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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);
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2 回答 2

5

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 的默认设置。准确地说，Java 默认为"AES/ECB/PKCS5Padding"for in Cipher.getInstance("AES")。

所以你应该可以接受这样的代码：

// 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 保存realIV，encryptedData并且您没有包含完整性保护，例如 MAC（对于密码，您可能不需要它）。

于 2012-11-03T12:26:17.343 回答
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我强烈怀疑你想要做的是 call ecipher.doFinal(stringToEncrypt)，可能在一系列doUpdate(...)if 你有更长的字符串之后。

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

当然，您需要在和 字节数组之间进行转换String。

于 2012-11-03T01:19:00.427 回答