Thanks. I think I can work from this. Just to let you know, it generates
a parse error ( I think its because of the quotes).
Cole
On Thu, 2004-06-17 at 18:09, Ng Hwee Hwee wrote:
> hmm... what about something like this??
>
> echo "<select name=\"dropdown\">;
>
> $query = "select code, name from table";
>
> $result = mysql_query($query);
>
> while($array = mysql_fetch_array($result))
> {
> echo "<option value=\"".$array["code"]."\">".$array["name"]."</option>";
> }
>
> echo "</select>";
>
> hth
>
> ----- Original Message -----
> From: "Cole Ashcraft" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Friday, June 18, 2004 7:50 AM
> Subject: [PHP-DB] Dropdown menus from DB query
>
>
> > How would you create a drop down menu from a database query? I have
> > figured how to do it with one field, but how could it be done with a
> > system where the value is different than the displayed value (ie.
> > numerical code as the value, name displayed)?
> >
> > Thanks,
> > Cole
>
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