java - sun.misc.BASE64 到 apache commons

Question

我这里有一个简单的代码，该代码使用sun.misc.BASE64Encoder和sun.misc.BASE64Decoder，它们在Eclipse Java 7.0中不可用，我想让代码使用Apache commons base 64并且仍然做同样的事情，

变量

private static final String ALGORITHM = "AES";
private static final byte[] keyValue = "2H5a1r5i6s3h8C1h".getBytes();

原始代码

public static String AESencrypt(String valueToEnc) throws Exception {
    Key key = new SecretKeySpec(keyValue, ALGORITHM);
    Cipher c = Cipher.getInstance(ALGORITHM);
    c.init(Cipher.ENCRYPT_MODE, key);
    byte[] encValue = c.doFinal(valueToEnc.getBytes());
    String encryptedValue = new BASE64Encoder().encode(encValue);
    return encryptedValue;
}

public static String AESdecrypt(String encryptedValue) throws Exception {
    Key key = new SecretKeySpec(keyValue, ALGORITHM);
    Cipher c = Cipher.getInstance(ALGORITHM);
    c.init(Cipher.DECRYPT_MODE, key);
    byte[] decordedValue = new BASE64Decoder().decodeBuffer(encryptedValue);
    byte[] decValue = c.doFinal(decordedValue);
    String decryptedValue = new String(decValue);
    return decryptedValue;
}

尝试修改apache commons代码

public static String AESencrypt(String valueToEnc) throws Exception {
    Key key = new SecretKeySpec(keyValue, ALGORITHM);
    Cipher c = Cipher.getInstance(ALGORITHM);
    c.init(Cipher.ENCRYPT_MODE, key);
    byte[] encValue = c.doFinal(valueToEnc.getBytes());
    String encryptedValue = new Base64().encodeBase64(encValue).toString();
    return encryptedValue;
}

public static String AESdecrypt(String encryptedValue) throws Exception {
    Key key = new SecretKeySpec(keyValue, ALGORITHM);
    Cipher c = Cipher.getInstance(ALGORITHM);
    c.init(Cipher.DECRYPT_MODE, key);
    byte[] decordedValue = new Base64().decodeBase64(encryptedValue);
    byte[] decValue = c.doFinal(decordedValue);
    String decryptedValue = new String(decValue);
    return decryptedValue;
}

原来的代码工作正常，新代码抛出以下异常，

Exception in thread "main" javax.crypto.IllegalBlockSizeException: Input length must be multiple of 16 when decrypting with padded cipher at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:750) at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:676) at com.sun.crypto.provider.AESCipher.engineDoFinal(AESCipher.java:313) at javax.crypto.Cipher.doFinal(Cipher.java:2087) at trial2.encrypt.AESdecrypt(encrypt.java:28) at trial2.encrypt.main(encrypt.java:37) Java Result: 1

如何在不改变初始代码太多的情况下解决这个问题，这里那里几行就可以了。是否有可能完全消除 Base64 编码步骤并使其仅适用于这样的密码:

public static String AESencrypt(String valueToEnc) throws Exception {
    Key key = new SecretKeySpec(keyValue, ALGORITHM);
    Cipher c = Cipher.getInstance(ALGORITHM);
    c.init(Cipher.ENCRYPT_MODE, key);
    byte[] encValue = c.doFinal(valueToEnc.getBytes());
    return encValue.toString();
}

public static String AESdecrypt(String encryptedValue) throws Exception {
    Key key = new SecretKeySpec(keyValue, ALGORITHM);
    Cipher c = Cipher.getInstance(ALGORITHM);
    c.init(Cipher.DECRYPT_MODE, key);
    byte[] decValue = c.doFinal(encryptedValue.getBytes());
    String decryptedValue = new String(decValue);
    return decryptedValue;
}

上面的代码与第二个代码有类似的异常:

Exception in thread "main" javax.crypto.IllegalBlockSizeException: Input length must be multiple of 16 when decrypting with padded cipher at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:750) at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:676) at com.sun.crypto.provider.AESCipher.engineDoFinal(AESCipher.java:313) at javax.crypto.Cipher.doFinal(Cipher.java:2087) at trial2.encrypt.AESdecrypt(encrypt.java:26) at trial2.encrypt.main(encrypt.java:35) Java Result: 1

请帮助 TYVM

Answer 1

错误出在您的加密方法中。以下代码是错误的:

byte[] encValue = c.doFinal(valueToEnc.getBytes());
String encryptedValue = new Base64().encodeBase64(encValue).toString();

您正在对字节数组调用toString()，这不会执行您想要的操作!相反，请尝试:

byte[] encValue = c.doFinal(valueToEnc.getBytes());
String encryptedValue = Base64.encodeBase64String(encValue);

这应该可以正常工作。