Codeigniter - Ajax Return Data from Controller/Model question

Question

This is my first foray into using ajax with a search input box. My ajax works fine however I do not know how to properly return the search data from the model and controller.

The controller: The "You have found twiiter ID's" works fine, however the $num does not come through

function search()
{         
    $this->form_validation->set_rules('twit', 'Search', 'required|trim|min_length[2]|max_length[25]');//check
    if ($this->form_validation->run() == TRUE) {
    $this->load->model('twit_model');
    $query = $this->twit_model->entity();
    if(!$query = $this->twit_model->entity())
    {  
    echo "The name does not exist";
        }else{     
        echo "<h3>You found $num Twitter ID's</h3>";    //this shows up in the view           
        echo "<li class=\"list1\">$twit_id - $name</li> ";   //the "-" shows up in the view        
        }               
    }
}

The model.

function entity()
{
    $twit = $this->input->post('twit');
    $this->db->select('id, name, state, party, twit_id, job');
    $this->db->like('name', $twit);
    $this->db->or_like('state', $twit);
    $this->db->or_like('party', $twit);
    $this->db->or_like('twit_id', $twit);
    $this->db->or_like('job', $twit);
    $query = $this->db->get('twit');
    $num = $query->num_rows();
    if($query->num_rows() > 0) {
         $data - array(
         $query,
         $num
         );
        return $data;
    }
  }

The basics of the controller and model work fine, I realize at the end of both I am not passing the data from the model to the controller correctly

The jquery

$(function() {
$('#display').hide();   
$('#name_error').hide(); 
    $('#submit').click(function() { // could be #form .submit

      var twit = $("#twitter_search").val();
        if (twit == "") {
          $("label#name_error").show();
          $("input#twitter_search").focus();
          return false;
        }  
var datastring = $('#form').serialize();         
$.ajax({
   url: "<?php echo site_url('twitter/search'); ?>",
   type: "POST",
   data: datastring, 
    success: function(msg) {
        $('#display').html(msg).show(3000);            
        }
    }); 
    return false;       

});
});

The jquery works just fine, alerts /firebug tell me I am passing the data to the controller. I just dont know how to write the passing of the variable data to the model->controller->view

Thanks for reading

icchanobot · Accepted Answer · 2011-04-14 01:52:47Z

3

It kinda just looks like you are using variables out of scope.

in your model you should change this:

$data - array(
     $query,
     $num
     );

to something like this:

$data = array('query' => $query, 'count' => $num, 'twit_id'=>$twit);

then in your controller change this:

echo "<h3>You found $num Twitter ID's</h3>";           
echo "<li class=\"list1\">$twit_id - $name</li> ";

into something like this:

echo "<h3>You found ".$query['count']." Twitter ID's</h3>"; 
echo "<li class=\"list1\">".$query['twit_id']." - ".$query['query']."</li> ";

answered Apr 14, 2011 at 1:52

icchanobot

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1 Comment

Brad Over a year ago

Thank you icchanobot, thats very helpful. I appreciate it

yangqi · Accepted Answer · 2011-04-13 16:31:29Z

1

$data - array(
     $query,
     $num
     );

did you make a typo here? should it be $data = array? besides, the variable $num inside the search function is neither b defined nor returned.

answered Apr 13, 2011 at 16:31

yangqi

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3 Comments

Brad Over a year ago

Yes that is a typo on my part. $num is $num = $query->num_rows(); in the model. But still have the same problem converting $query & $num to variables

yangqi Over a year ago

yeah, $num only in the model, but it's not returned to the search function. So you can not just use the $num in your search function directly.

Brad Over a year ago

I can pass data through the controller to a view, but I have never converted "data" to a variable in the controller. How is that done?

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