AES加密我嘗試使用AES加密

Question

加密我只想Ruby代碼遷移到Java

這裏是我的Ruby代碼

require 'openssl' 
require 'base64' 

key = '7c54367a45b37a192abc2cd7f45203042350406f8' 
cipher = OpenSSL::Cipher::Cipher.new('aes-128-ecb') 
cipher.encrypt() 


cipher = OpenSSL::Cipher::Cipher.new('aes-256-ecb') 
cipher.encrypt() 
cipher.key = key 

crypt = cipher.update('Rahul') 
crypt << cipher.final() 

puts (Base64.encode64(crypt)) 

以下是我在Java中

正在嘗試

String getDecodedString(String key,String encodedValue,SupportedEncryptionAlgorithm algoInfo) 
{ 
    Cipher cipher = getCipherInstancenew(algoInfo, key,Cipher.DECRYPT_MODE); 
    try 
    { 
     byte[] dec = new sun.misc.BASE64Decoder().decodeBuffer(encodedValue); 

     int ctLength = cipher.getOutputSize(dec.length); 
     byte[] plainText = new byte[cipher.getOutputSize(ctLength)]; 

     int ptLength = cipher.update(dec, 0, ctLength, plainText, 0); 
     ptLength += cipher.doFinal(plainText, ptLength); 

     return null; 
    } 
    catch (IllegalBlockSizeException e) 
    { 
     LoggerFactory.getLogger(EncryptionHelper.class).error("Security Alert",e); 
    } 
    catch (BadPaddingException e) 
    { 
     LoggerFactory.getLogger(EncryptionHelper.class).error("Security Alert",e); 
    } 
    return null; 
} 

public static byte[] stringToBytes(String s) { 
    byte[] b2 = new BigInteger(s, 36).toByteArray(); 
    return Arrays.copyOfRange(b2, 1, b2.length); 
} 

public static Cipher getCipherInstancenew(SupportedEncryptionAlgorithm algoInfo,String keyString,int mode) throws IOException 
{ 
    byte[] decodedBytes; 
    Cipher cipher=null;  
    try 
    { 
     decodedBytes = getBase64FromHEX(keyString).getBytes(); 
     SecretKeySpec skeySpec = new SecretKeySpec(decodedBytes, "AES"); 
     Security.addProvider(new BouncyCastleProvider()); 
     cipher = Cipher.getInstance("AES/ECB/PKCS5Padding", "BC"); 
     cipher.init(mode, skeySpec); 
    } 
    catch (java.security.GeneralSecurityException e) 
    { 
     /*Strictly no logging as it is security class 
     * There seems to be some issue with the Keys so alert it */ 
     //LoggerFactory.getLogger(EncryptionHelper.class).error("Security Alert",e); 
     throw new IOException("GetCipherInstance does not exsists"); 
    } 

    return cipher; 

} 

public static String getBase64FromHEX(String input) { 

    byte barr[] = new byte[16]; 
    int bcnt = 0; 
    for (int i = 0; i < 32; i += 2) { 
     char c1 = input.charAt(i); 
     char c2 = input.charAt(i + 1); 
     int i1 = intFromChar(c1); 
     int i2 = intFromChar(c2); 

     barr[bcnt] = 0; 
     barr[bcnt] |= (byte) ((i1 & 0x0F) << 4); 
     barr[bcnt] |= (byte) (i2 & 0x0F); 
     bcnt++; 
    } 

    BASE64Encoder encoder = new BASE64Encoder(); 
    return encoder.encode(barr); 
} 

private static int intFromChar(char c) { 
    char[] carr = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f' }; 

    char clower = Character.toLowerCase(c); 
    for (int i = 0; i < carr.length; i++) { 
     if (clower == carr[i]) { 
      return i; 
     } 
    } 

    return 0; 
} 

它適用於32字節的字符串，但不適用於Java中的41字節，但在Ruby中適用於大於32字節的任何長度奇怪請幫助

Answer 1

的下面Java代碼從作爲Ruby代碼加密輸出完全相同的基64編碼的結果，成功地將其解密：

final Cipher encryptCipher = Cipher.getInstance("AES/ECB/PKCS5Padding", "BC"); 
encryptCipher.init(Cipher.ENCRYPT_MODE, new SecretKeySpec("7c54367a45b37a192abc2cd7f4520304".getBytes(), "AES")); 
final byte[] encrypt = encryptCipher.doFinal("This is my text".getBytes()); 
System.out.println(new String(Base64.encode(encrypt))); 

final Cipher decryptCipher = Cipher.getInstance("AES/ECB/PKCS5Padding", "BC"); 
decryptCipher.init(Cipher.DECRYPT_MODE, new SecretKeySpec("7c54367a45b37a192abc2cd7f4520304".getBytes(), "AES")); 
final byte[] decrypt = decryptCipher.doFinal(encrypt); 
System.out.println(new String(decrypt)); 

紅寶石OpenSSL的API顯然僅使用前32個字節的關鍵的，因爲對於key以下值返回相同的值41字節版本：

key = '7c54367a45b37a192abc2cd7f4520304' 

另外，我不知道爲什麼cipher在Ruby代碼初始化兩次，因爲它是沒有必要的，只要我能告訴。

Answer 2

我懷疑你打給String.getBytes()的電話正在做你需要他們做的事情。

getBytes()方法使用平臺的默認字符編碼將String的字符轉換爲字節序列。默認的平臺字符編碼類似UTF-8，US-ASCII或ISO-8859-1。它不是base-64或十六進制。

大多數字符編碼不能處理加密操作中使用的隨機8位值。因此，例如，您通常不能從加密產生的字節中創建新的String。根據您的編碼，許多值將被替換爲 或？。而且，即使它在您的機器上正常工作，下一個桌面上的機器也可能配置不同，並且在嘗試解碼該字符串時會失敗。

如果您需要在二進制數據和文本之間進行轉換，請使用像Base-64這樣的編碼。