Question

我有一个类（名为C）来检查JSON的结构，如果结构错误，它将抛出异常。该类将从类B中调用。我可以在类B中使用try catch。实际上，类B由类A调用，它提供对类B的回调。如何从调用类C调用异常通过B类到A类的回调（OnFinishListener）？我想在A类的回调声明中提供try catch，而不是B类。

public class A {
    B b = new B();
    b.getResponseFromServer(new B.OnFinishListener {
        @Override
        public void onFinish(Object response) {
    });
}

public class B {
    public static interface OnFinishListener {
        void onFinish(Object response);
    }

    //other code

    public void getResponseFromServer(OnFinishListener onFinishListener) {
    Response.Listener<JSONObject> listener = new Response.Listener<>() {
            @Override
            public void onResponse(JSONObject response) {
                try {
                    onFinishListener.onFinish(C.someMethod(response);
                catch (JSONException e) {
                    Log.e("Error", e.getMessage());
                }
            }
        } 

        //do some Volley tasks
    }
}

public class C {
    public static Object someMethod(JSONObject json) throws JSONException {
        if (json.has("something")) {
            //return something
        } else {
            throw new JSONException("Error");
        }
    }
}

Answer 1

我得到了解决方案。我只需要在我的接口（onException）中添加一个方法，并在catch(Exception)上将此异常传递给该接口。

＆＃13;

public class A {
    B b = new B();
    b.getResponseFromServer(new B.OnFinishListener {
        @Override
        public void onFinish(Object response) {}

        @Override
        public void onException(JSONException response) {}
);
}

public class B {
    public static interface OnFinishListener {
        void onFinish(Object response);
        void onException(JSONException e);
    }

    //other code

    public void getResponseFromServer(OnFinishListener onFinishListener) {
	Response.Listener<JSONObject> listener = new Response.Listener<>() {
            @Override
            public void onResponse(JSONObject response) {
                try {
                    onFinishListener.onFinish(C.someMethod(response);
                catch (JSONException e) {
                    onFinishListener.onException(e);
                }
            }
        } 

        //do some Volley tasks
    }
}

＆＃13;

Answer 2

Try ur code to put in Try Catch...

try {
      if (json.has("something")) {

      } else {

      }
    }catch (JSONException e) {
        e.printStackTrace();
    }

如何在Android中向回调发送异常？

2 个答案: