无法隐藏错误消息

Question

如何隐藏这样的错误

You have an error in your SQL syntax; check the manual that corresponds 
to your MySQL server version for the right syntax to use near '\\\') as previousid,
(SELECT IFNULL(min(id),-1) FROM bikes WHERE' at line 2

我已经推迟off display_errors = Off和error_reporting = E_ALL & ~E_NOTICE也没关系，如果我放了error_reporting(0); 在我的代码或@之前的查询之前。 总是显示错误。

编辑：

$q = mysqli_query($con, 'SELECT *,
            (SELECT IFNULL(max(id),-1) FROM bikes WHERE `id` < '.($currentId).') as previousid,
            (SELECT IFNULL(min(id),-1) FROM bikes WHERE `id` > '.($currentId).') as nextid
                FROM bikes WHERE `id` = ' . ($currentId)))

如果我尝试使用URL，比如说bikes.php?id='或其他要进行sql注入的操作，我会在我的页面上看到。

查询工作正常，这是当我尝试在URL中操纵（sql注入）时。 然后显示它，我要隐藏它。 我不想显示我的桌子的信息。等等。

Answer 1

您的错误似乎来自您的SQL查询。请尝试以下操作：

$q = mysqli_query($con, 'SELECT *,
            (SELECT IFNULL(max(id),-1) FROM `bikes` WHERE `id` < '.$currentId.') as previousid,
            (SELECT IFNULL(min(id),-1) FROM `bikes` WHERE `id` > '.$currentId.') as nextid
                FROM bikes WHERE `id` = ' . $currentId))