AES 256位加密 - java.security.InvalidAlgorithmParameterException:错误的IV长度:必须是16个字节长

时间:2014-12-08 03:59:07

标签: java encryption cryptography aes salt

以下是我的加密逻辑。虽然我的IV是16字节长,但我仍然会收到无效IV长度的错误。非常感谢任何帮助

@Override
public String encrypt(String dataToEncrypt, String IV) throws Exception{
    if(encryptionKey.length() < 10){
        encryptionKey = generateEncryptionKey().toString();
    }
    System.out.println("number of IV bytes is "+IV.length()+" "+IV);
    Cipher cipher = Cipher.getInstance(encrpytionAlgo);
    SecretKey key = new SecretKeySpec(encryptionKey.getBytes(Charset.forName("UTF-8")), "AES");
    cipher.init(Cipher.ENCRYPT_MODE, key,new IvParameterSpec(IV.getBytes(Charset.forName("UTF-8"))));
    byte[] encryptedTextBytes = cipher.doFinal(dataToEncrypt.getBytes(Charset.forName("UTF-8")));
    return new Base64().encodeAsString(encryptedTextBytes);
}

IV和密钥生成逻辑

 @Override
public String generateRandomIV(){
        Random random = new SecureRandom();
        byte[] iv = new byte[16];
        random.nextBytes(iv);
        System.out.println("IV is "+Base64.encodeBase64(iv)+" "+ com.sun.jersey.core.util.Base64.base64Decode(new String(Base64.encodeBase64(iv)))+ " number of bytes is "+iv.length);
        return new String(Base64.encodeBase64(iv));
}

@Override
public SecretKey generateEncryptionKey(){
    KeyGenerator aesKey = null;
    try {
        aesKey = KeyGenerator.getInstance("AES");
    } catch (NoSuchAlgorithmException e) {
          e.printStackTrace();
    }
    aesKey.init(256);
    SecretKey secretKey = aesKey.generateKey();
    System.out.println("Encryption key is "+ new Base64().encode(secretKey.getEncoded()));
    return secretKey;
}

以下是异常的堆栈跟踪 例外情况符合:
cipher.init(Cipher.ENCRYPT_MODE,key,new IvParameterSpec(IV.getBytes(Charset.forName(&#34; UTF-8&#34;))));

java.security.InvalidAlgorithmParameterException: Wrong IV length: must be 16 bytes long
at com.sun.crypto.provider.SunJCE_f.a(DashoA13*..)
at com.sun.crypto.provider.AESCipher.engineInit(DashoA13*..)
at javax.crypto.Cipher.a(DashoA13*..)
at javax.crypto.Cipher.a(DashoA13*..)
at javax.crypto.Cipher.init(DashoA13*..)
at javax.crypto.Cipher.init(DashoA13*..)
at com.intuit.platform.publiccloudaccess.core.services.EncryptionServiceImpl.encrypt(EncryptionServiceImpl.java:47)

2 个答案:

答案 0 :(得分:3)

在从generateRandomIV返回之前,您已将IV编码为Base64。在将其用于加密和解密之前,您必须对其进行解码。

cipher.init(Cipher.ENCRYPT_MODE, key, new IvParameterSpec(java.util.Base64.Decoder.decode(IV.getBytes("UTF-8"))));

Java 8为获取不同的Base 64编码器和解码器提供了java.util.Base64类。

答案 1 :(得分:1)

按照Rob的评论,

System.out.println("number of IV bytes is "+IV.length()+" "+IV);

在这里,您可以根据字符串获得IV的长度。然而

cipher.init(Cipher.ENCRYPT_MODE, key,new IvParameterSpec(IV.getBytes(Charset.forName("UTF-8"))));    

这里提供的是IV字节数组,并且在字符串的长度中将IV的长度设置为16并不能保证其字节表示也是16字节。因此,Rob建议将IV保留在字节数组中并将其用作字节数组会更好。