Question

SecureRandom random = new SecureRandom(); // quite heavy, look into a lighter method.

String stringToEncrypt = "mypassword";
byte[] realiv = new byte[16];
random.nextBytes(realiv);
Cipher ecipher = Cipher.getInstance("AES");

SecureRandom random = new SecureRandom(); // quite heavy, look into a lighter method.

byte[] realiv = new byte[16];
random.nextBytes(realiv);       

byte[] secret = "somelongsecretkey".getBytes();
SecretKeySpec secretKey = new SecretKeySpec(secret, "AES");
ecipher.init(Cipher.ENCRYPT_MODE, secretKey, random);
byte[] encryptedData = ecipher.doFinal();

但init()只接受3个参数。我需要一种方法来做类似的事情：

ecipher.init(Cipher.ENCRYPT_MODE, stringToEncrypt, secretKey, random);

Answer 1

通常，您不需要为具有确定性行为的算法生成随机数的东西。此外，当您使用ECB块模式时，您不需要IV，这是Java默认的模式。确切地说，Java "AES/ECB/PKCS5Padding"中默认为Cipher.getInstance("AES")。

所以你应该可以使用这样的代码：

// lets use the actual key value instead of the platform specific character decoding
byte[] secret = Hex.decodeHex("25d6c7fe35b9979a161f2136cd13b0ff".toCharArray());

// that's fine
SecretKeySpec secretKey = new SecretKeySpec(secret, "AES");

// SecureRandom should either be slow or be implemented in hardware
SecureRandom random = new SecureRandom();

// first create the cipher
Cipher eCipher = Cipher.getInstance("AES/CBC/PKCS5Padding");

// filled with 00h characters first, use Cipher instance so you can switch algorithms
byte[] realIV = new byte[eCipher.getBlockSize()];

// actually fill with random
random.nextBytes(realIV);

// MISSING: create IvParameterSpec
IvParameterSpec ivSpec = new IvParameterSpec(realIV);

// create the cipher using the IV
eCipher.init(Cipher.ENCRYPT_MODE, secretKey, ivSpec);

// NOTE: you should really not encrypt passwords for verification
String stringToEncrypt = "mypassword";

// convert to bytes first, but don't use the platform encoding
byte[] dataToEncrypt = stringToEncrypt.getBytes(Charset.forName("UTF-8"));

// actually do the encryption using the data
byte[] encryptedData = eCipher.doFinal(dataToEncrypt);

现在看起来好多了。我已经使用Apache commons编解码器来解码十六进制字符串。

请注意，您需要使用realIV保存encryptedData，并且未包含完整性保护，例如MAC（对于密码，您可能不需要它）。

Answer 2

我强烈怀疑你想要做的是拨打ecipher.doFinal(stringToEncrypt)，如果你有更长的字符串，可能会在一系列doUpdate(...)之后致电。

.init()创建密码对象，update()和doFinal()填充加密输出并将明文作为输入。

当然，您需要convert between String and a byte array。

Android：使用带有iv和密钥的AES 256位加密来加密字符串

2 个答案: