Question

我试图解析这个Json响应。出于某种奇怪的原因，它不起作用。请多多包涵。我真吮吸Json。

这是用于解析它的代码：

      public class AsyncTaskParseJson extends AsyncTask<String, String, String> {

       final String TAG = "AsyncTaskParseJson.java";

    // set your json string url here


    @Override
    protected void onPreExecute() {

        Toast.makeText(getActivity(), "started", Toast.LENGTH_SHORT).show();

    }

    @Override
    protected String doInBackground(String... arg0) {

        String str = "";
        HttpResponse response;
        HttpClient myClient = new DefaultHttpClient();
        HttpPost myConnection = new   HttpPost("http://gdata.youtube.com/feeds/api/videos/iS1g8G_njx8?v=2&alt=jsonc");

        try {
            response = myClient.execute(myConnection);
            str = EntityUtils.toString(response.getEntity(), "UTF-8");

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


        try{
            JSONObject myJson = new JSONObject(str);

            String grande =  myJson.getString("title");
            Toast.makeText(getActivity(), ""+grande, Toast.LENGTH_SHORT).show();


        } catch ( JSONException e) {
            e.printStackTrace();
        }



         return null;
      }

        @Override
        protected void onPostExecute(String strFromDoInBg) {

        Toast.makeText(getActivity(), "done", Toast.LENGTH_SHORT).show();

       }
   }

我只需要解析标题和描述。感谢

Answer 1

它似乎你没有正确解析

使用以下代码更改您的try块

try{
            JSONObject myJson = new JSONObject(str);
            JSONObject entityObject = myJson.getJSONObject("entity");

            // Parsing title
            JSONOBject titleObject = entityObject.getJSONObject("title");
            String grande =  titleObject.getString("$t");

            // Do the same for Description as well (Like above two line)

            Toast.makeText(getActivity(), ""+grande, Toast.LENGTH_SHORT).show();


        } catch ( JSONException e) {
            e.printStackTrace();
   }

Answer 2

我认为你的toast信息（在doInBackground中）会出现问题，尝试将其评论然后再检查！

此外，最佳做法是将响应发送到onPostExecute作为参数，然后在onPostExecute中解析Json

Answer 3

尝试这样，

@Override
    protected String doInBackground(String... arg0) {

        String str = "";
        HttpResponse response;
        HttpClient myClient = new DefaultHttpClient();
        HttpPost myConnection = new   HttpPost("http://gdata.youtube.com/feeds/api/videos/iS1g8G_njx8?v=2&alt=jsonc");

        try {
            response = myClient.execute(myConnection);
            str = EntityUtils.toString(response.getEntity(), "UTF-8");
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

         return str;
      }

       @Override
        protected void onPostExecute(String result) {

        try{
            JSONObject myJson = new JSONObject(result);

             JSONObject entry =  results.getJSONObject("entry");
            JSONObject grande =  results.getJSONObject("title");
            String title =  grande.getString("$t"); 
            Toast.makeText(getActivity(), title, Toast.LENGTH_SHORT).show();


        } catch ( JSONException e) {
            e.printStackTrace();
        }

        Toast.makeText(getActivity(), "done", Toast.LENGTH_SHORT).show();

       }

解析简单的Json响应

3 个答案: