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

Question

So I get that error when I try to parse the JSON Object. I have google the problem and I am still confused and can't find a solution to it. It looks like my code is correct.

This is the Android Java file

InputStream is = null;
String result = "";
protected String doInBackground(String... params) {
    try {

        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(params[0]);
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();


        is = entity.getContent();
        Log.v("mysql", "Success");
    } catch (Exception e) {
        Log.v("log_tag", "Error in http connection " + e.toString());
    }
    return null;
}

protected void onPostExecute(String result) {
    try{
        BufferedReader reader = new BufferedReader(new InputStreamReader(is,"UTF-8"));
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
        }
        is.close();

        result=sb.toString();
        Log.v("result", result);
}catch(Exception e){
        Log.e("log_tag", "Error converting result "+e.toString());
}

    try{
        JSONArray jArray = new JSONArray(result);
        for(int i=0;i<jArray.length();i++){
                JSONObject json_data = jArray.getJSONObject(i);
                Log.i("log_tag","uID: "+json_data.getInt("uID")+
                        ", uName: "+json_data.getString("uName")+
                        ", uPass: "+json_data.getString("uPass")
                );
        }
}catch(JSONException e){
        Log.e("log_tag", "Error parsing data "+e.toString());

    }

}

The code for PHP file is this

    <?php
$username = "removed";
$password = "removed";
$database = "removed";
$user = "art";
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die ("it's dead jim");

$q = mysql_query("SELECT * FROM user WHERE uName = '$user'");
while($e=mysql_fetch_assoc($q)){
        print(json_encode($e));
        $output[]=$e;
}

?>

The JSON value I receive is

{"uID":"1","uName":"art","uPass":"password"}

When I try to parse it I get:

Error parsing data org.json.JSONException: Value {"uID":"1","uName":"art","uPass":"password"} of type org.json.JSONObject cannot be converted to JSONArray

What don't you understand about the exception message? Do you know the JSON format? — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Feb 27, 2014 at 1:58
I understand the error that it can't be converted but I don't understand exactly what I am doing wrong and how to fix it. I get that I am storing it into an JSONArray but what else can I store it in to parse it? — user2574427
– user2574427, Commented Feb 27, 2014 at 2:03
You're trying to parse it as a JSONArray when it is a JSONObject. — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Feb 27, 2014 at 2:04

Sushil · Accepted Answer · 2014-02-27 02:04:33Z

2

You data in the file is in JSONObject format and you are tryong to receive it in JSONArray format which is causing the problem. The line below is the culprit:

JSONArray jArray = new JSONArray(result);

Rather, you need to do something like this:

JSONObject jObject = null;
jObject = new JSONObject(mJsonString);
JSONArray jsonImageArray = jObject.getJSONArray("imageTarget");

Hope it helps.

answered Feb 27, 2014 at 2:04

Sushil

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4 Comments

user2574427 Over a year ago

Is mJsonString equivalient to my result string? Also can you explain the "imageTarget". Finally after I code that do I just go ahead with the for loop I had?

Sushil Over a year ago

Yes mJsonString is equivalent to "result" string. For imagetarget, I have attached my Json file. You can have a look at it : { "imageTarget": [ "image1" ], "videoUrls": [ "mylink.com/video1.mp4" ], "videoOrUrl": [ "video" ] }

user2574427 Over a year ago

Thanks. Just to make sure I am understanding this properly... For the JSON file you provided I would basically have jsonImageArray = jObject.getJSONArray("imageTarget"); followed by jsonImageArray = jObject.getJSONArray("videoUrls"); etc...?

Sushil Over a year ago

yes exactly.. just the name of variable needs to be changed :) .. rather than "jsonImageArray" I used "jsonUrlArray" to avoid overwriing

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