Question

我认为这是来自 androidsnippets.org 的片段 - 为什么这不是Android 2.3 上的功能？如何解决？

错误

03-05 23:19:17.479: WARN/System.err(3598): javax.crypto.BadPaddingException: pad block corrupted
03-05 23:19:17.518: WARN/System.err(3598):     at org.bouncycastle.jce.provider.JCEBlockCipher.engineDoFinal(JCEBlockCipher.java:715)
03-05 23:19:17.518: WARN/System.err(3598):     at javax.crypto.Cipher.doFinal(Cipher.java:1090)

我在这个主题上找到了几个答案，但我找不到一个体面的方法如何修复这个，添加 NoPadding 参数或不同的算法或..？

public static String code(String stringToCode) {
    try {
        stringToCode = encrypt("somekey",stringToCode);
    } catch (Exception e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }
    return stringToCode;
}

public static String decode(String stringToDecode) {
    try {
        stringToDecode = decrypt("somekey",stringToDecode);
    } catch (Exception e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }
    return stringToDecode;
}

public static String encrypt(String seed, String cleartext) throws Exception {
    byte[] rawKey = getRawKey(seed.getBytes());
    byte[] result = encrypt(rawKey, cleartext.getBytes());
    return toHex(result);
}

public static String decrypt(String seed, String encrypted) throws Exception {
    byte[] rawKey = getRawKey(seed.getBytes());
    byte[] enc = toByte(encrypted);
    byte[] result = decrypt(rawKey, enc);
    return new String(result);
}

private static byte[] getRawKey(byte[] seed) throws Exception {
    KeyGenerator kgen = KeyGenerator.getInstance("AES");
    SecureRandom sr = SecureRandom.getInstance("SHA1PRNG");
    sr.setSeed(seed);
    kgen.init(128, sr); // 192 and 256 bits may not be available
    SecretKey skey = kgen.generateKey();
    byte[] raw = skey.getEncoded();
    return raw;
}


private static byte[] encrypt(byte[] raw, byte[] clear) throws Exception {
    SecretKeySpec skeySpec = new SecretKeySpec(raw, "AES");
    Cipher cipher = Cipher.getInstance("AES");
    cipher.init(Cipher.ENCRYPT_MODE, skeySpec);
    byte[] encrypted = cipher.doFinal(clear);
    return encrypted;
}

private static byte[] decrypt(byte[] raw, byte[] encrypted) throws Exception {
    SecretKeySpec skeySpec = new SecretKeySpec(raw, "AES");
    Cipher cipher = Cipher.getInstance("AES");
    cipher.init(Cipher.DECRYPT_MODE, skeySpec);
    byte[] decrypted = cipher.doFinal(encrypted);
    return decrypted;
}

public static String toHex(String txt) {
    return toHex(txt.getBytes());
}
public static String fromHex(String hex) {
    return new String(toByte(hex));
}

public static byte[] toByte(String hexString) {
    int len = hexString.length()/2;
    byte[] result = new byte[len];
    for (int i = 0; i < len; i++)
        result[i] = Integer.valueOf(hexString.substring(2*i, 2*i+2), 16).byteValue();
    return result;
}

public static String toHex(byte[] buf) {
    if (buf == null)
        return "";
    StringBuffer result = new StringBuffer(2*buf.length);
    for (int i = 0; i < buf.length; i++) {
        appendHex(result, buf[i]);
    }
    return result.toString();
}

private final static String HEX = "0123456789ABCDEF";
private static void appendHex(StringBuffer sb, byte b) {
    sb.append(HEX.charAt((b>>4)&0x0f)).append(HEX.charAt(b&0x0f));
}

Answer 1

两点：

1）toByte（）方法不清楚它正在尝试做什么，但我敢打赌这是错误的，如行

int len = hexString.length()/2;

对于长度为6和7的字符串（比如说）

，

将给出3的相同结果

2）您不能依赖将stings转换为字节数组而不指定要使用的Charset。不同的区域设置和不同的运行机器可能具有不同的默认字符集。您应该在算法中使用str.getBytes（“UTF8”）。

Answer 2

NickT说：“对于长度为6和7的字符串（比如说），将给出3的相同结果” 我想在这种情况下，HEX字符串的长度总是偶数（参见函数appendHex）

Answer 3

这是BouncyCastle（BouncyCastle AES error when upgrading to 1.45）的问题。

我发现的是这个导致BC的两个不同值 1.34 vs 1.45。

我找不到解决方案，但是......

密码垫块在姜饼上损坏

3 个答案: