I want through the Callback to deserialize String to JSON , and I write this but i can't use

Question

public <T> void getJsonObject(String res, RequestCallBack<T> callBack) {
    T result;
    if (res != null) {
        ObjectMapper mapper = new ObjectMapper();
        try {
            result = mapper.readValue(res, new TypeReference<T>() {});
            callBack.onSuccess(result);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

it can't transmit T to the mapper the result is not what i want, what should I do?

T is not a type. You must specify actual class, otherwise it is same as TypeReference — Daniil Dubrovsky, Aug 25 '16 at 07:03
Actually it can be a duplicate of http://stackoverflow.com/questions/11664894/jackson-deserialize-using-generic-class — Daniil Dubrovsky, Aug 25 '16 at 07:04
thanks,but if I want to through the RequestCallBack to deserialize the string get the result ,how to write it? — H eyDacy, Aug 25 '16 at 07:16

score 0 · Accepted Answer · answered Aug 25 '16 at 14:01

A lot of libraries get around the type erasure by mandating the caller to pass the class in.

public <T> void getJsonObject(String res, RequestCallBack<T> callBack, Class<T> clazz) {
    T result;
    if (res != null) {
        ObjectMapper mapper = new ObjectMapper();
        try {
            result = mapper.readValue(res, clazz);
            callBack.onSuccess(result);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

On a side note, you probably shouldn't be creating a new ObjectMapper for each call (it can be an expensive operation). I would suggest creating a single static one.

Thank you , I used another method to solve it, I deserialize the string in the RequestCallback so i can get the type of T and deserialize it. — H eyDacy, Aug 26 '16 at 03:52

I want through the Callback to deserialize String to JSON , and I write this but i can't use

1 Answers1