Depends how you update you select box with your code, I think FireFox
need to replace the complete select.

Look here, I think it's mentioned somewhere :

http://www.devmoz.com/blog/2007/04/04/cakephp-update-a-select-box-using-ajax/

On Oct 2, 10:41 pm, Lir <[EMAIL PROTECTED]> wrote:
> Hello I want to submit a form that has a select which was updated
> using ajax.updater
> My Ajax.updater works fine. The problems is that in FIREFOX don't
> recognize or don't get the name of the select tag and don't submit the
> value or option.
> Inside the form I have a <Div> where ajax.updater will replace the
> select results. Once it's updated I submit the form but the select
> information isn't post. at least it works fine in IE
> Thanks in  advance for any help. This is the code
>
> HTML
> <form enctype="multipart/form-data" name="uploadt" method="post"
> action="' . $form_action . '" ">
> <tr><td>Loss Seq</td>
>       <td><div id="lossResult">
>              <select name="temp"><option></option></select> //AJAX
> RESULT HERE
>               </div>
>       </td></tr>
> </form>
>
> NEW SELECT - AJAX UPDATER
> echo "<select name='submit_loss' class='upload'>"; //IT LOOKS GOOD BUT
> IT'S NOT SUBMITTED
>      echo "<option value='000' ></option>";
>         foreach ($return as $row)
>         {
>             echo "<option value='$row[LOSS]' selected>$row[LOSS]</
> option></br>";
>         }
>     echo"</select>";
>
> AFTER I SUBMIT EVERYTHING I WANT TO DISPLAY THE $_POST[SUBMIT_LOSS]
> that right now I don't see it                            . Something
> like that
>  <tr>
>       <td>Loss Sequence Numbers:</td>
>        <td>'.$_POST['submit_loss'].'</td>//NOTHING APPEAR HERE
>  </tr>


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