Question

我们正在服务器上进行大量（所有必要的）调用以获取用户数据。为此，我们创建了一个公共类来发布数据，并将响应作为JSON对象进行检索。但是由此我们的应用程序变得太慢，所以我们在思考，我们可以通过异步任务检索数据。

所以我们修改了我们的代码，如下所示： -

UserFunction.java

public JSONObject getUserData(String userName,String pNumber){
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", getUser));
    params.add(new BasicNameValuePair("search_number_array", userName));
    params.add(new BasicNameValuePair("pNumber", pNumber));
    params.add(new BasicNameValuePair("id", pNumber));
    // getting JSON Object
    JSONParser jsonParser1=new JSONParser(loginURL, params);
  // jsonParser1.getJSONFromUrl(loginURL, params);
    jsonParser1.execute();
    Jsonresult j1=new Jsonresult();
    JSONObject json =j1.getjObjResult();
            Log.d("Response:", json.toString());   //line 45, on which getting error
    return json;
}

JSONParse.java

 public class JSONParser extends AsyncTask<String, String, JSONObject> {

static InputStream is = null;
static JSONObject jObj = null;
Jsonresult obj;
static String json = "";
List<NameValuePair> postparams= new ArrayList<NameValuePair>();
String URL=null;
// constructor
public JSONParser(String url, List<NameValuePair> params) {
    URL=url;
    postparams=params;
    Log.e("entered constructor", "entered in constructor ");

    // Making HTTP request


}

@Override
protected JSONObject doInBackground(String... strings) {
    Log.e("entered in bg", "entered in doinbackground ");

    try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(URL);
        httpPost.setEntity(new UrlEncodedFormEntity(postparams));

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        json = sb.toString();
        Log.e("JSON", json);
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

    // return JSON String
    return jObj;
}

public void getJSONFromUrl(String url, List<NameValuePair> params) {
    URL=url;
    postparams=params;
    Log.e("entered ", "entered in getjsonurl ");

    // Making HTTP request


}
@Override
protected void onPostExecute(JSONObject result) {
    this.obj.setjObjResult(result);
}
@Override
protected void onPreExecute() {
    // TODO Auto-generated method stub
    super.onPreExecute();
}
 }

Jsonresult.java

public class Jsonresult {
 JSONObject jObjResult = null;

public JSONObject getjObjResult() {
    return jObjResult;
}

public void setjObjResult(JSONObject jObjResult) {
    this.jObjResult = jObjResult;
}
}

但我们收到此错误

   java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String org.json.JSONObject.toString()' on a null object reference
        at com.keepAeye.gps.UserFunctions.getUserData(UserFunctions.java:45)

意味着控制在完成后台处理之前进行。请告诉我们如何解决这个问题？

Answer 1

当你打电话时

jsonParser1.execute();

异步任务发生在另一个线程中。这意味着这些行

Jsonresult j1=new Jsonresult();
JSONObject json =j1.getjObjResult();
Log.d("Response:", json.toString());
return json;

在JSONParser的doInBackground执行之前执行（实际上它们都是并行执行）。

这意味着当您在

中拨打j1.getjObjResult();代码时

@Override
protected void onPostExecute(JSONObject result) {
    this.obj.setjObjResult(result);
}

尚未执行。解决这个问题的简单方法是使用类似的界面。

public interface CallBack {
    void onSuccess(JSONObject json);
}

然后在UserFunction类链接

中实现它

class UserFunction implements CallBack {
   ...
   @Override
   public void onSuccess(JSONObject json) {
     Log.d("Response:", json.toString());
   }
}

执行AsyncTask时，会传递回调引用，如此

JSONParser jsonParser1=new JSONParser(loginURL, params, this);

并在AsyncTask中维护一个回调引用，如

CallBack mCallback;

public JSONParser(String url, List<NameValuePair> params, CallBack callBack) {
    URL=url;
    postparams=params;
    mCallback = callBack;        
    Log.e("entered constructor", "entered in constructor ");

    // Making HTTP request


}

在AsyncTask的onPostExecute中执行此操作。

@Override
protected void onPostExecute(JSONObject result) {
   mCallback.onSuccess(result);
}

您应该考虑使用网络库进行retrofit或volley等网络调用

在异步任务android之后返回JSON数据

1 个答案: