I am trying to create a drop down menu to set varaible $tablrname

Cheers

paul

"John W. Holmes" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> PAUL FERRIE wrote:
>
> > What am i overlooking?
>
> The part where you tell us what you're actually trying to achieve...
>
> > <?
> > include('common.php');
> > $result = mysql_list_tables("***_vinrev");
> > $i = 0;
> > while ($i < mysql_num_rows ($result)) {
> >     $tb_names[$i]= mysql_tablename ($result, $i);
> >  echo "<SELECT NAME='$tablename'>";
> >
> >   echo "<OPTION VALUE=\"".$tb_names[$i]."\">".
> >   $tb_names[$i]." </OPTION> ";
> >   echo $tb_names[$i] . "<BR>";
>
> Why are you echoing data and displaying <br> between <select> and
> </select>... That's bad HTML.
>
> >    $i++;
> >       }
> >    echo "</SELECT>";
> > ?>
> >
> >
> > When tested
> > heres what i get
> > http://thor.ancilenetworks.co.uk/~pferrie/vinrev/adm/table_list.php
> > There are 2 tables within the DB 'reviews', 'albums'
>
> So I see a select box with two options, 'reviews' and 'albums' that's
> made from badly formed HTML. What are you trying to accomplish??
>
> --
> ---John Holmes...
>
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