Question

I have a a form:

<form id="deletesubmit" style="display:inline" >
    <input  style="width:50px" type="text" id="delcustomerid" name="delcustomerid" value="'.$row['customersid'].'">
    <button type="submit" class="table-button ui-state-default ui-corner-all" title="delete"><span class="ui-icon ui-icon-trash"></span></button>
</form>

The form gets the customers id and inserts it as value. It shows the correct customer is for that row everything is fine. Then when i post the form via ajax somehow it posts the id of a diffent row. This is the script:

$("form#deletesubmit").submit(function () {

    var delcustomerid = $('#delcustomerid').attr('value');
    $.ajax({
        type: "POST",
        url: "delete/process.php",
        data: "delcustomerid=" + delcustomerid,
        success: refreshTable
    });
    return false;
});
});

And finally here is the php to post the form:

<?php include("../../config/config.php"); ?>
<?php          
    $deleteid  = htmlspecialchars(trim($_POST['delcustomerid']));
    mysql_send("DELETE FROM customers where id='$deleteid'");
?>

I have tested it without the ajax and it works fine. There must be something missing. It is not posting the correct value. Spent days trying to work it out.

Answer 1

By using attr('value') you are pulling the original value which may not what you want; use .val() instead. Also, you can setup your call a little easier:

$("form#deletesubmit").submit(function() {
  var delcustomerid  = $(this).find('#delcustomerid').val();
  $.post( "delete/process.php", { delcustomerid: delcustomerid }, refreshTable );
  return false;
});

Finally, make sure that your PHP is outputting the right id by viewing the source of the generated HTML.

Also, in the PHP: Don’t use htmlspecialchars to escape mysql. Since this supposed to be an integer, you can just use int:

$deleteid  = (int) trim($_POST['delcustomerid']);
mysql_send("DELETE FROM customers where id='$deleteid'");

Keep in mind, that an array (if someone passed deleteid[]= to this page) would evaluate to 1. So either build a test into it, or just make sure you don’t have an id = 1 left in that table.