[RELEASE] 9Dragons XSD Editor

• Saving uses a temp file to prevent loss of data when an error occurs while writing the XSD. A backup of the original file is also created
• Fixed an infinite recursion condition in Replace All
• Saving an encrypted XSD will create an XML file which is loaded next time you open the XSD. No longer have to type in the keys each time

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Originally Posted by saweet Update Saving uses a temp file to prevent loss of data when an error occurs while writing the XSD. A backup of the original file is also created Fixed an infinite recursion condition in Replace All Saving an encrypted XSD will create an XML file which is loaded next time you open the XSD. No longer have to type in the keys each time Thanks witek1992 and nervetnova for bug reports.

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Originally Posted by nervetnova Saweet, could you recheck the Replace all ? i've test on twice PCs (windows 8 and windows 7) it's still not not responding.

• Fixed save issue where you couldn't save across different volumes
• Rewrote Replace All to work better for large workloads. No longer piggybacks Find mechanics to reduce CPU I/O

• Auto-detects number of languages the XSD supports

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Originally Posted by jadovski512 is it possible to change the effect of things in game??

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Originally Posted by matmeoviet you can find the NDTEXTDB_UK.XSD cipher keys?

08 10

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Originally Posted by saweet Thanks everyone for sharing keys. 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