0
我試圖展示所有與當前所玩遊戲的類別相關的遊戲。但我得到一個解析錯誤在「if (!empty($_GET['gameId'])) $varGameId = $_GET['gameId'];
」解析錯誤子查詢
我該如何解決這個問題?
謝謝
<?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);
?>
添加分號;後'('') $ varGameId = 0'。 – BlitZ
**注意:**您的代碼受[*** SQL-Injection ***](http://en.wikipedia.org/wiki/SQL_injection)攻擊困擾。找到防止措施[** here **](http://stackoverflow.com/questions/60174/how-to-prevent-sql-injection-in-php)。 – BlitZ