JQUERY is not loading properly or is it my code that is wrong?

Question

I am trying to use jQuery methods for ajax to create a dropdown menu. Using this jQuery code:

JAVASCRIPT:

<SCRIPT>
$("select[name='carid']").on("change", function() {
  $.post(
    "execute.php",
    { carid: $("#carid").val() },
    function(data) {
      $("available").append(data);
    }
  );
});
</SCRIPT>

execute.php :

<?php
$carid = $_POST['carid'];
$result = mysql_query("SELECT mID, mName FROM Model WHERE cID = '$carid' ");
$select = "<select>";

while ($row = mysql_fetch_array($result))
{
    $mID = $row['mID'];
    $mName = $row['mName'];
    $select+= "<option value='".$mID."'>" .$mName. "</option>";
}
$select += "</select>"
echo $select;

?>

Unfortunately I am getting this error on loading in console event.returnValue is deprecated. Please use the standard event.preventDefault() instead. Any help is much appreciated.

add your script within a [dom ready handler](http://learn.jquery.com/using-jquery-core/document-ready/) like `$(function(){ //your script })`.... the script looks fine... whether there is any error in your console — Arun P Johny, Dec 22 '13 at 12:03
**Danger**: You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). You are also **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that a modern API would make it easier to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, Dec 22 '13 at 12:07
@user3104737 what is `available`? `
`'s `id` or `class` or what? — Kamran Ahmed, Dec 22 '13 at 12:11
@user3104737 please check my answer below, then.. It should solve your problem. — Kamran Ahmed, Dec 22 '13 at 12:15

score 0 · Answer 1 · answered Dec 22 '13 at 12:05

0

replace this line and try again

 $select += "</select>";

and if $("available").append(data); in this line available is class than use .available and if it an id than use #available

answered Dec 22 '13 at 12:05

Bhupendra

248
2
7

I tried this but it didn't work :S – user3104737 Dec 22 '13 at 12:09

Kamran Ahmed · Accepted Answer · 2013-12-22T12:18:09.720

The problem is here I think:

$("available").append(data);

What is available here? It won't append data any where.

Since you haven't shown us your html so assuming that it may be an element with id or class set to available so it should be $(".available").append(data); or $("available").append(data); depending upon the element to which you want to append.

UPDATE

since you specified that it's a <div id="available"> so try $("#available").append(data); instead of $("available").append(data);

Here is the complete script

<script>
$("select[name='carid']").on("change", function() {
  $.post(
    "execute.php",
    { carid: $("#carid").val() },
    function(data) {
      console.log(data); // just for testing
      $("#available").append(data);
    }
  );
});
</script>

If that doesn't work either, check your console and post the output here.

score 0 · Answer 3 · edited May 23 '17 at 11:56

0

See here

event.returnValue is deprecated. Please use the standard event.preventDefault() instead

This is just a warning and can be safely ignored for the time being. It has been changed in jQuery 1.11

edited May 23 '17 at 11:56

Community

1
1

answered Dec 22 '13 at 12:24

user888734

3,797
5
36
67

JQUERY is not loading properly or is it my code that is wrong?

3 Answers3