如何隐藏错误的,例如像这样无法隐藏的错误消息
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
我已经把off
上display_errors = Off
和error_reporting = E_ALL & ~E_NOTICE
也无所谓,如果我把error_reporting(0);
对我的代码或@
顶部查询之前。该错误始终显示。
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)))
,如果我尝试在URL可以说bikes.php?id='
或其他为sql注入我得到这个我的网页上。
查询工作正常,这是当我尝试操纵(sql注入)在URL中。然后它显示了这一点,我想隐藏它。我不想显示我的表信息..等
好了,会通过修复它! – Pankucins
这是当我尝试模拟SQL注入的URL .. – Goro
告诉我们你的php代码,你在查询这个 – SajithNair