Question

鉴于以下结构：

abstract class Message {
   Message anotherMessage;
   String attribute; //just random stuff
}

我想将以下json-string作为输出：

 {type=Class.getSimpleName(), data=gson(Message)}

因为Message是抽象的，可以有多个实现。问题是，“anotherMessage”不具有结构类型data。

我的序列化实现：

public JsonElement serialize(final Message src, final Type typeOfSrc,
    final JsonSerializationContext context) {
  Gson gson = new Gson();
  JsonObject elem = new JsonObject();
  elem.addProperty("type", src != null ? src.getClass().getSimpleName() : null);
  elem.addProperty("data", src != null ? gson.toJson(src) : null);
  return elem;
}

如何递归执行此操作？我无法获得已附加消息适配器（stackoverflow-exception）的Gson-Object

Answer 1

在序列化/反序列化期间可以使用JsonSerializationContext / JsonDeserializationContext来序列化/反序列化另一个对象。

Message.java

abstract class Message {
    Message anotherMessage;
    String theMessage;

    public Message getAnotherMessage() {
        return anotherMessage;
    }

    public String getTheMessage() {
        return theMessage;
    }
}

Info.java

public class InfoMessage extends Message {
    public InfoMessage(Message anotherMessage, String theMessage) {
        this.anotherMessage = anotherMessage;
        this.theMessage = theMessage;
    }
}

Alert.java

public class AlertMessage extends Message {
    public AlertMessage(Message anotherMessage, String theMessage) {
        this.anotherMessage = anotherMessage;
        this.theMessage = theMessage;
    }
}

ErrorMessage.java

public class ErrorMessage extends Message {
    public ErrorMessage(Message anotherMessage, String theMessage) {
        this.anotherMessage = anotherMessage;
        this.theMessage = theMessage;
    }
}

MessageSerializer.java

public JsonElement serialize(Message src, Type typeOfSrc, JsonSerializationContext context) {
    JsonObject elem = new JsonObject();

    if (src == null) {

    } else {
        elem.addProperty("type", src.getClass().getSimpleName());
        elem.addProperty("attribute", src.getTheMessage());
        elem.add("data", src.anotherMessage != null ? context.serialize(src.anotherMessage, Message.class): null);
    }

    return elem;
}

Test.java

public static void main(String[] args) {
        Gson gson = new GsonBuilder()
                            .registerTypeAdapter(Message.class, new MessageSerializer())
                            .setPrettyPrinting()
                            .create();

        String json = gson.toJson(
                            new InfoMessage(
                                    new AlertMessage(
                                            new ErrorMessage(null, "the error message"),
                                            "the alert message"), 
                                    "the info message"), 
                            Message.class);
        System.out.println(json);

}

递归类型适配器GSON

1 个答案: