Codeigniter and AJAX retrieving/getting data from database error

Question

I always land into the error function in ajax, what is the problem here?

I want to retrieve a img url stored in the database and append it using jquery to show my img gallery. But before I can do any of this, I have to make sure it always land on the success function, but I always land on error function... so I can not proceed and I am stuck here,

UPDATE:

check the console, it says SyntaxError: missing : after property id on success:function(result)

My script:

    function load_contents(track_page)
    {
        $('#loading').show();

        $.ajax({
            url:'<?php echo base_url('gallery/load_design');?>',
            type:'GET',
            dataType:'json',
            success:function(result)
            {
                alert("success");
            },
            error:function()
            {
                alert("failed"); //it always show the alert failed.

            }
        });
    }

My controller:

function load_design()
{
    $this->load->model('design');
    $this->load-model('profile');

    $user_id = $this->profile->retrieve_userid();

    $json = $this->design->load_gallery($user_id->id);
    echo json_encode($json);
}

My model:

    function load_gallery($user_id)
    {
        $query = $this->db->query("SELECT * from designs WHERE user_id = '".$user_id."' LIMIT 9");
        return $query->result_array();
    }

Alex K. · Accepted Answer · 2017-01-16 19:46:43Z

0

You will land in success function when server answer to you with 2xx http status.
If you always land on error function you have an error in codeigneter.
Try to get http status from you server (browser console will help you) and it will help you to understand where is problem.

answered Jan 16, 2017 at 19:46

Alex K.

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1 Comment

pidrowlist Over a year ago

It says SyntaxError: missing : after property id on success:function(result)

Urshit Sheth · Accepted Answer · 2017-06-21 07:01:59Z

0

You should create first global variable in footer

var base_url = "<?= base_url(); ?>";

And now you need to use like this

function load_contents(track_page)
    {
        $('#loading').show();
        var ref = new Date(); 
        var url = base_url + "gallery/load_design?=" + ref.getTime();
        $.ajax({
            url:url,
            type:'GET',
            cache: false,
            success:function(result)
            {
                alert("success");
            },
            error:function()
            {
                alert("failed"); //it always show the alert failed.

            }
        });
    }

answered Jun 21, 2017 at 7:01

Urshit Sheth

474 bronze badges

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