Ajax没有调用php文件？

Question

I am trying to get the price of Item after selecting the item in a dropdown. I have used ajax calls but my code is not going inside $.ajax({.... What is wrong with my code?

I have used alert in the php file but nothing displays, which means my code is not calling php file.

<script type='text/javascript'>
            $("#puja_name2").on('change',function()
            {   
                var id=$(this).val();
                var data = 'id='+ id;
                $.ajax({
                    type: "POST",
                    url: "../ajax_price.php",
                    data: data,
                    cache: false,
                    success: function(html)
                    {
                        $("#puja_price2").html(data);
                
                    }
                
                });
            });
    </script>

<div class="form-group">
  <label>
  Puja Name<span class="font-red">* </span>:
  </label>
                                
  <select class="form-control" name="puja_name2" id="puja_name2"  data-validetta="required">
   <option value="">Select Puja Name</option>
    <?php
                                                          
    $SQL_STATEMENT_puja = $DatabaseCo->dbLink->query("SELECT * FROM puja_type ");
   while($DatabaseCo->dbRow = mysqli_fetch_object($SQL_STATEMENT_puja)){
    ?>
 <option value="<?php echo $DatabaseCo->dbRow->puja_id; ?>" ><?php echo $DatabaseCo->dbRow->puja_name; ?></option>
   <?php                                 
    } 
                                ?>
 </select>
 </div>
                        
 <div class="form-group">
    <label> Price <span class="font-red">* </span>:</label>
  <select  id="puja_price2" >
                                    
    </select>                                      
 </div>

</div>

Answer 1

You must change in jquery puja_name to puja_id:

$("#puja_id").on('change',function()