Exception while parsing JSON data

Question

I am parsing JSON data from this link:

As the data is too large, I am posting only the (problem) relevant data here:

{ "info": "[{\"outlet\":{\"__v\":1,\"_id\":\"5316d59326b019ee59000026\",\"photos\":[],\"twyst_meta\":{\"recommend_list\":[],\"reviews\":[]},\"outlet_meta\":{\"links\":[],\"status\":\"active\",\"accounts\":[\"531574eeae738d654c00000a\",\"535a1a1e827651ca03000122\"]},\"attributes\":{\"dine_in\":true,\"outdoor\":true,\"air_conditioning\":\"Not Available\",\"parking\":\"Available\",\"reservation\":\"Not Required\",\"wifi\":\"Not Available\",\"tags\":[\"Pizza\",\"Pasta\",\"Italian\",\"Galleria\",\"DLF Phase 4\"],\"payment_options\":[\"cash\",\"amex\",\"visa\",\"master\"],\"cuisines\":[\"Italian\",\"Pizza\",\"Gelato\"],\"timings\":\"11 AM - 11 PM\",\"cost_for_two\":{\"min\":3,\"max\":4}},\"links\":{\"other_urls\":[],\"youtube_url\":\"\",\"twitter_url\":\"\",\"facebook_url\":\"https://www.facebook.com/pages/Crusty-gourmet-pizza-more/390843954384130\",\"website_url\":\"\"},\"contact\":{\"emails\":{\"email\":\"[email protected]\",\"person\":\"\"},\"phones\":{\"number\":\"\",\"landline\":\"02355156616\",\"reg_mobile\":[{\"_id\":\"4639320000e4532779b17de7\",\"num\":\"9134000410\"}

This is not the end of JSON data, it goes on...

But when the following line of code is executed, I am getting a JSONException. The e.printStackTrace() method tells me that there is typeMismatch error.

       JSONObject object = new JSONObject(str);        //1
       JSONArray array = object.getJSONArray("info");  //2

I have tried replacing the line 2 with the following code:

      JSONObject newObject = object.getJSONObject("info");

But here also I encountered the same problem.

I have checked the complete data, it is correct, no syntax error. Can anybody help me with the issue ? Can the presence of double quotes before starting of JSONArray be the root of this problem ?

Thanks.

What do you mean ? Presence of double quotes will be a problem, or it will parse, I am not doing it the correct way ? — Rajat
– Rajat, Commented May 24, 2014 at 15:54

Boann · Accepted Answer · 2014-05-24 15:53:55Z

2

info is not an array or an object. It's a string, which just happens to contain the JSON-encoded representation of another array. You will have to get that string and run it through a second JSON decoding pass:

JSONArray array = new JSONArray(object.getString("info"));

answered May 24, 2014 at 15:53

Boann

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3 Comments

Mgamerz Over a year ago

This is what I was trying to say in the above comment. Mobile is too slow :)

Boann Over a year ago

@Rajat The first element of the array is an object, so you can call getJSONObject(0) on the array to get the object, and then use its keys, if that's what you mean.

Rajat Over a year ago

@Boann: OK I get it. I misunderstood initially.

Boann · Accepted Answer · 2014-05-24 16:10:38Z

0

JSONObject json;
Object     info;
JSONArray  infoJsonArray;
JSONObject infoObject;
json = new JSONObject(str);

Object info= json.get("info");
if (info instanceof JSONArray) {
// It's an array
infoJsonArray = (JSONArray)info;
}
else if (info instanceof JSONObject) {
// It's an object
infoObject = (JSONObject)info;
} else {
// It's something else, like a string or number
}

edited May 24, 2014 at 16:10

Boann

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answered May 24, 2014 at 16:01

Emanuel

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