Re: A real problem

Yes, put your shop_models query and output within your shop_brands. > Hi > > I have this list that interogates the database and display the name of the > brand. > <select name="brand"> > <? > > $query = "SELECT id, name FROM shop_brands"; > > $result = mysql_query($query) or die("error in query - $query - > ".mysql_error()); > > while($row = mysql_fetch_row($result)) { > > list($brand_id, $brand_name) = $row; > > $query2 = "SELECT * FROM shop_models WHERE brand_id = $brand_id"; > > $result2 = mysql_query($query2) or die("error in query - $query2 - > ".mysql_error()); > > $rows = mysql_num_rows($result2); > > if($rows) { > > $brand_name = htmlspecialchars(stripslashes($brand_name)); > > if(!$save_brand_id) $save_brand_id = $brand_id; > > echo("<option value='$brand_id'>$brand_name</option>"); > > } > > } > ?> > > each brand has more than one model. > when i chose one of the brands i want to be displayed o form like this (in > real time): > > $query = "SELECT id, name FROM shop_models WHERE > brand_id=$save_brand_id"; > > $result = mysql_query($query) or die("error in query - $query - > ".mysql_error()); > > while($row = mysql_fetch_row($result)) { > > list($model_id, $model_name) = $row; > echo ("<input type='checkbox' > value=\"$model_id\">$model_name\n"); > > > }; > > > How can i do this? (maybe java could help me but i do not know how to do > it) > > Can someone plese help me? > > > Thanks for your reply > > >