Jquery click no triggering

Question

I am using a while loop to create a table and I am trying to pass a unique value stored in a php variable called $i to a jquery ajax call but all I get returned it 'undefined' .

The portion of the html table (whole table is wrapped in a div called audit_content which is not part of the while loop)

  <tr>
      <td colspan="3">
          <input type="button" class="delete_submit delete_from_audit" name="image_delete" value="" 
               onClick="delete_image('<? echo $image_id; ?>','<? echo $audit_id; ?>')" />
          <input type="button" class="reset_grade control_submit ie7_reset" 
               onclick="reset_image('<? echo $image_id; ?>','<? echo $audit_id; ?>')" value="" >
          <input type="hidden" class="form_id" value="<? echo $i; ?>" >
          <input type="button" name="audit_submit"
            class="audit_submit audit_submit_btn ie7_submit" value=""; />
      </td>
  </tr> 

the Jquery

<script>
  $(document).ready(function(){
    $('#audit_content').on("click", ".audit_submit", function(){
        var form_id = $(this).next('.form_id').val();
        alert(form_id);
           //$.ajax({
           //type: "POST",
           //url: "ajax/image_audit.php",
           //data: $('#audit_form').serialize()
        //});
    });
});
</script>

Answer 1

The .form_id element is before the .audit_submit button... Try prev() instead of next() :

var form_id = $(this).prev('.form_id').val();

You could also use siblings() :

var form_id = $(this).siblings('.form_id').val();

Answer 2

<script>
  $(document).ready(function(){
    $(document).on("click", ".audit_submit", function(){
        var form_id = $(this).next('.form_id').val();
        alert(form_id);
           //$.ajax({
           //type: "POST",
           //url: "ajax/image_audit.php",
           //data: $('#audit_form').serialize()
        //});
    });
});
</script>

Answer 3

 <input type="button" name="audit_submit"
        class="audit_submit audit_submit_btn ie7_submit" value=""**---> ; <----** />

Why is there a semi colon ?? Php will consider it as the end of the statement.