org.json.JSONArray cannot read my JSON object

Question

Im writing a client/server app for android that sends a username and password to the server and receives a status for that user account.

My php code is like this:

$return['login'] = 'success';   
echo json_encode($return);

and the text received by android is:

{"login":"success"}

but I still get an error when I try to decode the json string to read the parameters:

JSONArray jsonArray = new JSONArray(input);

exception:

02-10 11:54:37.743: WARN/System.err(332): 
    org.json.JSONException: 
         Value {"login":"success"} of type org.json.JSONObject 
         cannot be converted to JSONArray
02-10 11:54:37.779: WARN/System.err(332): 
    at org.json.JSON.typeMismatch(JSON.java:107)

So I guess Im missing something that php should send, but after reading on json.org I just can't see what it is. I have tried adding brackets before and after, as well as wrapping it in another array like this:

$parameters['login'] = 'success';   
    $return['parameters'] = $parameters;
echo json_encode($return);

The exception seems obvious but I think there is a misunderstanding because PHP Arrays are just ordered maps. — Che Jami
– Che Jami, Commented Feb 10, 2012 at 11:29

dacwe · Accepted Answer · 2012-02-10 11:04:05Z

2

You should be using JSONObject instead of JSONArray since you sending a key-value pair.

answered Feb 10, 2012 at 11:04

dacwe

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1 Comment

Thomas Over a year ago

That explains it, in my head I was under the impression that several key-value-pairs would be an array.

Lalit Poptani · Accepted Answer · 2012-02-10 11:04:08Z

1

Its an JSONObject not an JSONArray so it should be,

JSONObject json_obj = new JSONObject(your_string);

Then you can just use json_obj to get the value,

String login = json_obj.getString("login");
Log.d("login status", login);

answered Feb 10, 2012 at 11:04

Lalit Poptani

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Comments

WarrenFaith · Accepted Answer · 2012-02-10 11:06:45Z

0

Your json is not a JSONArray, it is a JSONObject.

try {
    JSONObject o = new JSONObject("{\"login\":\"success\"}");
    System.out.println(o.getString("login"));
} catch (JSONException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

Output:

System.out  I  success

answered Feb 10, 2012 at 11:06

WarrenFaith

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