How to convert result_array() to result() object in codeigniter

Question

I need JSON output like this,

{"restaurant":[{"id":"89","owner_id":"2161","name":"Bakso ",...}{"id":"90","owner_id":"3400","name":"Soto"}]}

And I do using this query and work perfectly:

$latest_rest = $this->Restaurants_model->getLatestRestaurants($limit)->result();
$jsonObject = array('restaurant'=>$latest_rest);

but i have query with result_array(), because I must filtering this data through function that return result_array();

how do I convert result_array() to result() in CodeIgniter?

if you want response in object then return $query->result(); only — Maksud Mansuri
– Maksud Mansuri, Commented Mar 30, 2017 at 12:56
Why don't you just call result() instead of result_array()? Also, if you are just encoding it as JSON anyway, why does it matter if you use result() or result_array()? What problem are you trying to solve here exactly? What output are you getting compared to the output you want? — gen_Eric
– gen_Eric, Commented Mar 30, 2017 at 14:28
I was solved the problem, result() or result_array(), json outputs are same like above, — Latief Anwar
– Latief Anwar, Commented Mar 30, 2017 at 15:39

Saroj Shrestha · Accepted Answer · 2017-03-30 14:24:52Z

3

If it is mandatory to get result in array in beginning then to get json data you can first encode data into json and later can decode as:

$object = json_decode(json_encode($array));

And, if it is not necessary in beginning then you can get result in object directly as:

$this->db->select('*')->from..........;
$query = $this->db->get();
return $query->result(); // instead of result_array();

answered Mar 30, 2017 at 14:24

Saroj Shrestha

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