Jquery return value from database

Question

I have a script which allows user to like(+1) a record. Everything is working, but I don't know how to show a new value from DB after button is pressed (without refresh). I'm using codeigniter and I'm not really know a lot about jquery.

My script:

    <script type="text/javascript">
  var base_url = '<?php echo base_url();?>';
 jQuery(document).ready(function(){
    $('.like').click(function() {
    var id = $(this).attr("uid");
    $.ajax({
            url: base_url+'quotes/quotes_rating',
            type: 'GET',
            data: {'id': id}
     }); 
        });  

});
</script>

My controller:

public function quotes_rating() {
    $id = $this->input->get('id');
    $this->quotes_model->write_likes($id);
 }

Model function write_likes() saves records into DB. This is my model function which shows how many likes has a record:

public function all_likes($id)
{
 $sql = "select COALESCE(sum(like),0) as likes from rating WHERE record_id = '{$id}'";
  $likes = $this->db->query($sql)->row_array();

  return $allikes['likes'];
}

What I have to change in this code?
When user will click .like button, he shoud see a new value from database. Thank you!!!

Senad Meškin · Accepted Answer · 2015-03-10 22:05:49Z

1

jQuery(document).ready(function(){
    $('.like').click(function() {
       var $this = $(this);
       $.ajax({
            url: base_url+'quotes/quotes_rating',
            type: 'GET',
            data: {'id': id},
            dataType: 'html',
            success: function(d){
              $this.html(d); //will set html of clicked link
            }, error: function(){ //code for error }
     }); 
});

answered Mar 10, 2015 at 22:05

Senad Meškin

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Juraj Carnogursky · Accepted Answer · 2015-03-10 22:12:02Z

0

<script type="text/javascript">
var base_url = '<?php echo base_url();?>';
jQuery(document).ready(function(){
    $('.like').click(function() {
        var id = $(this).attr("uid");
        $.ajax({
            url: base_url+'quotes/quotes_rating',
            type: 'GET',
            data: {'id': id}
            success: function(data) {
                $("someelementyouwanttoseelikesin").html(data);
                //data will contain anything you echo in your php script!
            }
        });
    });  
});
</script>

answered Mar 10, 2015 at 22:12

Juraj Carnogursky

3552 silver badges16 bronze badges

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