在LogCat中显示JSON Google搜索-Java 学习之路

我正在尝试从URL获取搜索结果

https://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=JSON

并使用 log.v() 将其显示在LogCat中 .

问题是，而不是在 LogCat 中显示搜索结果，程序不停地运行，我在 LogCat 中看到的是一个无穷无尽的列表

11-19 15：51：11.256：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％释放6651K / 7688K，暂停8ms，总计8ms 11-19 15:51:11.264：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％免费6651K / 7688K，暂停8ms，总计8ms 11-19 15:51:11.272：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％免费6651K / 7688K，暂停5ms，总计5ms 11-19 15：51：11.272：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％免费6651K / 7688K，暂停4ms，总计5ms 11-19 15:51:11.324：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14 ％免费6651K / 7688K，暂停45ms，总计45ms 11-19 15:51:11.332：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％免费6651K / 7688K，暂停8ms，总计8ms 11-19 15:51： 11.400：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％释放6651K / 7688K，暂停23ms，总计23ms 11-19 15:51:11.424：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％免费6651K / 7688K，暂停2ms，总计2ms 11-19 15:51:11.464：D / dalvikvm（1251）：GC_FOR_ALLOC释放480K，14％释放6651K / 7688K，暂停26ms，总计26ms

这永远不会停止并继续，直到我在模拟器中杀死应用程序 .

这是我的代码

公共类MainActivity扩展Activity {

public static final String TAG= MainActivity.class.getSimpleName();
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);


    if(isNetworkAvilable()){
        GetSearchData getSearchData = new GetSearchData();
        getSearchData.execute();

    }
    else{
        Toast.makeText(this, "The network is down", Toast.LENGTH_SHORT).show();
    }

}


private boolean isNetworkAvilable() {
    // TODO Auto-generated method stub

    boolean isAvailable= false;
    ConnectivityManager manager=  (ConnectivityManager) 
            getSystemService(CONNECTIVITY_SERVICE);

    NetworkInfo networkInfo=manager.getActiveNetworkInfo();

    if(networkInfo !=null && networkInfo.isConnected()){

        isAvailable=true;
    }

    return isAvailable;
}




private class GetSearchData extends AsyncTask<Object, Void, String>{

    protected String doInBackground(Object... params){

        int responseCode=-1;
        try{

            URL url=new URL("https://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=JSON");

            HttpURLConnection connection = (HttpURLConnection) url.openConnection();

            connection.connect();

             responseCode= connection.getResponseCode();

            // Log.i(TAG, "Code: "+responseCode);

             if(responseCode== HttpURLConnection.HTTP_OK){
                 InputStream inputStream= connection.getInputStream();
                 Reader reader= new InputStreamReader(inputStream);

                 int nextCharacter;
                 String responseData="";
                 while(true){
                     nextCharacter=reader.read();



                     if(responseCode==-1){
                         break;
                     }

                     responseData+= (char) nextCharacter;

                 }

                 JSONObject jsonObject= new JSONObject(responseData);



                JSONObject jsonObject2= jsonObject.getJSONObject("responseData");

                // JSONObject jsonObject3= jsonObject.getJSONObject("results");

                 JSONArray jsonArray= jsonObject2.getJSONArray("results");


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

                     String title= results.getString("title");

                     Log.v(TAG, "Result "+i+" : "+title);

                 }


             }
             else{
                 Log.i(TAG, "Wrong Response Code: " + responseCode);
             }



        }
        catch(MalformedURLException e){
            Log.e(TAG, "Exception: "+e);
        }
        catch(IOException e){
            Log.e(TAG, "Exception: "+e);
        }

        catch(Exception e){
            Log.e(TAG, "Exception: "+e);
        }

        return "Code: "+responseCode;
    }



}

}

1 回答

尝试使用此代码进行网络呼叫，它应该可以正常工作，至少从互联网上获取数据，然后将其发布到LogCat中 .

public class GetSearchData extends AsyncTask<Void,Void,Void> {

        @Override
        protected Void doInBackground(Void... params) {
            String url = "https://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=JSON";
            HttpClient httpClient = new DefaultHttpClient();
            HttpGet httpGet = new HttpGet(url);
            try{
                HttpResponse httpResponse = httpClient.execute(httpGet);
                String responseData = inputStreamToString(httpResponse.getEntity().getContent()).toString();
                Log.v("OUTPUT", responseData);

                JSONObject jsonObject= new JSONObject(responseData);
                JSONObject jsonObject2= jsonObject.getJSONObject("responseData");
                JSONArray jsonArray= jsonObject2.getJSONArray("results");

                for (int i = 0; i < jsonArray.length(); i++){
                   JSONObject results = jsonArray.getJSONObject(i);
                   String title = results.getString("title");
                   Log.v(TAG, "Result "+i+" : "+title);
                }

            } catch (IOException e){
                e.printStackTrace();
            } catch (JSONException e){
                e.printStackTrace();
            }
            return null;
        }

        public static StringBuilder inputStreamToString(InputStream is){
            String line;
            StringBuilder sb = new StringBuilder();
            BufferedReader rd = new BufferedReader(new InputStreamReader(is));
            try {
                while ((line = rd.readLine()) != null) {
                    sb.append(line);
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return sb;
        }
    }

回复于 2024-05-19T16:45:16+08:00

在LogCat中显示JSON Google搜索

1 回答

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