JAVA AES 256解密 [英] JAVA AES 256 Decrypt

问题描述

我正在尝试使用AES解密一些数据。我得到了一个256位密钥和16个字节的IV，如下所示：

I'm trying to decrypt with AES some data. I've been given a 256 bit key and 16 byte IV like these:

  String key = "Hh1s1f4T2mpN3yCh4ngeL8t3r\\.Thxpp";

  int[] v = {11, 1, 555, 222, 241, 21, 11, 33, 35, 91, 45, 6, 14, 30, 22, 234};

  String IV = Arrays.toString( v );

有人告诉我填充应该是PKCS7，但是当我用<$ c $初始化密码时c> AES / CBC / PKCS7PADDING 它表示：找不到任何支持AES / CBC / PKCS7PADDING的提供商

I've been told the padding should be PKCS7 but when I init the cipher with AES/CBC/PKCS7PADDING it says: Cannot find any provider supporting AES/CBC/PKCS7PADDING

如果我使用 AES / CBC / PKCS5PADDING 我会得到密钥大小非法，但是我已经检查过了密钥大小为32。

If I use AES/CBC/PKCS5PADDING I get Illegal key size but I've checked that the key size is 32.

  public static String decrypt(String key, String initVector, String encrypted) {
    try {
        System.out.println( "Key size: " + key.getBytes("UTF-8").length ); 

        IvParameterSpec iv = new IvParameterSpec(initVector.getBytes("UTF-8"));
        SecretKeySpec skeySpec = new SecretKeySpec(key.getBytes("UTF-8"), "AES");

        Cipher cipher = Cipher.getInstance("AES/CBC/PKCS5PADDING");
        cipher.init(Cipher.DECRYPT_MODE, skeySpec, iv);

        byte[] original = cipher.doFinal(Base64.decodeBase64(encrypted));

        return new String(original);
    } catch (Exception ex) {
        ex.printStackTrace();
    }

    return null;
 }

推荐答案

如果您得到了 IllegalKeySize 具有 AES-256 的异常，请检查以确保您具有 JCE无限加密强度策略文件。对于任何 128位的AES密钥长度，它们都是必需的。

If you are getting an IllegalKeySize exception with AES-256, check to ensure you have the JCE Unlimited Cryptographic Strength Policy files installed in your active JVM. They are required for any AES key length over 128 bits.

（PKCS＃7填充方案有效地等同于PKCS＃5在Java 中（块大小在规范定义中有所不同），但是Java从未在其列表中添加PKCS＃7名称，因此使用 PKCS5Padding 是正确的。）

(PKCS #7 padding scheme is effectively equivalent to PKCS #5 in Java (the block size differs in the spec definitions), but Java never added the PKCS #7 name to its list, so using PKCS5Padding is correct.)

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