Hi Jarratt, Thank you for your advice. I will try that and come back with the result. Alp "RaTT" <shaggie@xxxxxxxxx> wrote in message news:61f5e49905020802347d67bb4b@xxxxxxxxxxxxxxxxx > Hello Alp > > Try something like, > > <?php > > //DB fields as an array > > $db_fields = array('left' => 'left', > 'center' => 'center', > 'right' => 'right'); > > function drop($array,$sel_name='',$sel_field='',$css=''){ > $dropdown = "<select name=\"$sel_name\" $css>\n"; > foreach($arr as $key => $val){ > $sel = ($sel_field == $key)?' selected="selected"':''; > $dropdown .= "\t".'<option value="'.$key.'"'.$sel.'>'.$val.'</option>'."\n"; > } > $dropdown .= "</select>\n"; > return $dropdown; > } > usage: > echo drop($db_fields,'test_select','center'); > ?> > > HTH > Jarratt > > > > On Tue, 8 Feb 2005 17:14:24 +0800, Alp <alp.bekisoglu@xxxxxxxxxxxxxx> wrote: > > Is there an easier way to display/highlight the value stored in the database > > for a select option? Such as: > > Stored value is 'center'. The statement is: > > print '<select name="colalign">'; > > print '<option value="left">Left'; > > print '<option value="center">Center'; > > print '<option value="right">Right'; > > print '</select>'; > > > > I can have 3 sets of the above tied to 'if's but would rather ask for an > > easier or better way. > > > > Thanks in advance. > > > > Alp > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
