AsyncHttpResponseHandler class 必须声明为抽象或实现抽象方法 android

Question

Following this tutorial

但是我在下面的代码 AsyncHttpResponseHandler 中收到错误消息说 AsyncHttpResponseHandler class 必须声明为抽象的或实现抽象方法 android

client.get("http://192.168.2.2:9999/useraccount/login/dologin",params ,new AsyncHttpResponseHandler()

Answer 1

从official doc开始，正确的实现是：

AsyncHttpClient client = new AsyncHttpClient();
client.get("http://www.google.com", null, new AsyncHttpResponseHandler() {

    @Override
    public void onStart() {
        // called before request is started
    }

    @Override
    public void onSuccess(int statusCode, Header[] headers, byte[] response) {
        // called when response HTTP status is "200 OK"
    }

    @Override
    public void onFailure(int statusCode, Header[] headers, byte[] errorResponse, Throwable e) {
        // called when response HTTP status is "4XX" (eg. 401, 403, 404)
    }

    @Override
    public void onRetry(int retryNo) {
        // called when request is retried
    }
});