How do i split a string of jsonObjects into an array

Question

I'm getting some data from the Trello API over HTTP. So an example of the response would be:

'[{"name":"asd","desc":"yes"},{"name":"xyz","desc":"no"}]'

I'm using the volley library for making the request and getting the response. Is there a way for me to get the response in the form of json objects directly instead of in a string?

If not how should I proceed?

Thanks!

one minute googling for: volley json developer.android.com/training/volley/request.html#request-json — pskink
– pskink, Commented Oct 9, 2015 at 11:12
i could not really understand your question , but in Spring we create a model class with the fields "name" and "desc" with setter and getter. Ant We can use annotation to indicate to Spring to map json to the model. I dont know if it has helped but if u want i can give more information. — Albert Pinto
– Albert Pinto, Commented Oct 9, 2015 at 11:12
What strings? "name", "asd", "desc", "yes"? Please elaborate what you want to do. — Sufian
– Sufian, Commented Oct 9, 2015 at 11:13
Why dont you just parse the string in json array and get the json objects from it. — Shailesh Yadav
– Shailesh Yadav, Commented Oct 9, 2015 at 11:15

Sufian · Accepted Answer · 2015-10-09 11:39:31Z

4

You can use JSONArray(). And then you can use getString() so you can use all string function.

Example code:

JSONArray jsonArray = new JSONArray(responseString);
int i = 0;
while (i <jsonArray.length()) {
    JSONObject jsonObj = jsonArray.getJSONObject(i);
    String name = jsonObj.getString("name");
    String description = jsonObj.getString("desc");
    //TODO create your Java object and store these strings into it.
    i++;
}

edited Oct 9, 2015 at 11:39

Sufian

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answered Oct 9, 2015 at 11:13

alican akyol

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2 Comments

alican akyol Over a year ago

Sufian thanks:) JSONObject reader = new JSONObject(responseString); String res = reader.getString("TAG"); Like JSONArray.

alican akyol Over a year ago

Thank you :) It is better.

Egos Zhang · Accepted Answer · 2015-10-09 11:55:27Z

2

use volley can solve easily.

JsonArrayRequest jsonArrayRequest = new JsonArrayRequest(url,new Response.Listener<JSONArray>(){

        @Override
        public void onResponse(JSONArray response) {
             //the response is JsonArray       
        }
    },new Response.ErrorListener(){

        @Override
        public void onErrorResponse(VolleyError error) {

        }
});

answered Oct 9, 2015 at 11:55

Egos Zhang

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