Java AES解密BadPaddingException [英] Java AES decryption BadPaddingException

问题描述

注意：Java NOOB。

note: Java NOOB.

好吧，我知道这里已经回答了几十次，但解决方案似乎没有工作/直接应用到我理解他们的地方。 （是的，我知道我不完全理解加密/解密，AES等等，但这不是重点，我试图理解这一点）

Alright, I know this has been answered a few dozen times on here, but the solutions don't seem to work/apply directly to where I understand them. (Yes, I know I don't completely understand encryption/decryption, AES, etc. but that is not the point, I am trying to understand this)

我有一个实用程序api，我想传递一个字符串并返回一个加密的字符串。然后我想传递加密的字符串，并返回一个解密的字符串。简单。它适用于我传入的许多字符串，但在某些情况下，我得到异常 javax.crypto.BadPaddingException：给定最终块没有正确填充。

I have a utility api where I want to pass a string and return an encrypted string. Then I want to pass the encrypted string, and return a decrypted string. Simple. It works fine for many strings I pass in, but on some, I'm getting the exception javax.crypto.BadPaddingException: Given final block not properly padded.

例如，以下加密/解密罚款。

util / encrypt /？token = 123456789012wha = 4TR0CbCcQKqeRK73zr83aw ==

util / decrypt /？token = 4TR0CbCcQKqeRK73zr83aw == = 123456789012wha

For example, the following encrypts/decrypts fine.
util/encrypt/?token=123456789012wha = 4TR0CbCcQKqeRK73zr83aw==
util/decrypt/?token=4TR0CbCcQKqeRK73zr83aw== = 123456789012wha

以下加密，但不解密：

util / encrypt /？token = 123456789012what = NYaWmwnySoGNHyNmY9Jh + f3O2rqoLI1IAcnsl5V4OCE =

util / decrypt /？token = NYaWmwnySoGNHyNmY9Jh + f3O2rqoLI1IAcnsl5V4OCE = = exception

The following encrypts, but does not decrypt:
util/encrypt/?token=123456789012what = NYaWmwnySoGNHyNmY9Jh+f3O2rqoLI1IAcnsl5V4OCE=
util/decrypt/?token=NYaWmwnySoGNHyNmY9Jh+f3O2rqoLI1IAcnsl5V4OCE= = exception

这是我控制器中的代码：

Here is the code in my controller:

private static final String ALGO = "AES";


@RequestMapping(value = "/util/encrypt/", method = RequestMethod.GET)
@ResponseBody
public String encrypt(HttpServletResponse httpResponse,
        @RequestParam(value = "token", required=true) String token) throws Exception 
{
    Key key = generateKey();
    Cipher c = Cipher.getInstance(ALGO);
    c.init(Cipher.ENCRYPT_MODE, key);
    byte[] encVal = c.doFinal(token.getBytes());
    String encryptedValue = Base64.encodeBase64String(encVal);
    return encryptedValue.trim();
}



@RequestMapping(value = "/util/decrypt/", method = RequestMethod.GET)
@ResponseBody
public String decrypt(HttpServletResponse httpResponse,
        @RequestParam(value = "token", required=true) String token) throws Exception 
{
    token = URLDecoder.decode(token, "UTF-8");
    Key key = generateKey();
    Cipher c = Cipher.getInstance(ALGO);
    c.init(Cipher.DECRYPT_MODE, key);
    byte[] decordedValue = Base64.decodeBase64(token);
    byte[] decValue = c.doFinal(decordedValue);
    String decryptedValue = new String(decValue);
    return decryptedValue.trim();
}



private Key generateKey() throws Exception 
{
    Key key = new SecretKeySpec(getAesKey().getBytes(), ALGO);
    return key;
}

我认为它必须是通话电话 Cipher .getInstance（）我尝试使用 Cipher.getInstance（AES / CBC / PKCS5Padding），但在解密时总是失败。我很想知道这里发生了什么以及如何解决它。

I figure it must be something with the call Cipher.getInstance() and I've tried using Cipher.getInstance("AES/CBC/PKCS5Padding") but that always fails when decrypting. I would love to really understand what is happening here and how to fix it.

推荐答案

使用函数 encodeBase64URLSafeString 。 javadoc说

Use the function encodeBase64URLSafeString. the javadoc says

使用base64算法
的URL安全变体对二进制数据进行编码，但不会对输出进行分块。 url-safe变体发出 - 和_
而不是+和/字符。注意：没有添加填充。

Encodes binary data using a URL-safe variation of the base64 algorithm but does not chunk the output. The url-safe variation emits - and _ instead of + and / characters. Note: no padding is added.

这应该可以解决问题。

这篇关于Java AES解密BadPaddingException的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！