0
我试图展示所有与当前所玩游戏的类别相关的游戏。但我得到一个解析错误在“if (!empty($_GET['gameId'])) $varGameId = $_GET['gameId'];
”解析错误子查询
我该如何解决这个问题?
谢谢
<?php
$varCategoria_GameData = "0";
$varGameId = 0
if (!empty($_GET['gameId'])) $varGameId = $_GET['gameId'];
if (isset($_GET["cat"])) {
$varCategoria_GameData = $_GET["cat"];
}
$sql_categoria = "SELECT * FROM jogos WHERE intCategoria =
(SELECT intCategoria FROM jogos WHERE idGames = $varGameId)";
$query_categoria = mysql_query($sql_categoria, $gameconnection) or die(mysql_error());
$categoria = mysql_fetch_assoc($query_categoria);
?>
添加分号;后'('') $ varGameId = 0'。 – BlitZ
**注意:**您的代码受[*** SQL-Injection ***](http://en.wikipedia.org/wiki/SQL_injection)攻击困扰。找到防止措施[** here **](http://stackoverflow.com/questions/60174/how-to-prevent-sql-injection-in-php)。 – BlitZ