FlexJson序列化和反序列化接口

Question

我遇到一个错误，试图反序列化我的数据结构，这是一个条目列表，每个项都实现了一个接口。此外，接口的一个字段是object，每个继承都将该对象视为不同的字段。

在这个问题上花了这么多时间之后，任何答案都将不胜感激。

，这是我收到的错误：

线程"main“中的异常:不能在flexjson.factories.BeanObjectFactory.instantiate(BeanObjectFactory.java:17) at flexjson.ObjectBinder.bind(ObjectBinder.java:86) at flexjson.ObjectBinder.bindIntoObject(ObjectBinder.java:139) at flexjson.factories.ClassLocatorObjectFactory.instantiate(ClassLocatorObjectFactory.java:38) at flexjson.ObjectBinder.bind(ObjectBinder.java:86)上将java.lang.String转换为java.util.Map在flexjson.ObjectBinder.bindIntoCollection(ObjectBinder.java:101) at flexjson.factories.ListObjectFactory.instantiate(ListObjectFactory.java:13) at flexjson.ObjectBinder.bind(ObjectBinder.java:86) at flexjson.ObjectBinder.bind(ObjectBinder.java:65) at flexjson.JSONDeserializer.deserialize(JSONDeserializer.java:158) at testSerizlizeDeserializeInterface.entryPointForTestingSerialize.main(entryPointForTestingSerialize.java:34)

我举了个例子如果有人也想玩的话.

1. 接口
2. The EPersonType
3. 继承
4. 主修班
5. 输出量

谢谢!

接口

    public interface IPerson {
        EPersonType getPersonType();
        String getName();
        void setName(String name);
            int getAge();
        void setAge(int age);
        Object getValue();
            void setValue(Object value);
    }

这是一个非常简单的界面。正如我已经提到的，棘手的部分是表示为对象的值将包含基于接口实现的不同值。

EPersonType

public enum EPersonType {
    Father,
    Mother,
}

继承

    public class Father implements IPerson {

    private String name;
    private int age;
    private String value;

    @Override
    public String getName() {
        return name;
    }

    @Override
    public void setName(String name) {
        this.name = name;
    }

    @Override
    public int getAge() {
        return age;
    }

    @Override
    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public Object getValue() {
        return value;
    }

    @Override
    public void setValue(Object value) {
        this.value = (String) value;
    }
    @Override
public EPersonType getPersonType() {
    return EPersonType.Father;
}


}

和另一个实例

public class Mother implements IPerson {
    private String name;
    private int age;
    private boolean value;

    @Override
    public String getName() {
        return name;
    }

    @Override
    public void setName(String name) {
        this.name = name;
    }

    @Override
    public int getAge() {
        return age;
    }

    @Override
    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public Object getValue() {
        return value;
    }

    @Override
    public void setValue(Object value) {
        this.value = (boolean) value;
    }
    @Override
public EPersonType getPersonType() {
    return EPersonType.Mother;
}


}

主类

public class entryPointForTestingSerialize {

    public static void main(String[] args) {
        List<IPerson> family = new ArrayList<IPerson>();

        IPerson father = new Father();
        father.setAge(50);
        father.setName("Oz");
        father.setValue("Hello");

        IPerson mother = new Mother();
        mother.setAge(50);
        mother.setName("Mother");
        mother.setValue(false);

        family.add(father);
        family.add(mother);

        String serialized = new JSONSerializer().deepSerialize(family);

        System.out.println(serialized);

    List<IPerson> deserialized = (List<IPerson>) new flexjson.JSONDeserializer<List<IPerson>>()
            .use("values", new TypeLocator<String>("personType")
                    .add("Mother", Mother.class).add("Father", Father.class))
            .deserialize(serialized);

        System.out.println(deserialized);
    }
}

输出

[{"age":50,"class":"testSerizlizeDeserializeInterface.Father","name":"Oz","personType":"Father","value":"Hello"},{"age":50,"class":"testSerizlizeDeserializeInterface.Mother","name":"Mother","personType":"Mother","value":false}]

谢谢!

奥兹拉德。

Answer 1

从我的角度来说，我通过把基础设施改造成更好的基础设施来解决这个问题。它的名字是XStream，它处理一切事情都很顺利和迅速。这些代码行都完成了：

    XStream xstream = new XStream(new DomDriver()); // does not require XPP3 library
    String xml = xstream.toXML(family);

为了拿回数据：

    List<IPerson> familyAfterSerialize = (List<IPerson>)xstream.fromXML(xml);