根据jsonarray请求填充java Arraylist

Question

我有一个返回arraylist的方法，它是从jsonarray填充的 jsonrequest工作正常，但当我调用该方法时，arraylist返回大小为0的值，即使它已填充。 下面是代码，我做错了什么

private ArrayList<String> queryRes(String url, String searchString) {


    final ArrayList<String> mylist = new ArrayList<>();

    AsyncHttpClient client = new AsyncHttpClient();


    client.get(url + searchString,
            new JsonHttpResponseHandler() {

                @Override
                public void onSuccess(JSONArray jsonArray) {

                    for(int i =0; i < jsonArray.length(); i++) {

                   mylist.add(i, jsonArray.optJSONObject(i).optString("id"));

                        //Check if it is added to the list
                        Log.d("Try: ", mylist.get(i));
                    }

                }

                @Override
                public void onFailure(int statusCode, Throwable throwable, JSONObject error) {

                }
            });

    return mylist;
}

Answer 1

当得到响应时调用onSuccess（）意味着AsynHttp调用在另一个线程中完成，所以在onSuccess之前调用return语句，这就是大小为0的原因

您可以在调试器上验证它

我的建议是从onSuccess（）调用一个方法，这样就可以得到Arraylist

Answer 2

如果你在一节课上做每件事

public void populateList(String url, String searchString)
{
 final ArrayList<String> mylist = new ArrayList<>();

    AsyncHttpClient client = new AsyncHttpClient();


    client.get(url + searchString,
            new JsonHttpResponseHandler() {

                @Override
                public void onSuccess(JSONArray jsonArray) {

                    for(int i =0; i < jsonArray.length(); i++) {

                   mylist.add(i, jsonArray.optJSONObject(i).optString("id"));

                        //Check if it is added to the list
                        Log.d("Try: ", mylist.get(i));
                    }

              onListPopulated(myList);

                }

                @Override
                public void onFailure(int statusCode, Throwable throwable, JSONObject error) {

                }
            });

}

一种处理列表的方法

public void onListPopulated(ArrayList<String> list)
{
//do something with the list
}