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我在这里有一个简单的代码，该代码使用了 Eclipse Java 7.0 中不可用的 sun.misc.BASE64Encoder 和 sun.misc.BASE64Decoder，我想让代码使用 Apache commons base 64 并且仍然可以完全一样的东西，

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;
}

尝试修改为 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;
}

原始代码工作正常，新代码抛出以下异常，

线程“main”javax.crypto.IllegalBlockSizeException 中的异常：在 com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:750) 在 com.sun.crypto 使用填充密码解密时，输入长度必须是 16 的倍数。 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 结果：1

如何在不过多更改初始代码的情况下解决这个问题，这里和那里的几行就可以了。是否可以完全消除 Base64encoding 步骤并使其仅使用如下密码：

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：

线程“main”javax.crypto.IllegalBlockSizeException 中的异常：在 com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:750) 在 com.sun.crypto 使用填充密码解密时，输入长度必须是 16 的倍数。 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 结果：1

请帮助和TYVM

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1 回答 1

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错误在您的加密方法中。以下代码是错误的：

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

你正在调用toString()一个字节数组，它不会做你想要的！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。

于 2013-03-05T12:38:10.863 回答