Код (Text): <?php define ("ITEMS_IN_SET", 10); $id = intval($_POST["id"]); $auth_key = $_POST["hash"]; $set = intval($_POST["set"]); if ($auth_key != h($id)) exit("req=false&error=auth_error"); $str = "SELECT names.name, names.all_sh, names.id, buildings.palace AS level, users.viewer_id FROM names JOIN buildings JOIN users WHERE names.id = buildings.id AND names.id = users.id ORDER BY buildings.palace DESC LIMIT ".($set * ITEMS_IN_SET).",".ITEMS_IN_SET; $req = mysql_query($str, $db); if (!$req) exit ("<xml>\n\t<req>false</req>\n\t<error>mysql_error</error>\n\t<s>0</s>\n</xml>"); echo "<xml>\n\t<req>true</req>"; while ($arr = mysql_fetch_assoc($req)) { echo "\n\t<item>"; echo "\n\t\t<name>".$arr["name"]."</name>"; echo "\n\t\t<all_sh>".$arr["all_sh"]."</all_sh>"; echo "\n\t\t<level>".$arr["level"]."</level>"; echo "\n\t\t<viewer_id>".$arr["viewer_id"]."</viewer_id>"; echo "\n\t</item>"; } echo "\n</xml>"; ?> Как засунуть в echo столбец из buildings.farm ?
SELECT names.name, names.all_sh, names.id, buildings.palace AS level, users.viewer_id , buildings.farm FROM names ... и echo "\n\t\t<farm>".$arr["farm"]."</farm>"; не?