使用 AES 和 Base64 编码加密和解密 [英] Encrypt and decrypt with AES and Base64 encoding

问题描述

我有以下加密数据的程序.

I have following program for encrypting data.

import java.security.Key;

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

import org.apache.commons.codec.binary.Base64;

public class Test {

    private static final String ALGORITHM = "AES";
    private static final byte[] keyValue = "ADBSJHJS12547896".getBytes();

    public static void main(String args[]) throws Exception {
        String encriptValue = encrypt("dude5");
        decrypt(encriptValue);

    }

    /**
     * @param args
     * @throws Exception
     */

    public static String encrypt(String valueToEnc) throws Exception {

        Key key = generateKey();
        Cipher c = Cipher.getInstance(ALGORITHM);
        c.init(Cipher.ENCRYPT_MODE, key);

        System.out.println("valueToEnc.getBytes().length "+valueToEnc.getBytes().length);
        byte[] encValue = c.doFinal(valueToEnc.getBytes());
        System.out.println("encValue length" + encValue.length);
        byte[] encryptedByteValue = new Base64().encode(encValue);
        String encryptedValue = encryptedByteValue.toString();
        System.out.println("encryptedValue " + encryptedValue);

        return encryptedValue;
    }

    public static String decrypt(String encryptedValue) throws Exception {
        Key key = generateKey();
        Cipher c = Cipher.getInstance(ALGORITHM);
        c.init(Cipher.DECRYPT_MODE, key);

        byte[] enctVal = c.doFinal(encryptedValue.getBytes());
        System.out.println("enctVal length " + enctVal.length);

        byte[] decordedValue = new Base64().decode(enctVal);

        return decordedValue.toString();
    }

    private static Key generateKey() throws Exception {
        Key key = new SecretKeySpec(keyValue, ALGORITHM);
        return key;
    }

}

这里我得到了以下内容，但有例外?

Here I am getting the following out put with exception?

valueToEnc.getBytes().length 5
encValue length16
encryptedValue [B@aa9835
Exception in thread "main" javax.crypto.IllegalBlockSizeException: Input length must be multiple of 16 when decrypting with padded cipher
    at com.sun.crypto.provider.SunJCE_f.b(DashoA13*..)
    at com.sun.crypto.provider.SunJCE_f.b(DashoA13*..)

有人能解释一下原因吗?为什么它在解密时只说那个长度应该是 16.它不会像使用 doFinal 方法加密一样转换为 16.

Can some one explain me the cause? Why its only saying when decrypting that length should be 16. Doesn't it convert to 16 as like encrypting with the doFinal method.

正如例外所说如何在没有填充密码的情况下解密?"

推荐答案

您的加密命令: getBytes, encrypt, encode, toString
您的解密顺序(错误*): getBytes、decrypt、decode、toString

两个问题:

1. 正如有人已经提到的，您应该颠倒解密操作的顺序.你没有这样做.
   
2. encrypt 给你 16 个字节，编码 24 个字节，但 toString 给你 106 个字节.与占用额外空间的无效字符有关.

注意:另外，您不需要调用 generateKey() 两次.

Note: Also, you don't need to call generateKey() twice.

解决问题 #1 使用相反的解密顺序.
解密的正确顺序: getBytes，decode，decrypt，toString

Fix problem #1 by using the reverse order for decryption.
Correct order for decrypt: getBytes, decode, decrypt, toString

修复问题 #2，将 xxx.toString() 替换为 new String(xxx).在加密和解密函数中执行此操作.

Fix problem #2 by replacing xxx.toString() with new String(xxx). Do this in both the encrypt and decrypt functions.

您的解密应如下所示:

c.init(Cipher.DECRYPT_MODE, key)
val decodedValue = new Base64().decode(encryptedValue.getBytes())
val decryptedVal = c.doFinal(decodedValue)
return new String(decryptedVal)

这应该会给你回dude5"

This should give you back "dude5"

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