php - 在选择上显示结果

Question

我正在使用下面的代码，它显示???仅代替结果。您能帮我纠正下面表格中的错误吗？消息显示空白结果，如 ??格式。有什么帮助吗？ --

<table width="100%" border="0" cellspacing="2" cellpadding="2">
  <tr>
    <td width="12%">&nbsp;</td>
    <td width="2%">&nbsp;</td>
    <td width="2%">&nbsp;</td>
    <td width="2%">&nbsp;</td>
    <td width="16%">&nbsp;</td>
    <td width="66%">&nbsp;</td>
  </tr>
  <tr>
    <td><img src="images/australia1.jpg" width="132" height="70"></td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td><select name="select">
      <option>Select Course Type</option>
      <option>Under Graduate</option>
      <option>Post Graduate</option>
    </select>
    </td>
    <td><div id="message"></div>​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​</td>
  </tr>
  <tr>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
  </tr>

</table>
<script type="text/javascript" language="javascript">
var messages = {
    Under Graduate: 'UG text is here',
    Post Graduate: 'PG text is here'
}
$('select').change(function() {
    $('#message').text(messages[$(this).val()]);
});​
</script>

Answer 1

<script type="text/javascript" language="javascript">
var messages = {
    under: 'UG text is here',
    post: 'PG text is here'
}
$('select').change(function() {
    if($(this).val() != '')
        $('#message').text(messages[$(this).val()]);
});​
</script>

<select name="select">
  <option value="">Select Course Type</option>
  <option value="under">Under Graduate</option>
  <option value="post">Post Graduate</option>
</select>