why this jquery script is not working Code igniter

Question

i am working in a code igniter .. i am making five rows ..actually i am doing is that the options which are display in the 2nd select box based on the first select box .. if i dont make five rows with the loop then script is working fine but if i put them in a loop ..selection dont work .. in firebug its give me response false and saying that

  localhost/......./controller/get_items/undefined...

i dont know whats wrong in that code

<?php 
for ($i = 0; $i < 5; $i++) {
?>
<?php echo form_dropdown('cat_id', $records2, '#', "id='category_".$i."'");?>

<?php echo form_dropdown('item_id', $records3, '#', "id='items_".$i."'"); ?>

<script type="text/javascript">
// <![CDATA[
$(document).ready(function()
{  
    for (var i= 0; i<5; i++)
    {
        $('#category_'+ i).change(function()
        {
            $('#items_'+ i > option").remove(); 
            var category_id = $('#category_'+ i).val(); 

            $.ajax({
                type    : "POST",
                url     : "stockInController/get_Items/"+category_id, 
                success : function(items)
                {
                    $.each(items,function(item_id,item_name) 
                    {
                        var opt = $('<option />'); 
                        opt.val(item_id);
                        opt.text(item_name);
                        $('#items_'+ i).append(opt); 
                    });
                }

            });
        });
    }
}

@andy why it is not getting a value ..i think i am doing all right .. see my updated question — user1972143, Jan 18 '13 at 13:33
your script is totally wrong because you are initializing function in a loop — Muhammad Raheel, Jan 18 '13 at 13:40
what is this ` $('#items_'+ i > option").remove(); ` quotations alert!! — mamdouh alramadan, Jan 18 '13 at 13:41
@raheel shan .. so what should i do ? please can u rearrange it or so i get to know how to make it right — user1972143, Jan 18 '13 at 13:41
@raheel shah i have a form where i display 5 rows in a loop n in each row there r 2 dropdown fields n three simple text input boxes so when i was working in simple one simple i mean when the form was not in a loop then the two dropdown boxes working fine which is what that "the second dropdown options based on the selection of previous dropdown select box" so then i put them in a loop..and attach "$i" with the id of every select box because i have to given each of the row a different ids because if i dont do so then only the select boxes of first row is successfully working not other four ? — user1972143, Jan 18 '13 at 13:54
so then as i was getting data from controller through jquery ... so i have to given different ids in jquery also in order for jquery to catch the values and send to controller and then give back the response ..which i did ..did u get my problem ? — user1972143, Jan 18 '13 at 13:57
Unless your doing XHTML you can get rid of the cdata wrapper.http://stackoverflow.com/questions/66837/when-is-a-cdata-section-necessary-within-a-script-tag — jholloman, Jan 18 '13 at 13:59
@raheel shah actually it is like that categories ->products->price->quantity->total — user1972143, Jan 18 '13 at 14:01
so it means on category selection you are fetching products then on products selection you are selecting price , on price you are selecting quantity etc.. — Muhammad Raheel, Jan 18 '13 at 14:02
@raheel no no ... i am just fetching products based on the categories ... the price and other things are for the user .. he manually type it — user1972143, Jan 18 '13 at 14:09
`for ($i = 0; $i < 5; $i++) {` where this `for` ends? I don't see the `` — Tomás, Jan 18 '13 at 14:50

Muhammad Raheel · Answer 1 · 2013-01-18T18:12:48.530

As far as i understand your problem here is the solution.

Yor function should look like this

function getCategories(obj)
{

    var category_id = obj.id      ;
    $.ajax({
         type    : "POST",
         url     : "stockInController/get_Items/"+category_id, 
         success : function(data)
         {
             $('#'+id).html(data);// here you should provide 2nd dropdown id so that it can be replaced
         }
    });
}

Controller method

function get_Items()
{
    $id =   $this->uri->segment(3);
    $data['result'] = $this->somemodel->getProducts($id);
        $this->load->view('options',$data);
}

View

options.php

foreach($result as $row){
?> <option value = "<?php echo $row->item_id?>"><?php echo $row->item_name?></option><?php    
}

And on each dropdown of categories you should call this function.

<?php echo form_dropdown('cat_id', $records2, '#', "id='category_".$i."' onclick = "getCategories(this)"");?>

why this jquery script is not working Code igniter

1 Answers1