How to get value from controller with jquery

Controller :
function auto_barang(){
$id=$this->input->post('id');
$res= $this->barang_model->get_autocomplete($id);
//foreach ($res as $rows){
// $hasil['stock']=$rows['stock'];
//}
//$this->output->set_content_type('application/json');
$this->output->set_output(json_encode($res));
}

Script :
<script>
$(document).ready(function(){
    $("#barang").change(function(){
        $.post("<?php echo base_url()."index.php/pembelian/auto_barang"; ?>",
        {
            id:$("#barang optionelected").val()
        }, function(data){
            alert(data);
        });        
});
</script>

Output :
[{"id":"2","nama":"Lenovo ideapad 200","harga_beli":"2000000","harga_jual":"2500000","stock":"5"}]

How to if I just want to take stock value? Thank you !

Try this:

PHP Code:

$this->output->set_content_type('application/json')
echo json_encode($res); 


You already have the stock in your array just assign the array stock key and value to the res array then json_encode it.

What did you Try? What did you Get? What did you Expect?

Joined CodeIgniter Community 2009.  ( Skype: insitfx )

Oh! I did not know that (set_content_type).

Is that important?

When I ajax with a json response in the controller I just json_encode the array and echo it. It has all worked fine up till now.

Should I be setting the content type as well?

Paul.

PS the gaps in my knowledge continue to amaze me :-)

The newer browser should know what type it is but I use it just to be safe.

What did you Try? What did you Get? What did you Expect?

Joined CodeIgniter Community 2009.  ( Skype: insitfx )

If you just need one field then just output that one field

$this->output->set_output(json_encode($res['stock']));

On the package it said needs Windows 7 or better. So I installed Linux.

I always do things like get result from query as array such as result_array() for multi dimension or single array like row_array() then simply
For example , in your case, I would do this
$res= $this->barang_model->get_autocomplete($id); //assume you get row_array as return
then
echo json_encode($res);
OR I usually wrap it up with success and message, data like this:
echo json_encode(array('success'=>1, 'message'=>'anything you want to read', 'data'=>$res));

My frontend at ajax,  I often using

Code:

$.post(yoururl,{optionaldata},function(result){
//you can check result data here
},'json')    //make sure you have 'json' to indicate that you need json as content type !
.done(function(result){
//I often do this here
if(result.success===1){
   console.log(JSON.stringify(result.data));//show all your json string as data
   console.log(JSON.stringify(result.data.stock)); //will show your stock value !
}
})
.fail(function(result){
//action for failures so on
});