JSONObject not being correctly created (possible JSON formatting error?)

Question

I am getting this JSON string:

[
{
    "id": 135,
    "date": "2013-08-30 19:00:29",
    "timestamp": "2013-08-30 19:00:29",
    "lat": "54.328274",
    "long": "-2.747215",
    "strap": "annual International Festival of Street Arts",
    "link": "http://dev.website.co.uk//?p=135",
    "title": "Title"
}
]

Which is correct JSON syntax im sure (works fine in the iOS app), however when parsing to a JSONObject it catches an error. Java:

public static JSONObject getJSONfromURL(String url){

    //initialize
    InputStream is = null;   
    String result = "";   
    JSONObject jArray = null;

    //http post
    try {

        HttpClient httpclient = new DefaultHttpClient();   
        HttpPost httppost = new HttpPost(url);    
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();

        is = entity.getContent();    

    } catch (Exception e) {
        Log.e("log_tag", "Error in http connection "+e.toString());   
    }
    //convert response to string

    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 + "\n");
        }

        is.close();
        result=sb.toString();

    } catch (Exception e) {    
        Log.e("log_tag", "Error converting result "+e.toString());
    }
    //try parse the string to a JSON object

    try {
        Log.d("log_tag", "jresult: " + result + "finish");
        jArray = new JSONObject(result);

    } catch (JSONException e) {
        Log.e("log_tag", "Error parsing data "+e.toString());
    }
    return jArray;
}

Is there an error somewhere in the JSON?

Raghunandan · Accepted Answer · 2013-06-28 14:03:14Z

4

[ represents a json array node

{ represents a json object node

    JSONArray jArray  = new JSONArray(result);
    return jArray;

Also you can have one try block instead of many.

answered Jun 28, 2013 at 14:03

Raghunandan

134k26 gold badges232 silver badges261 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Josh Boothe Over a year ago

Thanks! Best explanation here, so thank you! I will accept it when it allows :)

Raghunandan Over a year ago

@JoshBoothe you are welcome. also you can have one try block and have many catch block based on exception hierarchy

faffaffaff · Accepted Answer · 2013-06-28 14:03:04Z

0

It's an array, so you need to do "new JSONArray()" instead of "new JSONObject()".

answered Jun 28, 2013 at 14:03

faffaffaff

3,57918 silver badges28 bronze badges

Comments

nsfyn55 · Accepted Answer · 2013-06-28 14:04:03Z

0

when a json string starts with [ the it will be treated as a JSONArray you need to do new JSONArray()

answered Jun 28, 2013 at 14:04

nsfyn55

15.4k9 gold badges54 silver badges79 bronze badges

Collectives™ on Stack Overflow

JSONObject not being correctly created (possible JSON formatting error?)

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related