Error parsing data android with subarray

Question

This is the JSON result of my query.

{
    "_id":{"$id":"551fb585ecba12c819000032"},
    "nome":"Google","loc":[-122.083983,37.422969],
    "icona":1,
    "istituzione_id":{"$id":"551fb556ecba12c819000031"}
    }
    {
    "_id":{"$id":"5520fe2becba12c003000029"},
    "nome":"Oracle","loc":[-122.262168,37.531595],
    "icona":1,
    "istituzione_id":{"$id":"551fb556ecba12c819000031"}
    }

I've tried to parse JSON result in this manner:

try {
BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;

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

           is.close();
            result = sb.toString();
                    Log.e("log_a_result", result);

        } catch (Exception e)
                {
                    Log.e("log_tag_convert", "Error converting result" + e.toString());
                }

        //-----PARSER----
        try {
            JSONArray jArray = new JSONArray(result);
            for (int i = -1; i < jArray.length() - 1; i++) {
                JSONObject json_data = jArray.getJSONObject(i);
                Log.i("log_tag", "id:" + json_data.getString("_id"));
            }
        }
        catch (JSONException e)
             {
            Log.e("log_tag_parsing", "Error parsing data" + e.toString());
             }

But after run the app, in my logcat i found error:

log_tag_parsing﹕ Error parsing dataorg.json.JSONException: Value {"icona":1,"loc":[-122.083983,37.422969],"nome":"Google","istituzione_id":{"$id":"551fb556ecba12c819000031"},"_id":{"$id":"551fb585ecba12c819000032"}} of type org.json.JSONObject cannot be converted to JSONArray

I have thought that the problem was about the subarray loc:[....] maybe.

I can't reach the solution. Someone can help me ?

Trinimon · Accepted Answer · 2015-04-07 12:00:12Z

1

You're missing a comma between objects; and use [] for arrays and {} for objects.

Next time use jsonlint.com to validate your JSON first.

edited Apr 7, 2015 at 12:00

Trinimon

14k9 gold badges46 silver badges61 bronze badges

answered Apr 7, 2015 at 11:55

MetalDragon

365 bronze badges

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

Comments

Fahim · Accepted Answer · 2015-04-07 11:54:34Z

0

org.json.JSONObject cannot be converted to JSONArray

The answer is in your exception. You are trying to convert jsonobject to jsonArray. Convert your result into JsonObject

answered Apr 7, 2015 at 11:54

Fahim

12.4k5 gold badges42 silver badges57 bronze badges

Comments

Sachin Gupta · Accepted Answer · 2015-04-07 11:55:08Z

0

see here you are getting JSONObject as a result. but while parsing, you are expecting JSONArray.

Before parsing First add one check to the result.

E.g.

if (resultObj instanceof JSONArray) {
    // process Array
}
else if (resultObj instanceof JSONObject) {
    // processObj
}

answered Apr 7, 2015 at 11:55

Sachin Gupta

8,4684 gold badges36 silver badges47 bronze badges

Comments

Pankaj · Accepted Answer · 2015-04-07 12:17:07Z

0

Why your for loop have starting value as -1 and change the Log printing line to. See code below:

//-----PARSER----
        try {
            JSONArray jArray = new JSONArray(result);
            for (int i = -1; i < jArray.length() - 1; i++) {   //this line
                JSONObject json_data = jArray.getJSONObject(i);
                Log.i("log_tag", "id:" + json_data.getString("_id"));//this line
            }
        }

change both mentioned line to as:

//-----PARSER----
            try {
                JSONArray jArray = new JSONArray(result);
                for (int i = 0; i < jArray.length(); i++) {   //this line
                    JSONObject json_data = jArray.getJSONObject(i);
                    Log.i("log_tag", "id:" + json_data.getString("$id"));
                }
            }

edited Apr 7, 2015 at 12:17

answered Apr 7, 2015 at 12:03

Pankaj

8,0787 gold badges46 silver badges69 bronze badges

1 Comment

Pankaj Over a year ago

You have to change at 2 different lines. For the Log printing line the "_id" you have to change it to "$id"

Mina Fawzy · Accepted Answer · 2015-04-07 12:38:58Z

0

Json array syntax should be like this one

employees is array name

[] // for array that contains objects {} between one object and another ,

"employees":[
     {"firstName":"John", "lastName":"Doe"}, 
     {"firstName":"Anna", "lastName":"Smith"}, 
     {"firstName":"Peter","lastName":"Jones"}
 ]

create correct json array

EDIT

to generate array of json

$a = array(
        array('foo' => 'bar'),
        array('foo' => 'baz'));
$json = json_encode($a);

to given

[{"foo":"bar"},{"foo":"baz"}]

2 Comments

Motosega Over a year ago

the json is generated in my php file with this "echo json_encode($row);"

Mina Fawzy Over a year ago

make array and then encode it echo json_encode($arr); which arr is your array

Collectives™ on Stack Overflow

Error parsing data android with subarray

5 Answers 5

Comments

Comments

Comments

1 Comment

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

Comments

1 Comment

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related