如何创建Gson封送和解封的通用方法？

Question

我对泛型是个新手。

public class Message {
    private MessageBody messageBody;
    ....
}
public class MessageBody {
    private List<BodyParam> bodyParams;
}
public class MessageBodyParam {
    private String name;        
    private Object value;
}

public class JsonUtil {

    //This is not working
    public static <T> String marshal(Object object, Class<T> jtype) {
        Gson gson = new Gson();

        Type type = new TypeToken<T>() {}.getType();
        return gson.toJson(object, type);
    }

    //this is working as expected
    public static <T> T unmarshal(String jsonStr, Class<T> jtype) {
        Gson gson = new Gson();
        Type type = new TypeToken<T>() {}.getType();
        return gson.fromJson(jsonStr, type);
    }
}

//Message is my class
String jsonMessage = JsonUtil.marshal(message, Message.class);
Message message = JsonUtil.unmarshal(message, Message.class);

Answer 1

为时已晚,但这个解决方案可以为gson创建一个泛型类：

public class JSONConversion<T>{

    private  T t;


    public T getT() {
        return t;
    }

    public void setT(T t) {
        this.t = t;
    }

    public String ObjectToString(){

        Gson gson=new Gson();
        return gson.toJson(t);

    }
    public T StringToObject(String jsonString ,Class<T> tt){
        Gson gson=new Gson();
        return gson.fromJson(jsonString, tt);
    }

}

您可以像这样使用这个泛型gson类：

LoginResponse loginResponse = response.body();
JSONConversion<LoginResponse> jsonConversion=new JSONConversion<>();
jsonConversion.setT(loginResponse);
String createdJson=jsonConversion.ObjectToString());
Log.d("toJson",createdJson);            
LoginResponse convertedObject=jsonConversion.StringToObject(createdJson,LoginResponse.class);