How to parse json and check the value of it?

Question

I'm trying to get a value of my json. First here is my json.

{
    "status": "true",
    "userData": {
        "user_id": "USER001",
        "username": "boby",
        "password": "c83e4046a7c5d3c4bf4c292e1e6ec681",
        "fullname": "Boby Kurniawan",
        "phone": "089688xxxxxxx",
        "profilpic": null,
        "status": null,
        "flag": 1,
        "tipe": "TP001",
        "Deskripsi": "Ini tentang gua",
        "Email": "boby@mail.com"
    }
}

Ok, for the first there will be 2 json, if the login success the json will look like above, but when it fails it will return

 { "status" : "false" }

So I need to parse it right? so I add this line

JSONObject jsonObj = new JSONObject(result);
JSONArray status = jsonObj.getJSONArray("status");

but I get this error

unhandled exception : org.json.JSONException

how can I fix it?

Possible duplicate of [How to parse JSON in Android](https://stackoverflow.com/questions/9605913/how-to-parse-json-in-android) — ADM, Dec 27 '17 at 09:17

Ratilal Chopda · Accepted Answer · 2017-12-27T09:31:17.737

unhandled exception : org.json.JSONException

ANS : you are getting that error Because status is string not a JSON Array

You should use jsonObj.getString("status") instead of jsonObj.getJSONArray("status");

Try this

try 
{


    JSONObject jsonObj = new JSONObject(result);
    String status = jsonObj.getString("status");

    Log.i("status", "->" + status);

    JSONObject jsonObject = jsonObj.getJSONObject("userData");

    Log.i("user_id", "->" + jsonObject.getString("user_id"));
    Log.i("username", "->" + jsonObject.getString("username"));
    Log.i("password", "->" + jsonObject.getString("password"));
    Log.i("fullname", "->" + jsonObject.getString("fullname"));
    Log.i("phone", "->" + jsonObject.getString("phone"));
    Log.i("profilpic", "->" + jsonObject.getString("profilpic"));
    Log.i("status", "->" + jsonObject.getString("status"));
    Log.i("flag", "->" + jsonObject.getString("flag"));
    Log.i("tipe", "->" + jsonObject.getString("tipe"));
    Log.i("Deskripsi", "->" + jsonObject.getString("Deskripsi"));
    Log.i("Email", "->" + jsonObject.getString("Email"));

} 
catch (JSONException e) 
{
    // TODO Auto-generated catch block
    e.printStackTrace();
}

OUTPUT

@YVS1102 Happy to help you. – Ratilal Chopda Dec 27 '17 at 09:37 — Ratilal Chopda, Dec 27 '17 at 09:37

score 1 · Answer 2 · answered Dec 27 '17 at 09:13

1

unhandled exception : org.json.JSONException

=> Because you are trying to get JSON Array, where as it's not JSON array but a field only.

Try:

String status = jsonObj.getString("status");

answered Dec 27 '17 at 09:13

Paresh Mayani

127,700
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295

score 1 · Answer 3 · answered Dec 27 '17 at 09:15

   try {
            JSONObject jsonObj = new JSONObject(result);
            String status = jsonObj.getString("status");
            JSONObject userData = jsonObj.getJSONObject("userData");
        } catch (JSONException e) {
            e.printStackTrace();
        }

score 0 · Answer 4 · answered Dec 27 '17 at 09:22

Here "status" is not an Array, its String. Json Array should enclosed by [ ]. To get value of "status" you should use,

String status = jsonObj.getString("status");

Then you need to check status value. Remember,

Json object enclosed by { }

Json array enclosed by [ ]

Also use try-catch block while fetching data,

try{
    String status = jsonObj.getString("status");
} catch (JSONException e) {
     e.printStackTrace();
}

fuzzKitty · Answer 5 · 2017-12-27T09:26:07.520

-1

you need to surround with a try/catch:

    try {
            JSONObject jsonObj = new JSONObject(result);
            String status = jsonObj.optString("status");
        } catch (JSONException e) {
            e.printStackTrace();
        }

edited Dec 27 '17 at 09:26

answered Dec 27 '17 at 09:19

fuzzKitty

334
3
7

wrong ans check it **status** is a string not a json array – AskNilesh Dec 27 '17 at 09:20
1

you're right... I forgot to change it after the copy paste :) fixed... also, best to use optString() and not getString() – fuzzKitty Dec 27 '17 at 09:26

How to parse json and check the value of it?

5 Answers5