为什么这种简单的AES加密不起作用？

Question

为什么这种AES加密不起作用？我用Java编写它来测试,但我无法解密.解密后我得到了垃圾.为什么？它如此简单 - 在主要方法中,打印纯文本,加密,打印密文,解密,再次打印纯文本.难道我做错了什么？请帮我弄清楚问题所在.

import javax.crypto.Cipher;
import javax.crypto.spec.IvParameterSpec;
import javax.crypto.spec.SecretKeySpec;

public class AESTest {
    public static void main(String [] args) {
        try {
            String plainText = "Hello World!!!!!";
            String encryptionKey = "E072EDF9534053A0B6C581C58FBF25CC";

            System.out.println("Before encryption - " + plainText);

            String cipherText = encrypt(plainText, encryptionKey);

            System.out.println("After encryption - " + cipherText);

            String decrypted = decrypt(cipherText, encryptionKey);

            System.out.println("After decryption - " + decrypted);
        } catch (Exception e) {
            e.printStackTrace();
        } 
    }

    public static String encrypt(String plainText, String passkey) throws Exception {
        Cipher cipher = Cipher.getInstance("AES/CBC/NoPadding", "SunJCE");
        SecretKeySpec key = new SecretKeySpec(hexStringToByteArray(passkey), "AES");
        cipher.init(Cipher.ENCRYPT_MODE, key, new IvParameterSpec(new byte[cipher.getBlockSize()]));
        String cipherText = new String(cipher.doFinal(plainText.getBytes()));
        return cipherText;
    }

    public static String decrypt(String cipherText, String passkey) throws Exception{
        Cipher cipher = Cipher.getInstance("AES/CBC/NoPadding", "SunJCE");
        SecretKeySpec key = new SecretKeySpec(hexStringToByteArray(passkey), "AES");
        cipher.init(Cipher.DECRYPT_MODE, key, new IvParameterSpec(new byte[cipher.getBlockSize()]));
        String plainText = new String(cipher.doFinal(cipherText.getBytes()));
        return plainText;
    }

    public static byte[] hexStringToByteArray(String s) {
        int len = s.length();
        byte[] data = new byte[len / 2];
        for (int i = 0; i < len; i += 2) {
            data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4)
                                 + Character.digit(s.charAt(i+1), 16));
        }
        return data;
    }
}

Answer 1

密码的输出是一系列随机字节.您不能保证这些字节对于系统的默认编码中的字符串是有效的编码.所以这一行:

 String cipherText = new String(cipher.doFinal(.....));

可能会丢失您需要解密的信息.

因此,您将无法在decrypt操作中重建正确的字节.例如,如果您的默认编码是UTF-8,则绝对不可能正确的密文String.getBytes()是甚至能够生成的密文.