Grabbing JSON content with Java

Question

Here is a sample of my json_encode in PHP:

print(json_encode($row));

leads to {"AverageRating":"4.3"} which is good.

But in Java, I can not seem to grab this 4.3 value. Here it is (for an Android project) I have edited non-relevant data.

 public class Rate extends ListActivity {

   JSONArray jArray;
   String result = null;
   InputStream is = null;
   StringBuilder sb = null;
   String Item, Ratings, Review, starAvg;
   RatingBar ratingsBar;
   ArrayList<NameValuePair> param;

  public void onCreate(Bundle savedInstanceState) {

      starAvg = "0"; // Sets to 0 in case there are no ratings yet.
      new starRatingTask().execute();
      ratingsBar = (RatingBar) findViewById(R.id.theRatingBar);


  class starRatingTask extends AsyncTask<String, String, Void> {

    InputStream is = null;
    String result = "";


    @Override
    protected Void doInBackground(String... params) {
        String url_select = "http://www.---.com/---/average_stars.php";

        ArrayList<NameValuePair> param = new ArrayList<NameValuePair>();
        param.add(new BasicNameValuePair("item", Item));


        HttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url_select);


        try {
            httpPost.setEntity(new UrlEncodedFormEntity(param));
            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();

            // read content
            is = httpEntity.getContent();

        } catch (Exception e) {

            Log.e("log_tag", "Error in http connection " + e.toString());
        }
        try {
            BufferedReader br = new BufferedReader(
                    new InputStreamReader(is));
            StringBuilder sb = new StringBuilder();
            String line = "";
            while ((line = br.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            result = sb.toString();

        } catch (Exception e) {

            Log.e("log_tag", "Error converting result " + e.toString());
        }

        return null;

    }

    protected void onPostExecute(Void v) {

        String starAvgTwo = null;
        try {
            jArray = new JSONArray(result);
            JSONObject json_data = null;
            for (int i = 0; i < jArray.length(); i++) {
                json_data = jArray.getJSONObject(i);
                starAvg = json_data.getString("AverageRating");

                   starAvgTwo = starAvg;

            }
        } catch (JSONException e1) {
            Toast.makeText(getBaseContext(), "No Star Ratings!",
                    Toast.LENGTH_LONG).show();
        } catch (ParseException e1) {
            e1.printStackTrace();
        }


                Toast.makeText(getBaseContext(), starAvgTwo,
                        Toast.LENGTH_LONG).show();

        ratingsBar.setRating(Float.valueOf(starAvg));



    }
}

That second toast produces a blank (I assume a "" - empty string?). If I change the toast variable back to starAvg, then it toasts "0".

How can I retrieve the value of 4.3.

What is inside "result" after you get it from sb.toString()? — Carlos Barcellos
– Carlos Barcellos, Commented Jun 22, 2012 at 21:04
I toasted result; it is actually shows {"AverageRating":"4.3"} That is very interesting... so it goes back to the JSON parsing section I imagine? — TheLettuceMaster
– TheLettuceMaster, Commented Jun 22, 2012 at 21:14
That json describes a JSONObject, not a JSONArray. If it were [{"AverageRating":"4.3"},{"AverageRating":"2.3"},{"AverageRating":"5.0"}] then you'd be working with an array. — nEx.Software
– nEx.Software, Commented Jun 22, 2012 at 21:25
So does that mean the problem is on the PHP end in how it is encoding? — TheLettuceMaster
– TheLettuceMaster, Commented Jun 22, 2012 at 21:29
Well, you should be able to parse it as a JSONObject if you will always be getting it this way, or if you will be getting a list of star ratings, then the PHP side needs to json_encode an array of rows instead of just one. — nEx.Software
– nEx.Software, Commented Jun 22, 2012 at 21:37

nEx.Software · Accepted Answer · 2012-06-22 22:15:22Z

1

As we discussed in the comments on the original question, the PHP is sending down as single JSONObject rather than an array. Parsing as a JSONObject is required in it's present state; however, if you begin sending down an array of your value objects, then you'd use JSONArray to parse it.

answered Jun 22, 2012 at 22:15

nEx.Software

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1 Comment

TheLettuceMaster Over a year ago

Thanks, marked you correct since you got to the bottom of it first!

yosafatade · Accepted Answer · 2012-06-22 22:17:40Z

I think your JSON doesn't contain array. so just do this:

JSONObject jsonObject = new JSONObject(result); //to convert string to be a JSON object
String averageRating = jsonObject.getString("AverageRating"); //get the value of AverageRating variable

and try toast the averageRating.

and how to get the array from JSON object?

if you have JSON:

{"employees": [
    { "firstName":"John" , "lastName":"Doe" }, 
    { "firstName":"Anna" , "lastName":"Smith" }, 
    { "firstName":"Peter" , "lastName":"Jones" }]
}

then use this code

JSONArray jsonArray = new JSONArray(result);
for (int i = 0; i < jsonArray.length(); i++) {
    JSONObject jsonObject = jsonArray.getJSONObject(i);
    Log.i(Rate.class.getName(), jsonObject.getString("firstName"));
}

that code will produce

John Anna Peter

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