Re: A real problem

I don't quite understand... Can you show me on this example? PS. I need java todo this, because i do not know java. ----- Original Message ----- From: "Daniel Clark" <dclark@nwlink.com> To: "Phpu" <phpu@go.ro> Cc: <php-general@lists.php.net> Sent: Monday, April 26, 2004 9:53 PM Subject: Re: [PHP] 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 > > > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >