CODEWITHSUNDEEP
javaandroidjsonjsonexception
Halim
Halim

Reputation: 11

Android JSON WebRequest - JSONException: End of input at character 0

I have this piece of code here which I have grabbed from http://mobilesiri.com/json-parsing-in-android-using-android-studio/ and modified it to my own use. 

    public String makeWebServiceCall(String addr, int requestMethod) {
    URL url;
    String response = "";
    try {
        url = new URL(addr);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        conn.setReadTimeout(15001);
        conn.setConnectTimeout(15001);
        conn.setDoInput(true);
        conn.setDoOutput(true);

        if (requestMethod == Constant.GET) {
            conn.setRequestMethod("GET");
        } else if (requestMethod == Constant.DELETE) {
            conn.setRequestMethod("DELETE");
        }

        int reqResponseCode = conn.getResponseCode();

        if ((requestMethod == Constant.GET || requestMethod == Constant.DELETE) && reqResponseCode == HttpURLConnection.HTTP_OK) {
            String line;
            BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            while ((line = br.readLine()) != null) {
                response += line;
            }
        } else {
            response = "";
        }

    } catch (MalformedURLException murle) {
        Log.e(MalformedURLException.class.getName(), murle.getMessage());
    } catch (IOException ioe) {
        Log.e(IOException.class.getName(), ioe.getMessage());
    }

    return response;
}

Put this in a normal Java environment, along with the libraries, I'm able to get the output in this form

{"status":"FOUND","data":[{"category_id":3,"category_name":"Experience Sharing Area"},{"category_id":4,"category_name":"Frequently Asked Questions"},{"category_id":1,"category_name":"General Pain Advice"},{"category_id":2,"category_name":"Pain Categorization Section"}]}

Put this in Android to parse data, I get the JSONException - org.json.JSONException: End of input at character 0 of

Here is where I implement it.

protected Void doInBackground(Void... voids) {
    WebRequest webReq = new WebRequest();
    String jsonStr = webReq.makeWebServiceCall(Constant.WEB_SERVER_ADDR + Constant.GET_CATEGORY, Constant.GET);
    Log.d("URL:", Constant.WEB_SERVER_ADDR + Constant.GET_CATEGORY);
    Log.d("Response:", " > " + jsonStr);

    categoryList = parseJSON(jsonStr);

    return null;
}

Is there any issues with this?

Upvotes: 0

Views: 103

Answers (1)

IntelliJ Amiya
IntelliJ Amiya

Reputation: 75788

You are getting JSONException: End of input at character 0.

Post your parseJSON Class .

{"status":"FOUND","data":[{"category_id":3,"category_name":"Experience Sharing Area"},{"category_id":4,"category_name":"Frequently Asked Questions"},{"category_id":1,"category_name":"General Pain Advice"},{"category_id":2,"category_name":"Pain Categorization Section"}]}

Your Parsing will be

try {
JSONObject reader = new JSONObject("YOUR_JSON_STRING");
JSONArray jsonArray = reader.getJSONArray("data");


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

    try {
        jsonItem = jsonArray.getJSONObject(i);
    } catch (JSONException e1) {
        e1.printStackTrace();
    }


    String cate_id= jsonItem.getString("category_id");
    String category_name= jsonItem.getString("category_name");

}
// Add adapter

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

}

Upvotes: 1

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