Question

我是Android Development的新手，我正在尝试开发我的第一个Android应用，该应用使用public APIs从某些Android Volley获取数据。

我正在使用在启动器活动中初始化的singleton Volley Request Queue。当我在Fragment layout/view (uses RecyclerView & CardView)内设置RecyclerView个适配器时，我成功地能够解析JSON内容并在Volley's JsonObjectRequest上显示它们。

以下代码确实显示数据，但遭遇时间竞争条件。注意：RvJoiner是一个库，它合并多个适配器，并按照FIRST COME FIRST SERVE基础订购一个适配器。

My Fragment类如下：

public class Fragment1 extends Fragment {

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {
    View v = inflater.inflate(R.layout.recylcer_main, container, false);
    ParseJSON parseJSON = new ParseJSON(v);
    parseJSON.makeRequest1();
    parseJSON.makeRequest2();

    return v;
   }
}

我的ParseJSON类如下

public class ParseJSON {

private static final String URL1 = "some url";
private static final String URL2 = "some other url";
private static final String TAG = "ParseJSON";
private RequestQueue requestQueue;

private boolean FLAG_REQUEST1_FETCHED;
private boolean FLAG_REQUEST2_FETCHED;

private ArrayList<status1> status1ArrayList;
private ArrayList<status2> status2ArrayList;

private Context context;
private RvJoiner rvJoiner;
private View view;

ProgressDialog pd;

ParseJSON (View v){
    this.view= v;
    this.context=v.getContext();
    pd = ProgressDialog.show(v.getContext(), "Please Wait", "Getting Data from APIs", true);
    requestQueue = AppController.getInstance(v.getContext()).getRequestQueue();
    rvJoiner = new RvJoiner();
}


public void makeRequest1() {
    JsonObjectRequest request1 = new JsonObjectRequest(Request.Method.GET, URL1,
            null, new Response.Listener<JSONObject>() {
        @Override
        public void onResponse(JSONObject response) {

            try {
                /* Parsing Stuff and storing it in status1ArrayList */ 

        FLAG_REQUEST1_FETCHED=true;

                    Status1Adapter status1Adapter = new Status1Adapter(status1ArrayList);
                    RecyclerView recList = (RecyclerView) view.findViewById(R.id.cardList);
                    recList.setLayoutManager(new LinearLayoutManager(context));

                    rvJoiner.add(new JoinableAdapter(status1Adapter));
                    recList.setAdapter(rvJoiner.getAdapter());
                    pd.dismiss();
                }

            } catch (JSONException e) {}
        }

    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {}
    });
    AppController.getInstance(context).addToRequestQueue(request1);
}

public void makeRequest2() {

    JsonObjectRequest request2 = new JsonObjectRequest(Request.Method.GET, URL2,
            null, new Response.Listener<JSONObject>() {
        @Override
        public void onResponse(JSONObject response) {
            try {
        /* Parsing stuff and storing it inside ArrayList status2ArrayList */

FLAG_REQUEST2_FETCHED=true;

                    Status2Adapter status2Adapter = new Staus2Adapter(status2ArrayList);

                    RecyclerView recList = (RecyclerView) view.findViewById(R.id.cardList);
                    recList.setLayoutManager(new LinearLayoutManager(context));

                    rvJoiner.add(new JoinableAdapter(status2Adapter));
                }

            } catch (JSONException e) {}
        }

    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {}
    });
    AppController.getInstance(context).addToRequestQueue(request2);
}

public boolean isStatusFetched(){
    return FLAG_REQUEST1_FETCHED && FLAG_REQUEST2_FETCHED;
}

public ArrayList<status1> getstatus1ArrayList() {
    return status1ArrayList;
}


public ArrayList<status2> getstatus2ArrayList() {
    return status2ArrayList;
}
}

在上面的代码中，我遇到了竞争条件。由于Volley网络调用本质上是异步的，因此我无法控制哪个请求将首先完成并显示在Fragment CardView上。（即任何rvJoiner.add()个请求都可以先执行）

我想让我的UI保持一致，即我希望首先将Request1适配器添加到RvJoiner然后再添加到Request2。

如果可能的话，我想移动所有设置适配器的代码，并将它们从JsonObjectRequest连接到我的Fragment的onCreateView方法。所以，通过这种方式，我可以控制适配器的顺序。但是，我需要一种方法，通过FLAG_REQUEST1_FETCHED方法连续检查FLAG_REQUEST2_FETCHED和isStatusFetched的值。

Fragment类的代码将是

public class Fragment1 extends Fragment {

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {
    View v = inflater.inflate(R.layout.recylcer_main, container, false);
    ParseJSON parseJSON = new ParseJSON(v);
    parseJSON.makeRequest1();
    parseJSON.makeRequest2();

while(!parseJSON.isDataFetched()){
/* I want to wait here till both status1ArrayList and status2ArrayList gets     populated with data in ParseJSON. In this way I can control the order in which adapters are added inside RvJoiner. If I don't wait here I will get NullPointerException on runtime since Volley calls are asynchronous and getStatus1ArrayList/getStatus2ArrayList will most probably return null. But how to wait here without consuming too much CPU power? */
}

    ArrayList<status1> status1ArrayList = parseJSON.getstatus1ArrayList();
    ArrayList<status2> status2ArrayList = parseJSON.getstatus2ArrayList();  

    Status1Adapter status1Adapter = new Status1Adapter(status1ArrayList);
    Status2Adapter status2Adapter = new Status2Adapter(status2ArrayList);

    RecyclerView recList = (RecyclerView) v.findViewById(R.id.cardList);
    recList.setLayoutManager(new LinearLayoutManager(v.getContext()));

RvJoiner rvJoiner = new RvJoiner();
/* Problem solved as I'm adding adapters in the order I want */
    rvJoiner.add(new JoinableAdapter(status1Adapter)); 
    rvJoiner.add(new JoinableAdapter(status2Adapter));

    recList.setAdapter(rvJoiner.getAdapter());
    return v;
}

}

一种解决方案可以使用回调。我在某处读到了关于它们的信息，但我不确定它是否能解决我在保持秩序的同时“多次请求”的问题。

另一个解决方案是限制我的Volley Queue一次只处理一个请求，但这会增加获取和提供数据所需的时间。这是我的最后选择。

我几乎没有想法，希望有人帮助我，这样我就可以控制设置适配器的顺序并保持一致的UI。如果您需要任何其他信息，请告诉我。

感谢。

Answer 1

这就是避免两个请求的竞争条件一般工作的方式。你应该使用回调。 onResponse方法的实现是回调，因为在完成一个请求后会调用这些方法。响应处理适用于UI线程吗？因此，响应可以一个接一个地处理。这意味着你只需维持那里的秩序。在收到一个回复后，提取您想要做的工作。您需要一些布尔标志来指示您的请求是否已完成。伪代码看起来像这样：

request1Done = false;
request2Done = false;
doRequest1();
doRequest2();
onResponse1() {
    doWorkForRequest1();  // always start handling the response
    request1Done = true;
    if (request2Done) {   // if this is true, request2 was faster than request1
        doWorkForRequest2();
    }
};
onResponse2() {
     request2Done = true;
     if (request1Done) {  // request1 did its work, no its request2's turn
         doWorkForRequest2();
     }
};

所以基本上你应该修复你的onReponse方法。希望这会帮助你。：）

避免在Android应用中使用Android Volley获得竞争条件

1 个答案: