Re: A real problem

If I was displaying it. :-) Here is a "ruff" guess. <? $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>"); $query = "SELECT id, name FROM shop_models WHERE brand_id=$save_brand_id"; $result = mysql_query($query) or die("error in query - $query - ".mysql_error()); while($row2 = mysql_fetch_row($result)) { list($model_id, $model_name) = $row2; echo "$model_id -- $model_name" . "<BR>"; }; } } ?> > 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 >> >> > >> >> > >> >> > >> >> >> > >> > >> > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >