JSONException: Value of type java.lang.String cann

Question

I have a JSON file with 2 JSON-Arrays in it: One Array for routes and one Array for sights.

A route should consist of several sights where the user gets navigated to. Unfortunately I am getting the error:

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

Here are my variables and the code that parses the JSON-File:

private InputStream is = null;
private String json = "";
private JSONObject jObj = null;

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();
    // hier habe ich das JSON-File als String
    json = sb.toString();
    Log.i("JSON Parser", 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;
}

Log.i("JSON Parser", json); shows me that at the beginning of the generated string there is a strange sign:

but the error happens here:

try {
    jObj = new JSONObject(json);
} catch (JSONException e) {
    Log.e("JSON Parser", "Error parsing data " + e.toString());
}

04-22 14:01:05.043: E/JSON Parser(5868): Error parsing data org.json.JSONException: Value //STRANGE SIGN HERE // of type java.lang.String cannot be converted to JSONObject

anybody has a clue on how to get rid of these signs in order to create the JSONObject?

Answer 1

Had the same problem for few days. Found a solution at last. The PHP server returned some unseen characters which you could not see in the LOG or in System.out.

So the solution was that i tried to substring my json String one by one and when i came to substring(3) the error went away.

BTW. i used UTF-8 encoding on both sides. PHP side: header('Content-type=application/json; charset=utf-8');

JAVA side: BufferedReader reader = new BufferedReader(new InputStreamReader(is, "utf-8"), 8);

So try the solution one by one 1,2,3,4...! Hope it helps you guys!

try {
            jObj = new JSONObject(json.substring(3));
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data [" + e.getMessage()+"] "+json);
        }

Answer 2

Reason is some un-wanted characters was added when you compose the String. The temp solution is

return new JSONObject(json.substring(json.indexOf("{"), json.lastIndexOf("}") + 1));

But try to remove hidden characters on source String.

Answer 3

if value of the Key is coming as String and you want to convert it to JSONObject,

First take your key.value into a String variable like

 String data = yourResponse.yourKey;

then convert into JSONArray

JSONObject myObj=new JSONObject(data);

Answer 4

I think the problem may be in the charset that you are trying to use. It is probably best to use UTF-8 instead of iso-8859-1.

Also open whatever file is being used for your InputStream and make sure no special characters were accidentally inserted. Sometimes you have to specifically tell your editor to display hidden / special characters.

Answer 5

In my case the problem occured from php file. It gave unwanted characters.That is why a json parsing problem occured.

Then I paste my php code in Notepad++ and select Encode in utf-8 without BOM from Encoding tab and running this code-

My problem gone away.

Answer 6

see this http://stleary.github.io/JSON-java/org/json/JSONObject.html#JSONObject-java.lang.String-

JSONObject

public JSONObject(java.lang.String source)
           throws JSONException

Construct a JSONObject from a source JSON text string. This is the most commonly used` JSONObject constructor.

Parameters:
    source - `A string beginning with { (left brace) and ending with } (right brace).` 
Throws:
    JSONException - If there is a syntax error in the source string or a duplicated key.

you try to use some thing like:

new JSONObject("{your string}")