Done. Searches tables for rows matching the Name column. Find Next wraps around when it reaches the end. Find Next table search order can be altered by manually selecting a table.
Long time no see, saweet. Nice to see you again on ePvP's 9D section. You told it's easy to figure out the encryption keys. Can you tell - how?
Done. Searches tables for rows matching the Name column. Find Next wraps around when it reaches the end. Find Next table search order can be altered by manually selecting a table.
Long time no see, saweet. Nice to see you again on ePvP's 9D section. You told it's easy to figure out the encryption keys. Can you tell - how?
By the way - thanks for great update.
The 2nd key is obvious because it's only used in 2 byte characters and only ever used in the 2nd byte position. They use XOR encryption so it becomes very obvious when null is XOR'd, because 0 XOR N = N. Thus N is the 2nd key.
The 1st key can be learned by knowing the most common character in a string, with that you can guess the possible keys. What I did was grab all the table names in both an unencrypted and an encrypted XSD and compared string length to find matches as possible candidates for the real value. MobModel_XMS was a good choice because we have 3 Ms in a single short string. With that I found ZxuZxsr{HOZD of the same string length in the encrypted file. Notice the pattern, 3 Zs in the same position?
If you know Z (5A) in a string is really M (4D), all we have to do is figure out what possible XORs of 4D will equal 5A to get a list of possible keys. I did this in python and had the key in a fraction of a second:
Code:
>>> [hex(i) for i in range(256) if 0x4d ^ i == 0x5a]
['0x17']
Lets see what happens (ciphertext is "ZxuZxsr{HOZD" in hex)
Code:
>>> ciphertext = bytearray([0x5A, 0x78, 0x75, 0x5A, 0x78, 0x73, 0x72, 0x7B, 0x48, 0x4F, 0x5A, 0x44])
>>> bytearray([ciphertext[i] ^ 0x17 for i in range(len(ciphertext))])
bytearray(b'MobModel_XMS')
>>>
The 2nd key is obvious because it's only used in 2 byte characters and only ever used in the 2nd byte position. They use XOR encryption so it becomes very obvious when null is XOR'd, because 0 XOR N = N. Thus N is the 2nd key.
The 1st key can be learned by knowing the most common character in a string, with that you can guess the possible keys. What I did was grab all the table names in both an unencrypted and an encrypted XSD and compared string length to find matches as possible candidates for the real value. MobModel_XMS was a good choice because we have 3 Ms in a single short string. With that I found ZxuZxsr{HOZD of the same string length in the encrypted file. Notice the pattern, 3 Zs in the same position?
If you know Z (5A) in a string is really M (4D), all we have to do is figure out what possible XORs of 4D will equal 5A to get a list of possible keys. I did this in python and had the key in a fraction of a second:
Code:
>>> [hex(i) for i in range(256) if 0x4d ^ i == 0x5a]
['0x17']
Lets see what happens (ciphertext is "ZxuZxsr{HOZD" in hex)
Code:
>>> ciphertext = bytearray([0x5A, 0x78, 0x75, 0x5A, 0x78, 0x73, 0x72, 0x7B, 0x48, 0x4F, 0x5A, 0x44])
>>> bytearray([ciphertext[i] ^ 0x17 for i in range(len(ciphertext))])
bytearray(b'MobModel_XMS')
>>>
Viola
greate work brb ... but can you explain how to program that 2 cipher XOR encrytion ?
[RELEASE] 9Dragons Bot Spammer! 03/19/2011 - 9Dragons - 12 Replies Old Version.Hi all, some time ago im was looking for a 9Dragons Bot Spammer and i find Nothing, so i decide to learn to make my own. and i want to share with you!
http://img52.imageshack.us/img52/8607/spammeri.jp g
How to use:
1.first u need to have the game runing
2.the chat box muts be Open befor u start the bot
3.open the bot and pres start don`t forget to type wath u want to bot type
![[RELEASE] 9Dragons XSD Editor](https://www.elitepvpers.com/forum/iconimages/9dragons/release-9dragons-xsd-editor_ltr.gif){kind=small}
it's hella easier now.
