org.json.JSONException: Value of type java.lang.String cannot be converted to JSONObject

Question

Hi i got a problem with Android JSON Parser in 2.3, all is ok in 4.0>.

I look other topic they are talking about encoding (server) or other stuff server side, but i tried to put "test" in all my JSON field an the problem still persist.

Here is my code :

URL url = new URL(c.getString(R.string.url_ws) + url_ws);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestMethod("POST");
OutputStreamWriter request = new OutputStreamWriter(connection.getOutputStream());
request.write("&test=test");
request.flush();
request.close();


String line;
InputStreamReader isr = new InputStreamReader(connection.getInputStream());
BufferedReader reader = new BufferedReader(isr);
StringBuilder sb = new StringBuilder();

while ((line = reader.readLine()) != null){
    sb.append(line + "\n");
}
String json = sb.toString().trim();
try {
     JSONObject obj = new JSONObject(json);
} catch (JSONException e) {
    e.printStackTrace();
    Log.d(Tools.TAG+"/debug JSONException", e.toString());
    return null;
}

Here is my exception

debug JSONException(4184): org.json.JSONException: Value  of type java.lang.String cannot be converted to JSONObject

i saw double space between "Value" and "of" so i guess the problem is a empty string but i dont get it.

thanks.

EDIT

it was a encode mistake in my JSON,

char a = json.substring(0, 1).charAt(0);
int ascii = (int)a;
Tools.myLog(">"+ascii+"<");

i found 65279 char

SOLUTION

Why is  appearing in my HTML?

encoding UTF-8 without BOM.

http://stackoverflow.com/questions/17792779/what-is-the-error-denotes-in-androidjson-error-parsing-data-org-json-jsonexcep/17793283#17793283. similar!. Post your json. — Raghunandan, Jul 23 '13 at 08:08
JSONArray obj = new JSONArray(json); use this instead of JSONObject obj = new JSONObject(json); — Giant, Jul 23 '13 at 08:11
my JSON isn't an array : {"bHasError":false,"sSessionId":"5e10be14c408732179aa4731899882fd","iMembreId":7510064,"iCurrentDate":1374565720 .... — Jeremy Piednoel, Jul 23 '13 at 08:13
Log the string before creating the JSONObject so you can see if it's a json string. — Mugur, Jul 23 '13 at 08:18
{ "bHasError": false, "sSessionId": "5e10be14c408732179aa4731899882fd", "iMembreId": 7510064, "iCurrentDate": 1374565720, "iExpirationDate": 1374567520, "aData": { "aVentesEnCours": [ — Jeremy Piednoel, Jul 23 '13 at 08:18
You can check the validity of the json string here: http://jsonformatter.curiousconcept.com — Mugur, Jul 23 '13 at 08:18

Mohammod Hossain · Accepted Answer · 2013-07-23T08:35:37.947

while ((line = reader.readLine()) != null){
    sb.append(line + "\n");
}

instead try

while ((line = reader.readLine()) != null){
    sb.append(line);
   json = sb.toString().substring(0, sb.toString().length()-1);
}

Follow this method for getting json object:

 public JSONObject getJSONObjFromUrl(String url, List<NameValuePair> params) {      
            System.out.println("url:: "+url );
            System.out.println("params:: "+ params +" " +params.get(0) );
            // Making HTTP request
            try {
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

                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);
                }
                is.close();

                json = sb.toString().substring(0, sb.toString().length()-1);
               // 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;

        }

i delete the \n but your trick with substring make an error : debug JSONException(4461): org.json.JSONException: Unterminated object at character 242 of — Jeremy Piednoel, Jul 23 '13 at 08:24
Nice very good site i found the mistake it was a space between [" "] in my JSON, i update my question. TYVM ! — Jeremy Piednoel, Jul 23 '13 at 08:36

org.json.JSONException: Value of type java.lang.String cannot be converted to JSONObject

SOLUTION

1 Answers1

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