EditPlus查找PHP源码简单数字型注入的正则表达式
佚名
今天看一个项目代码,文件不多,不过每个文件中都N多注入,一个一个看实在太累,索性花了点时间,弄了个正则表达式,搜索出来,然后再将安全的筛选出去。省了不少时间的说。
1.查找select、update、delete语句
((select|SELECT|update|UPDATE|delete|DELETE) .*(from|FROM|set|SET) .*(where|WHERE) .*) 查询语句,对于没有条件判断的基本不存在注入问题,因而仅搜索此语句即可
例子:
select * from user where 2.简单的数字型注入
((select|SELECT|update|UPDATE|delete|DELETE) .*(from|FROM|set|SET) .*(where|WHERE) .*=[ ]?["]?["]?\$) 能找到select、update delete三种语句,5种格式的整形注入,如:
直接变量传入
select * from guess where id=$subject_id
update guess set is_valid=0 where id=$subject_id
delete from guess where id=$subject_id
=与变量之间存在空格
select * from guess where id= $subject_id
update guess set is_valid=0 where id= $subject_id
delete from guess where id= $subject_id
变量双引号
select * from guess where id="$subject_id"
update guess set is_valid=0 where id="$subject_id"
delete from guess where id="$subject_id"
=与双引号之间存在空格
select * from guess where id= "$subject_id"
update guess set is_valid=0 where id= "$subject_id"
delete from guess where id= "$subject_id"
=与引号、双引号之间存在空格
select * from guess where id= " $subject_id"
update guess set is_valid=0 where id= " $subject_id"
delete from guess where id= " $subject_id"
((select|SELECT|update|UPDATE|delete|DELETE) .*(from|FROM|set|SET) .*(where|WHERE) .*) 查询语句,对于没有条件判断的基本不存在注入问题,因而仅搜索此语句即可
例子:
select * from user where 2.简单的数字型注入
((select|SELECT|update|UPDATE|delete|DELETE) .*(from|FROM|set|SET) .*(where|WHERE) .*=[ ]?["]?["]?\$) 能找到select、update delete三种语句,5种格式的整形注入,如:
直接变量传入
select * from guess where id=$subject_id
update guess set is_valid=0 where id=$subject_id
delete from guess where id=$subject_id
=与变量之间存在空格
select * from guess where id= $subject_id
update guess set is_valid=0 where id= $subject_id
delete from guess where id= $subject_id
变量双引号
select * from guess where id="$subject_id"
update guess set is_valid=0 where id="$subject_id"
delete from guess where id="$subject_id"
=与双引号之间存在空格
select * from guess where id= "$subject_id"
update guess set is_valid=0 where id= "$subject_id"
delete from guess where id= "$subject_id"
=与引号、双引号之间存在空格
select * from guess where id= " $subject_id"
update guess set is_valid=0 where id= " $subject_id"
delete from guess where id= " $subject_id"