Re: A real problem

I understand. I think the only solution is to be displayed in html. Please show me on this example where to put shop_models query and output within your shop_brands. Sorry for disturbing you... ----- Original Message ----- From: "Daniel Clark" <dclark@nwlink.com> To: "Phpu" <phpu@go.ro> Cc: <dclark@nwlink.com> Sent: Monday, April 26, 2004 10:20 PM Subject: Re: [PHP] A real problem > My mistake. I didn't realize it was a drop down SELECT. I thought it > was displaying the data in HTML. > > > A Javascript function could do it in the client's browser, without another > trip to the server. But I couldn't tell you how at this point. > > > > > 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 > >> > > >> > > >> > > >> > > > > >