I think this is the line which is the cause of your coding woes...
The quote (") doesn't have any matching pair. There's one quote in
front, but there's none at the back.
echo "<select name=\"dropdown\">;
* *
----- Original Message -----
From: Cole Ashcraft <[EMAIL PROTECTED]>
Date: Friday, June 18, 2004 9:23 am
Subject: Re: [PHP-DB] Dropdown menus from DB query
> Thanks. I think I can work from this. Just to let you know, it
> generatesa 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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> >
>
>
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