Question

我在ActionScript 3中有一个有效的解密，现在我想在Java中解密时获得相同的结果。（我知道OFB模式和NullPadding可能不是首选，但那是我当时使用的，这就是我现在需要解密的内容......）

（非常古老）Adobe ActionScript 3代码：

static public function decryptTest(): Boolean {
    var iv: String = "0df1eff724d50157ab048d9ff214b73c";
    var cryptext: String = "2743be20314cdc768065b794904a0724e64e339ea6b4f13c510e2d2e8c95dd7409aa0aefd20daae80956dd2978c98d6e914d1d7b5b5be47b491d91e7e4f16f7f30d991ba80a81bafd8f0d7d83755ba0ca66d6b208424529c7111bc9cd6d11786f3f604a0715f";
    var kkey: String = "375f22c03371803ca6d36ec42ae1f97541961f7359cf5611bbed399b42c7c0be";

    var kdata: ByteArray = Hex.toArray(kkey);
    var data: ByteArray = Hex.toArray(cryptext);
    var name: String = 'aes-256-ofb';
    var pad:IPad = new NullPad();
    var mode: ICipher = Crypto.getCipher(name, kdata, pad);
    pad.setBlockSize(mode.getBlockSize());
    trace("mode block size: " + mode.getBlockSize());

    if (mode is IVMode) {
        var ivmode:IVMode = mode as IVMode;
        ivmode.IV = Hex.toArray(iv);
    }
    mode.decrypt(data);

    var res: String = data.toString();
    trace("result: " + res);

    return res == "01020506080b10131c22292d313536393b464c535466696d6e7d7f808a8e9899a2adb1b8babcbebfc1c6c7c8cecfd8e0e4e8ef";
}

trace("decryption test: " + netplay.decryptTest());

Flash输出是：

mode block size: 16
result: 01020506080b10131c22292d313536393b464c535466696d6e7d7f808a8e9899a2adb1b8babcbebfc1c6c7c8cecfd8e0e4e8ef
decryption test: true

我尝试了什么？

我在Java中尝试了两种不同的方法，一种使用内置的Cipher类，另一种使用this code/class。但是，第一种方法给了我一个IllegalKeyException而另一种方法给了我垃圾。此外，第二种方法没有明确规定如何输入解密的IV数据，也没有让我指定OFB模式或填充。

java.security.InvalidKeyException: Illegal key size
    at javax.crypto.Cipher.checkCryptoPerm(Cipher.java:1023)
    at javax.crypto.Cipher.implInit(Cipher.java:789)
    at javax.crypto.Cipher.chooseProvider(Cipher.java:848)
    at javax.crypto.Cipher.init(Cipher.java:1347)
    at javax.crypto.Cipher.init(Cipher.java:1281)
    at test.net.zomis.ZomisTest.decryptCipher(ZomisTest.java:112)
@Test
public void decryptCipher() throws UnsupportedEncodingException, NoSuchAlgorithmException, NoSuchPaddingException, InvalidKeyException, InvalidAlgorithmParameterException, IllegalBlockSizeException, BadPaddingException {
    String iv = "0df1eff724d50157ab048d9ff214b73c";
    String cryptext = "2743be20314cdc768065b794904a0724e64e339ea6b4f13c510e2d2e8c95dd7409aa0aefd20daae80956dd2978c98d6e914d1d7b5b5be47b491d91e7e4f16f7f30d991ba80a81bafd8f0d7d83755ba0ca66d6b208424529c7111bc9cd6d11786f3f604a0715f";
    String key = "375f22c03371803ca6d36ec42ae1f97541961f7359cf5611bbed399b42c7c0be"; // Hexadecimal String, will be converted to non-hexadecimal String
    String expectedResult = "01020506080b10131c22292d313536393b464c535466696d6e7d7f808a8e9899a2adb1b8babcbebfc1c6c7c8cecfd8e0e4e8ef";

    byte[] kdata = Util.hex2byte(key);

    Assert.assertEquals(32, kdata.length); // 32 bytes = 256-bit key

    String result;

    Cipher cipher;
    cipher = Cipher.getInstance("AES/OFB/NoPadding");
    // Below line is 112, which is causing exception
    cipher.init(Cipher.DECRYPT_MODE, new SecretKeySpec(kdata, "AES"), new IvParameterSpec(iv.getBytes("UTF-8")));
    byte[] cryptData = Util.hex2byte(cryptext);
    byte[] ciphertext = cipher.doFinal(cryptData);
    result = new String(ciphertext);


    Assert.assertEquals(expectedResult, result);
}

@Test
public void decryptAES() {
    String iv = "0df1eff724d50157ab048d9ff214b73c"; 
    // Problem: Where should I specify the IV ???? Currently it is an unused variable...

    String cryptext = "2743be20314cdc768065b794904a0724e64e339ea6b4f13c510e2d2e8c95dd7409aa0aefd20daae80956dd2978c98d6e914d1d7b5b5be47b491d91e7e4f16f7f30d991ba80a81bafd8f0d7d83755ba0ca66d6b208424529c7111bc9cd6d11786f3f604a0715f";
    String key = "375f22c03371803ca6d36ec42ae1f97541961f7359cf5611bbed399b42c7c0be"; // Hexadecimal String, will be converted to non-hexadecimal String
    String expectedResult = "01020506080b10131c22292d313536393b464c535466696d6e7d7f808a8e9899a2adb1b8babcbebfc1c6c7c8cecfd8e0e4e8ef";

    Assert.assertEquals(64, key.length());

    AES aes = new AES();
    aes.setKey(Util.hex2byte(key));
    byte[] byteCryptedData = Util.hex2byte(cryptext);
    String byteCryptedString = new String(byteCryptedData);

    while (byteCryptedString.length() % 16 != 0) byteCryptedString += " ";


    String result = aes.Decrypt(byteCryptedString);
    Assert.assertEquals(expectedResult, result); // Assertion Failed
}

问题： 如何以与ActionScript 3相同的方式使Java解密？当然，我想在两者上得到相同的结果。

Answer 1

第一种方法是给您一条Illegal key size错误消息，因为您没有安装不受限制的策略文件。如果没有这些密钥，Java将拒绝使用“强”密钥长度（例如256位AES）。

如果在您所在的司法辖区内这样做是合法的，Google可以使用“ Unlimited Strength Jurisdiction Policy Files ”并下载适用于您的Java安装的版本。您最终将有两个文件转储到JRE中的lib/security。

Answer 2

是的，有这样的图书馆，请看http://www.bouncycastle.org/。这有点具体Java Bouncy Castle Cryptography - Encrypt with AES

Java AES-256解密 - 从ActionScript 3转换代码

2 个答案: