我在 ActionScript 3 中可以正常解密,现在我想在用 Java 解密时得到相同的结果。 (我知道 OFB 模式和 NullPadding 可能不是首选,但这就是我当时使用的,这就是我现在需要解密的......)
(非常旧)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());
闪光输出为:
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 。然而,第一种方法给了我一个 IllegalKeyException,另一种方法给了我垃圾。另外,第二种方法没有明确指定如何输入 IV 数据进行解密,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: 如何使 Java 以与 ActionScript 3 相同的方式解密?当然,我希望两者都能得到相同的结果。
最佳答案
第一种方法是向您提供非法 key 大小
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,Java 将拒绝使用“强” key 长度(例如 256 位 AES)。
如果在您所在的司法管辖区这样做是合法的,请 Google 搜索“无限强度司法管辖区策略文件”并下载适用于您的 Java 安装的版本。您最终将得到两个文件,可以转储到 JRE 中的 lib/security
中。
关于Java AES-256 解密 - 从 ActionScript 3 翻译代码,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12897260/