.JSONReader.parseRoot(JSONReader.java:161)
    at org.eclipse.persistence.internal.oxm.record.json.JSONReader.parse(JSONReader.java:118)
    at org.eclipse.persistence.internal.oxm.record.SAXUnmarshaller.unmarshal(SAXUnmarshaller.java:827)
    at org.eclipse.persistence.internal.oxm.record.SAXUnmarshaller.unmarshal(SAXUnmarshaller.java:350)
    at org.eclipse.persistence.internal.oxm.record.SAXUnmarshaller.unmarshal(SAXUnmarshaller.java:334)
    at org.eclipse.persistence.oxm.XMLUnmarshaller.unmarshal(XMLUnmarshaller.java:407)
    at org.eclipse.persistence.jaxb.JAXBUnmarshaller.unmarshal(JAXBUnmarshaller.java:133)
    ... 2 more

还有一个关于序列化 JSON 的问题：有没有办法让 JSON 序列化程序发布“@type”而不是“type”。目前，它看起来像具有“类型”属性的对象。如果我们可以用“@”来装饰它，那么它更像是一个类型信息而不是一个属性会更明显。

谢谢，贝扎德

4

1 回答 1

19

下面是我对你的两个问题的回答：

问题 #1 - 例外

当您使用该MarshallerProperties.JSON_INCLUDE_ROOT属性关闭根元素时，您需要使用其中一种方法，该unmarshal方法需要一个Class参数来告诉 MOXy 您希望解组的对象类型。

StreamSource json = new StreamSource("src/forum14246033/input.json");
Zoo zoo = unmarshaller.unmarshal(json, Zoo.class).getValue();

问题2

还有一个关于序列化 JSON 的问题：有没有办法让 JSON 序列化程序发布“@type”而不是“type”。目前，它看起来像具有“类型”属性的对象。如果我们可以用“@”来装饰它，那么它更像是一个类型信息而不是一个属性会更明显。

@前缀表示字段/属性映射到 XML 属性。您可以使用该JAXBContextProperties.JSON_ATTRIBUTE_PREFIX属性指定前缀来限定映射到 XML 属性的数据。

properties.put(JAXBContextProperties.JSON_ATTRIBUTE_PREFIX, "@");

完整示例

演示

package forum14246033;

import java.util.*;
import javax.xml.bind.*;
import javax.xml.transform.stream.StreamSource;
import org.eclipse.persistence.jaxb.JAXBContextProperties;

public class Demo {

    public static void main(String[] args) throws Exception {
        Map<String, Object> properties = new HashMap<String, Object>(2);
        properties.put(JAXBContextProperties.MEDIA_TYPE, "application/json");
        properties.put(JAXBContextProperties.JSON_INCLUDE_ROOT, false);
        properties.put(JAXBContextProperties.JSON_ATTRIBUTE_PREFIX, "@");
        JAXBContext jc = JAXBContext.newInstance(new Class[] {Zoo.class}, properties);

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        StreamSource json = new StreamSource("src/forum14246033/input.json");
        Zoo zoo = unmarshaller.unmarshal(json, Zoo.class).getValue();

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(zoo, System.out);
    }

}

输入.json/输出

{
   "animals" : [ {
      "@type" : "Bird",
      "name" : "bird-1",
      "wingSpan" : "6 feets",
      "preferredFood" : "food-1"
   }, {
      "@type" : "Cat",
      "name" : "cat-1",
      "favoriteToy" : "toy-1"
   }, {
      "@type" : "Dog",
      "name" : "dog-1",
      "breed" : "bread-1",
      "leashColor" : "black"
   } ]
}

领域模型

我不建议在您的域模型中使用公共字段，但如果您这样做，您可以将元数据减少到以下内容：

动物园

import java.util.Collection;

class Zoo {
    public Collection<? extends Animal> animals;
}

动物

import javax.xml.bind.annotation.XmlSeeAlso;
import org.eclipse.persistence.oxm.annotations.XmlDiscriminatorNode;

@XmlSeeAlso({Bird.class, Cat.class, Dog.class})
@XmlDiscriminatorNode("@type")
abstract class Animal {

    public String name; 

}

鸟

import org.eclipse.persistence.oxm.annotations.XmlDiscriminatorValue;

@XmlDiscriminatorValue("Bird")
class Bird extends Animal {
    public String wingSpan;
    public String preferredFood;
}

jaxb.properties

要将 MOXy 指定为您的 JAXB (JSR-222) 提供程序，您需要包含一个jaxb.properties在与您的域模型相同的包中调用的文件，其中包含以下条目：

javax.xml.bind.context.factory=org.eclipse.persistence.jaxb.JAXBContextFactory
于 2013-01-09T21:05:49.987 回答