java - 每次执行的 AES 加密 key 生成是否一致？

Question

我发现了这个关于如何使用 AES 加密的写得很好的示例，我承认一些更高级的方面超出了我的能力范围。只要我使用相同的实例对象，该类就可以正常工作。如果我使用完全相同的密码创建另一个对象 - 该对象无法再正确解码前一个对象创建的任何类型的字符串或数据。我只能得出这样的结论:由于这段代码采用了相当弱的 passPhrase 字符串，混合了 SALT，并构建了一个更强的 128 位 key - 这个 key 构建过程每次都以某种方式随机化。其意义在于:

new AESEncrypter("MyPassword") <> new AESEncrypter("MyPassword")

有人可以帮我修改下面的类以获得所需的行为吗:

AESEncrypter a = new AESEncrypter("MyPassword")
String encoded = a.encrypt("my message")

AESEncrypter b = new AESEncrypter("MyPassword")
b.decrypt(encoded) == "my message"

import java.security.spec.KeySpec;
import javax.crypto.Cipher;
import javax.crypto.SecretKey;
import javax.crypto.SecretKeyFactory;
import javax.crypto.spec.IvParameterSpec;
import javax.crypto.spec.PBEKeySpec;
import javax.crypto.spec.SecretKeySpec;
import sun.misc.BASE64Decoder;
import sun.misc.BASE64Encoder;

public class AESEncrypter {

    private static final byte[] SALT = {
            (byte) 0xA9, (byte) 0x9B, (byte) 0xC8, (byte) 0x32,
            (byte) 0x56, (byte) 0x35, (byte) 0xE3, (byte) 0x03
    };
    private static final int ITERATION_COUNT = 65536;
    private static final int KEY_LENGTH = 128;
    public Cipher ecipher;
    public Cipher dcipher;

    AESEncrypter(String passPhrase) throws Exception {

    SecretKeyFactory factory = SecretKeyFactory.getInstance("PBKDF2WithHmacSHA1");
        KeySpec spec = new PBEKeySpec(passPhrase.toCharArray(), SALT, ITERATION_COUNT, KEY_LENGTH);
        SecretKey tmp = factory.generateSecret(spec);

        // I Think the problem is here???

        SecretKey secret = new SecretKeySpec(tmp.getEncoded(), "AES");

        ecipher = Cipher.getInstance("AES/CBC/PKCS5Padding");
        ecipher.init(Cipher.ENCRYPT_MODE, secret);

        dcipher = Cipher.getInstance("AES/CBC/PKCS5Padding");
        byte[] iv = ecipher.getParameters().getParameterSpec(IvParameterSpec.class).getIV();
        dcipher.init(Cipher.DECRYPT_MODE, secret, new IvParameterSpec(iv));
    }

    public String encrypt(String encrypt) throws Exception {
        byte[] bytes = encrypt.getBytes("UTF8");
        byte[] encrypted = encrypt(bytes);
        return new BASE64Encoder().encode(encrypted);
    }

    public byte[] encrypt(byte[] plain) throws Exception {
        return ecipher.doFinal(plain);
    }

    public String decrypt(String encrypt) throws Exception {
        byte[] bytes = new BASE64Decoder().decodeBuffer(encrypt);
        byte[] decrypted = decrypt(bytes);
        return new String(decrypted, "UTF8");
    }

    public byte[] decrypt(byte[] encrypt) throws Exception {
        return dcipher.doFinal(encrypt);
    }

   public static void main(String[] args) throws Exception {

        String message = "MESSAGE";
        String password = "PASSWORD";

        AESEncrypter encrypter1 = new AESEncrypter(password);
        AESEncrypter encrypter2 = new AESEncrypter(password);

        String encrypted1 = encrypter1.encrypt(message);
        String encrypted2 = encrypter2.encrypt(message);

        System.out.println("Display Encrypted from object 1 and 2..why do they differ?" );

        System.out.println(encrypted1) ;
        System.out.println(encrypted2) ;

        System.out.println("Display Each object decrypting its own encrypted msg. Works as expected" );

        System.out.println(encrypter1.decrypt(encrypted1)) ;
        System.out.println(encrypter2.decrypt(encrypted2)) ;

        System.out.println("Attempt to decrypt the each others msg.. will fail" );

        System.out.println(encrypter1.decrypt(encrypted2)) ;
        System.out.println(encrypter2.decrypt(encrypted1)) ;

    }

}

显示对象 1 和 2 加密的内容......为什么它们不同？

drGy+BNSHPy34NWkkcNqLQ==9p06VfBgTuh7TizZSbvKjw==

Display Each object decrypting its own encrypted msg. Works as expected

MESSAGEMESSAGE

Attempt to decrypt the each others msg.. will fail

Error:

Exception in thread "main" javax.crypto.BadPaddingException: Given final block not properly padded
    at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:966)
    at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:824)
    at com.sun.crypto.provider.AESCipher.engineDoFinal(AESCipher.java:436)
    at javax.crypto.Cipher.doFinal(Cipher.java:2165)

Answer 1

问题是，当您在 CBC 模式下初始化新的 Cipher 时，它会为您生成一个新的随机 IV。该初始化 vector 不必是 secret 的，但它必须是不可预测的以提供语义安全性。你可以简单地将IV放在密文前面并用它来解密。

public byte[] encrypt(byte[] plain) throws Exception {
    byte[] iv = ecipher.getParameters().getParameterSpec(IvParameterSpec.class).getIV();
    byte[] ct = ecipher.doFinal(plain);
    byte[] result = new byte[ct.length + iv.length];
    System.arraycopy(iv, 0, result, 0, iv.length);
    System.arraycopy(ct, 0, result, iv.length, ct.length);
    return result;
}

public byte[] decrypt(byte[] encrypt) throws Exception {
    byte[] iv = new byte[dcipher.getBlockSize()];
    byte[] ct = new byte[encrypt.length - dcipher.getBlockSize()];
    System.arraycopy(encrypt, 0, iv, 0, dcipher.getBlockSize());
    System.arraycopy(encrypt, dcipher.getBlockSize(), ct, 0, ct.length);

    dcipher.init(Cipher.DECRYPT_MODE, secret, new IvParameterSpec(iv));
    return dcipher.doFinal(ct);
}

您需要在初始化步骤中将 secret 存储在私有(private)变量中，以便解密工作。

<小时/>

请记住，PBDKF2 的盐也应该是随机的且长度为 16 个字节。您可以将其与 IV 一起存储。