23

我可以像这样使用 Ruby 解密 AES 加密消息：

require 'openssl'
require 'base64'

data = "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"
key = "2e35f242a46d67eeb74aabc37d5e5d05"

aes = OpenSSL::Cipher::Cipher.new('aes-128-cbc')
aes.decrypt
aes.key = key.scan(/../).collect{ |x| x.hex }.pack('c*')
aes.iv = Base64.decode64(data)[0...16]
puts aes.update(Base64.decode64(data)[16..-1]) + aes.final
# => JSON data...

作为 CryptoJS 的新手，我无法组合出可行的替代方案。这是我到目前为止所做的：

data = "IYkyGxYaNgHpnZWgwILMalVFmLWFgTCHCZL9263NOcfSo5lBjAzOZAtF5bF++R0Bi+9c9E+p3VEr/xvj4oABtRWVJ2wlWzLbYC2rKFk5iapFhb7uZCUpO4w4Su3a5QFa2vInjYueziRoqySZd/DpstMJ8rsJ94VGizFFFZ1l0sw1ax+wfBAv5+wHs/hlnHi/ea66KBO3rgXKahvV28h+4bh5etc8RCrmiiNbfg6Oj0jQJDjdYIdW8T9YPOI9E1hih8lbfRnMWcOFJgYekfLpoy5LI525UGnlM46J1k6ekLqsn9FqvbiOOoLgqa4YqBm1i9P0ePyjkME+t+RiL8xXX+ItgOYr9G7kM64wlTJPCW8B/crmUdmGzQNC/hD/u/8wfHBS2f8u6OtQMG/+Kpk1oju8lcUZGI/4S8A6/OuktvQr2zgnbs2aADMrM37Oait/pJ3G73S7NwVT8EaK+X43c0C/fUvW2/bD/rqCNpAh9WQlz4Cj6JHwjbmwuind6aCimF1tHjXuR9FXu+g17sPT4ZkKZ6aeBG+m170XdCGn2hVM0wH1rh3VeCG2u/JFqfuGKGSoqeHeNY/icu9pEhtZDzHd7aPoaMXcWvXC9PjooBf7GM1EPacSdnon1kBobjtKSt1l15DjO5TMrJoX7VO7GotQwo+uI/u5Kop01hBXxyxyggl1/8N0ESohPJoqLDrIwvbGK5kW4B49FVPnx9CMvjZDdSsoxPAh+hx6SPe8Hj0Nx4bRs06cbtOkte/V8QSYIqjiJDleEqPrdiKlvgToZz9L29ZR/3Ln65qU1sq7q9c0SEYxIopV7TdTjFS7y76zDPFZkhzc3DjfLtJo/M1hdtt648APcZdmAIgWH6fh3eJZ0qbiPh8RStYH7I2COmnlMw4+t/B5mlhYVSgwPK2Ir736Mh+P9Bw0fF8r9Ghhs4AJzpU0RiK9d1tCsrLfK/hSRwTXhtsSB6eDWEGkO7oeEIz43mgn0sv3SrA9JNHzYkg=";
key = "2e35f242a46d67eeb74aabc37d5e5d05";

CryptoJS.AES.decrypt(atob(data).substring(16), key, {
    iv: atob(cipher).substring(0, 16),
    mode: CryptoJS.mode.CBC,
    padding: CryptoJS.pad.Pkcs7
});

CryptoJS 是否以某种不同的方式期望数据、密钥和初始化向量？据我了解，IV 是数据的前 16 个字节。

4

5 回答 5

59

这适用于使用 javascript 进行解密。

<script src="https://cdnjs.cloudflare.com/ajax/libs/crypto-js/3.1.9-1/aes.js"></script>
var key = "2e35f242a46d67eeb74aabc37d5e5d05";
var data = CryptoJS.AES.encrypt("Message", key); // Encryption Part
var decrypted = CryptoJS.AES.decrypt(data, key).toString(CryptoJS.enc.Utf8); // Message

我想我参加聚会有点晚了。

于 2014-03-24T05:05:10.337 回答
36

这似乎对我有用（将解码的字符串写入控制台）：

data = "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";
key = "2e35f242a46d67eeb74aabc37d5e5d05";

// Decode the base64 data so we can separate iv and crypt text.
var rawData = atob(data);
var iv = btoa(rawData.substring(0,16));
var crypttext = btoa(rawData.substring(16));

// Decrypt...
var plaintextArray = CryptoJS.AES.decrypt(
  {
    ciphertext: CryptoJS.enc.Base64.parse(crypttext),
    salt: ""
  },
  CryptoJS.enc.Hex.parse(key),
  { iv: CryptoJS.enc.Base64.parse(iv) }
);

// Convert hex string to ASCII.
// See https://stackoverflow.com/questions/11889329/word-array-to-string
function hex2a(hex) {
    var str = '';
    for (var i = 0; i < hex.length; i += 2)
        str += String.fromCharCode(parseInt(hex.substr(i, 2), 16));
    return str;
}

console.log(hex2a(plaintextArray.toString()));

正如 jlvaquero 所说，重要的一点是在您的输入中指定无盐并将密钥作为 WordArray 传递。可能需要进行一些效率改进，例如，我猜有一种将字符串转换为 WordArray 的方法比转到 base64 并返回更简单，但至少它可以成功解码。我将 hex2a 函数从 StackOverflow 问题Word Array 提升到 String。

编辑：我想出了如何使用 CryptoJS 将字符串转换为 WordArrays，反之亦然，因此不需要额外的 base64 编码/解码和十六进制到 ASCII 函数。事实证明，不需要指定 salt 属性。这更加紧凑和高效：

data = "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";
key = "2e35f242a46d67eeb74aabc37d5e5d05";

// Decode the base64 data so we can separate iv and crypt text.
var rawData = atob(data);
var iv = rawData.substring(0,16);
var crypttext = rawData.substring(16);

// Decrypt...
var plaintextArray = CryptoJS.AES.decrypt(
  { ciphertext: CryptoJS.enc.Latin1.parse(crypttext) },
  CryptoJS.enc.Hex.parse(key),
  { iv: CryptoJS.enc.Latin1.parse(iv) }
);

console.log(CryptoJS.enc.Latin1.stringify(plaintextArray));
于 2013-02-28T00:03:07.440 回答
2 
encryptWithCryptoJS(plainText: string): string {
    const key = CryptoJS.enc.Utf8.parse("hf8685nfhfhjs9h8");
    const iv1 = CryptoJS.enc.Utf8.parse("hf8685nfhfhjs9h8");
    const encrypted = CryptoJS.AES.encrypt(plainText, key, {
        keySize: 16,
        iv: iv1,
        mode: CryptoJS.mode.ECB,
        padding: CryptoJS.pad.Pkcs7
    });

    return encrypted + "";
}

decryptionWithCryptoJS(cipher: string): string {
    const key = CryptoJS.enc.Utf8.parse("hf8685nfhfhjs9h8");
    const iv1 = CryptoJS.enc.Utf8.parse("hf8685nfhfhjs9h8");
    const plainText = CryptoJS.AES.decrypt(cipher, key, {
        keySize: 16,
        iv: iv1,
        mode: CryptoJS.mode.ECB,
        padding: CryptoJS.pad.Pkcs7
    });

    return plainText.toString(CryptoJS.enc.Utf8);
}
于 2019-01-19T07:01:31.087 回答
0 
    var key = CryptoJS.enc.Utf8.parse('8080808080808080');
    var iv = CryptoJS.enc.Utf8.parse('8080808080808080');
    var _enid = CryptoJS.AES.decrypt(data, key,
  {
      keySize: 128 / 8,
      iv: iv,
      mode: CryptoJS.mode.CBC,
      padding: CryptoJS.pad.Pkcs7
  }).toString(CryptoJS.enc.Utf8);
    return _enid;
于 2018-10-08T12:49:01.873 回答
0

对于其他用户提供的 hex2a，如果 ascii 码超过 128（即文本包含中文等），它可能无法正常工作

您可以使用以下内容返回正确的 unicode

function hex2a(hex) {
    var str = '';
    for (var i = 0; i < hex.length; i += 2){

        var dec = parseInt(hex.substr(i, 2), 16);
        character = String.fromCharCode(dec);


        if (dec > 127)
            character = "%"+hex.substr(i,2);

        str += character;

    }

    return decodeURI(str);
}
于 2016-04-22T01:43:40.217 回答