通过加密数据生成MAC [英] Generating MAC by encrypting data
本文介绍了通过加密数据生成MAC的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我希望这个问题不会因为没有建设性而被封闭。我一直在敲我的头到墙上来解决这个问题，仍然无处可去。所以，这是我最后的希望：
问题基本上是关于生成ISO 8583-Rev 93消息的MAC字段。
我有一些转储的有效消息，我尝试生成正确的MAC。
- 加密算法为
DES，Mode = CBC，IV = new byte [] {0，0，0，0，0，0，0，0} - 加密金钥是：
new byte [] {0x11，0x11，0x11，0x11，0x11，0x11，0x11，0x11} - 要加密的数据是（消息）（208字节长）：
新字节[ ] [49,50,48,48,70,54,51,52,56,52,48,49,56,56,69,49,49,48,48,48,48,49,48,48,48,49,48,48， 48，48，48，48，48，48，48，48，22，54，50，56，48，50，51，49，49，49，48，48，48， 48，48，48，48，48，48，48，48，48，48，48，48，48，48，48，48，48，48，48， 48，48，48，48，48，48，49，49，54，48，56，48，56，52，53，51，57，55，57， 49，49，49，51，56，52，53，48，48，48，48，48，49，49，54，54，49，48，53，48， 48，54，49，51，49，55，67，6，54，50，56，48，50，51，6，54，50，56，48，50，51，53，49，52，50， 52，49，52，51，50，53，52，51，48，57，57，57，57，52，48，50，48，48，48，48，48，57，57， 52，48，50，32，32，32，7，84，101，108，66，97，110，107，0，0，51，
- 我应该加密数据和（可能在将结果转换为十六进制后）获得进入消息的第128个字段的有效
BF327C0CED48F26B值。
以下是我尝试过的（还有更多实现）：
var mac = Encrypt（dataToEncrypt）;
//转换为十六进制表示。
var hexMac = IsoUtils.ByteArrayToHex（mac）;
//没有BF327C0CED48F26B在hexMac！有可能是错误的！
///< summary>
///加密输入数据。尝试了许多其他方式具有相同的结果。
///< / summary>
public byte [] Encrypt（byte [] input）
{
DESCryptoServiceProvider desProvider = new DESCryptoServiceProvider（）;
desProvider.Padding = PaddingMode.None;
desProvider.Mode = CipherMode.CBC;
ICryptoTransform cryptoTransform = desProvider.CreateEncryptor（this._key，this._iv）;
MemoryStream encryptedStream = new MemoryStream（）;
CryptoStream cryptStream = new CryptoStream（encryptedStream，cryptoTransform，CryptoStreamMode.Write）;
cryptStream.Write（input，0，input.Length）;
cryptStream.FlushFinalBlock（）;
encryptedStream.Position = 0;
byte [] result = new byte [encryptedStream.Length];
encryptedStream.Read（result，0，（int）encryptedStream.Length）;
cryptStream.Close（）;
return result;
}
///< summary>
///获取字节数组的十六进制表示。长度为2xdata.Length
///< / summary>
public static string ByteArrayToHex（byte [] data）
{
string result = string.Empty;
foreach（数据中的字节ascii）
{
var n =（int）ascii;
result + = n.ToString（X）。PadLeft（2，'0'）;
}
返回结果;
}
BF327C0CED48F26B 通过加密 dataToEncrypt 变量？
任何帮助。
UPDATE ：这是邮件详情，以便它可以帮助任何人：
--------------------------------------- ---------------------------------
加密数据
------ -------------------------------------------------- ----------------
十六进制
31 32 30 30 46 36 33 34 38 34 30 31 38 38 45 31 1200F634840188E1
31 30 30 30 30 31 30 30 30 30 30 30 30 30 30 30 1000010000000000
30 30 30 30 16 36 32 38 30 32 33 31 31 31 30 30 000062802311100
30 30 36 34 32 33 31 30 30 30 30 30 30 30 30 30 0064231000000000
30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 0000000000000000
30 30 30 30 31 31 36 30 38 30 38 34 35 33 39 37 0000116080845397
39 38 36 31 32 30 31 31 36 31 31 33 38 34 35 30 9861201161138450
30 30 30 30 31 31 36 36 31 30 35 30 30 36 31 33 0000116610500613
31 37 43 60 36 32 38 30 32 33 60 36 32 38 30 32 17c62802362802
33 35 31 34 32 34 31 34 33 32 35 34 33 30 39 39 3514241432543099
39 39 34 30 32 30 30 30 30 30 39 39 39 39 34 30 9940200000999940
32 20 20 20 70 54 65 6C 42 61 6E 6B 00 00 33 37 2 TelBank 37
36 45 37 31 43 44 45 31 44 38 36 36 33 36 00 00 6E71CDE1D86636
------------------------------------------------- -----------------------
消息准备发送
-------------- -------------------------------------------------- --------
十六进制
30 32 32 34 31 32 30 30 46 36 33 34 38 34 30 31 02241200F6348401
38 38 45 31 31 30 30 30 30 31 30 30 30 30 30 30 88E1100001000000
30 30 30 30 30 30 30 31 16 36 32 38 30 32 33 31 000000016280231
31 31 30 30 30 30 36 34 32 33 31 30 30 30 30 30 1100006423100000
30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 0000000000000000
30 30 30 30 30 30 30 30 31 31 36 30 38 30 38 34 0000000011608084
35 33 39 37 39 38 36 31 32 30 31 31 36 31 31 33 5397986120116113
38 34 35 30 30 30 30 30 31 31 36 36 31 30 35 30 8450000011661050
30 36 31 33 31 37 43 60 36 32 38 30 32 33 60 36 061317C6280236
32 38 30 32 33 35 31 34 32 34 31 34 33 32 35 34 2802351424143254
33 30 39 39 39 39 34 30 32 30 30 30 30 30 39 39 3099994020000099
39 39 34 30 32 20 20 20 70 54 65 6C 42 61 6E 6B 99402电话银行
00 00 33 37 36 45 37 31 43 44 45 31 44 38 36 36 376E71CDE1D866
33 36 00 00 42 46 33 32 37 43 30 43 45 44 34 38 36 BF327C0CED48
46 32 36 42 00 00 00 00 F26B
----------------------- -------------------------------------------------
Fields
------------------------------------------ ------------------------------
[LLVar n ..19 0016] 002 [6280231110000642]
[固定n 6 0006] 003 [310000]
[固定n 12 0012] 004 [000000000000]
[固定n 12 0012] 006 [000000000000]
[固定n 10 0010] 007 [ 0116080845]
[固定n 6 0006] 011 [397986]
[固定n 12 0012] 012 [120116113845]
[固定n 4 0004] 014 b $ b [固定
[固定的12 0012] 022 [61050061317C]
[LLVar n ..11 0006] 032 [628023]
[LLVar n。11 0006] 033 [628023]
[固定12 0012] 037 [514241432543]
[固定8 0008] 041 [09999402]
[固定ans 15 0015] 042 [000009999402]
[LLVar ans ..40 0007] 043 [TelBank]
[LLLVar ans ..999 0000] 048 []
[固定16 0016] 052 [376E71CDE1D86636]
[LLLVar ans .. 999 0000] 072 []
[固定16 0016] 128 [BF327C0CED48F26B]
em> Update 2：
那么，那些白痴忘记在他们的文档中提到字段128应该填充 0 / em>发送进行加密。
解决方案
-
MAC只是最后的8
-
您应该只计算部分消息的MAC。我不认为ISO 8583指定哪些部分，因此您需要查看您应该由银行提供的规范。
I hope this question will not get closed as not constructive. I've been banging my head to the wall to solve this and still got nowhere. so, this is my last hope:
Question is basically about generating the MAC field of an ISO 8583-Rev 93 message.
I have some dumps of valid messages and I'm trying to generate the correct MAC.
- Encryption Algorithm is
DES,Mode=CBC,IV = new byte[] {0, 0, 0, 0, 0, 0, 0, 0} - Encryption key is:
new byte[] { 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11 } - Data To encrypt is (the message) (208 bytes long):
new byte[] { 49, 50, 48, 48, 70, 54, 51, 52, 56, 52, 48, 49, 56, 56, 69, 49, 49, 48, 48, 48, 48, 49, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 22, 54, 50, 56, 48, 50, 51, 49, 49, 49, 48, 48, 48, 48, 54, 52, 50, 51, 49, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 49, 49, 54, 48, 56, 48, 56, 52, 53, 51, 57, 55, 57, 56, 54, 49, 50, 48, 49, 49, 54, 49, 49, 51, 56, 52, 53, 48, 48, 48, 48, 48, 49, 49, 54, 54, 49, 48, 53, 48, 48, 54, 49, 51, 49, 55, 67, 6, 54, 50, 56, 48, 50, 51, 6, 54, 50, 56, 48, 50, 51, 53, 49, 52, 50, 52, 49, 52, 51, 50, 53, 52, 51, 48, 57, 57, 57, 57, 52, 48, 50, 48, 48, 48, 48, 48, 57, 57, 57, 57, 52, 48, 50, 32, 32, 32, 7, 84, 101, 108, 66, 97, 110, 107, 0, 0, 51, 55, 54, 69, 55, 49, 67, 68, 69, 49, 68, 56, 54, 54, 51, 54, 0, 0 };
- I should encrypt the data and (probably after converting the result to hex) get to the valid
BF327C0CED48F26Bvalue which goes into the message's 128th field.
Here's what I have tried (along many more implementations):
var mac = Encrypt(dataToEncrypt);
// Convert to hex representation.
var hexMac = IsoUtils.ByteArrayToHex(mac);
// No BF327C0CED48F26B is in hexMac! something is probably wrong!
/// <summary>
/// Encrypts the input data. Tried many other ways with the same result.
/// </summary>
public byte[] Encrypt(byte[] input)
{
DESCryptoServiceProvider desProvider = new DESCryptoServiceProvider();
desProvider.Padding = PaddingMode.None;
desProvider.Mode = CipherMode.CBC;
ICryptoTransform cryptoTransform = desProvider.CreateEncryptor(this._key, this._iv);
MemoryStream encryptedStream = new MemoryStream();
CryptoStream cryptStream = new CryptoStream(encryptedStream, cryptoTransform, CryptoStreamMode.Write);
cryptStream.Write(input, 0, input.Length);
cryptStream.FlushFinalBlock();
encryptedStream.Position = 0;
byte[] result = new byte[encryptedStream.Length];
encryptedStream.Read(result, 0, (int)encryptedStream.Length);
cryptStream.Close();
return result;
}
/// <summary>
/// Gets the hex representation of the byte array. The length is 2xdata.Length
/// </summary>
public static string ByteArrayToHex(byte[] data)
{
string result = string.Empty;
foreach (byte ascii in data)
{
var n = (int)ascii;
result += n.ToString("X").PadLeft(2, '0');
}
return result;
}
In short, How can I achieve BF327C0CED48F26B by encrypting dataToEncrypt variable?
I'd really appreciate any help.
UPDATE: Here's the message details in case it could be of any help:
------------------------------------------------------------------------
Data For Encryption
------------------------------------------------------------------------
HEX NORMAL
31 32 30 30 46 36 33 34 38 34 30 31 38 38 45 31 1200F634840188E1
31 30 30 30 30 31 30 30 30 30 30 30 30 30 30 30 1000010000000000
30 30 30 30 16 36 32 38 30 32 33 31 31 31 30 30 000062802311100
30 30 36 34 32 33 31 30 30 30 30 30 30 30 30 30 0064231000000000
30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 0000000000000000
30 30 30 30 31 31 36 30 38 30 38 34 35 33 39 37 0000116080845397
39 38 36 31 32 30 31 31 36 31 31 33 38 34 35 30 9861201161138450
30 30 30 30 31 31 36 36 31 30 35 30 30 36 31 33 0000116610500613
31 37 43 60 36 32 38 30 32 33 60 36 32 38 30 32 17C62802362802
33 35 31 34 32 34 31 34 33 32 35 34 33 30 39 39 3514241432543099
39 39 34 30 32 30 30 30 30 30 39 39 39 39 34 30 9940200000999940
32 20 20 20 70 54 65 6C 42 61 6E 6B 00 00 33 37 2 TelBank 37
36 45 37 31 43 44 45 31 44 38 36 36 33 36 00 00 6E71CDE1D86636
------------------------------------------------------------------------
Message ready to be sent
------------------------------------------------------------------------
HEX NORMAL
30 32 32 34 31 32 30 30 46 36 33 34 38 34 30 31 02241200F6348401
38 38 45 31 31 30 30 30 30 31 30 30 30 30 30 30 88E1100001000000
30 30 30 30 30 30 30 31 16 36 32 38 30 32 33 31 000000016280231
31 31 30 30 30 30 36 34 32 33 31 30 30 30 30 30 1100006423100000
30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 0000000000000000
30 30 30 30 30 30 30 30 31 31 36 30 38 30 38 34 0000000011608084
35 33 39 37 39 38 36 31 32 30 31 31 36 31 31 33 5397986120116113
38 34 35 30 30 30 30 30 31 31 36 36 31 30 35 30 8450000011661050
30 36 31 33 31 37 43 60 36 32 38 30 32 33 60 36 061317C6280236
32 38 30 32 33 35 31 34 32 34 31 34 33 32 35 34 2802351424143254
33 30 39 39 39 39 34 30 32 30 30 30 30 30 39 39 3099994020000099
39 39 34 30 32 20 20 20 70 54 65 6C 42 61 6E 6B 99402 TelBank
00 00 33 37 36 45 37 31 43 44 45 31 44 38 36 36 376E71CDE1D866
33 36 00 00 42 46 33 32 37 43 30 43 45 44 34 38 36 BF327C0CED48
46 32 36 42 00 00 00 00 F26B
------------------------------------------------------------------------
Fields
------------------------------------------------------------------------
[LLVar n ..19 0016] 002 [6280231110000642]
[Fixed n 6 0006] 003 [310000]
[Fixed n 12 0012] 004 [000000000000]
[Fixed n 12 0012] 006 [000000000000]
[Fixed n 10 0010] 007 [0116080845]
[Fixed n 6 0006] 011 [397986]
[Fixed n 12 0012] 012 [120116113845]
[Fixed n 4 0004] 014 [0000]
[Fixed n 4 0004] 017 [0116]
[Fixed an 12 0012] 022 [61050061317C]
[LLVar n ..11 0006] 032 [628023]
[LLVar n ..11 0006] 033 [628023]
[Fixed an 12 0012] 037 [514241432543]
[Fixed an 8 0008] 041 [09999402]
[Fixed ans 15 0015] 042 [000009999402 ]
[LLVar ans ..40 0007] 043 [TelBank]
[LLLVar ans ..999 0000] 048 []
[Fixed an 16 0016] 052 [376E71CDE1D86636]
[LLLVar ans ..999 0000] 072 []
[Fixed an 16 0016] 128 [BF327C0CED48F26B]
Update 2:
Well, those idiots forgot to mention in their documents that field 128 should be filled with 0 and then be sent for encryption. Found this out by try and fault!
解决方案
The MAC is only the last 8 bytes of the CBC-encryption.
You should only calculate the MAC over part of the message. I do not believe ISO 8583 specifies which parts, so you will need to look at the specification that you should have been provided by the bank.
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