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blind sql injection 대응 코드 수정

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chicpro
2014-03-21 10:18:47 +09:00
parent e9e1f9713f
commit 1e84bc6f1d
2 changed files with 11 additions and 5 deletions
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8
common.php
View File

@ -80,10 +80,10 @@ if (file_exists($dbconfig_file)) {
}
// sql_escape_string 적용
$_POST = array_map_deep('sql_escape_string', $_POST);
$_GET = array_map_deep('sql_escape_string', $_GET);
$_COOKIE = array_map_deep('sql_escape_string', $_COOKIE);
$_REQUEST = array_map_deep('sql_escape_string', $_REQUEST);
$_POST = array_map_deep(G5_ESCAPE_FUNCTION, $_POST);
$_GET = array_map_deep(G5_ESCAPE_FUNCTION, $_GET);
$_COOKIE = array_map_deep(G5_ESCAPE_FUNCTION, $_COOKIE);
$_REQUEST = array_map_deep(G5_ESCAPE_FUNCTION, $_REQUEST);
//==============================================================================
// PHP 4.1.0 부터 지원됨

8
lib/common.lib.php
View File

@ -1451,8 +1451,14 @@ function sql_select_db($db, $connect)
// mysql_query 와 mysql_error 를 한꺼번에 처리
function sql_query($sql, $error=G5_DISPLAY_SQL_ERROR)
{
// Blind SQL Injection 취약점 해결
$sql = trim($sql);
// union의 사용을 허락하지 않습니다.
$sql = preg_replace("#^select.*from.*union.*#i", "select 1", $sql);
// `information_schema` DB로의 접근을 허락하지 않습니다.
$sql = preg_replace("#^select.*from.*where.*`?information_schema`?.*#i", "select 1", $sql);
if ($error)
$result = @mysql_query($sql) or die("<p>$sql<p>" . mysql_errno() . " : " . mysql_error() . "<p>error file : {$_SERVER['PHP_SELF']}");
$result = @mysql_query($sql) or die("<p>$sql<p>" . mysql_errno() . " : " . mysql_error() . "<p>error file : $_SERVER[PHP_SELF]");
else
$result = @mysql_query($sql);
return $result;