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有没有人幸运地使用任何 javascript 库加密外国字符?

我正在使用 jCryption v. 1.4 库(RSA 算法)加密外来字符。看起来该库不适用于外来字符(我使用的是 cp1251)。加密适用于 ASCII/英文字符,但我在解密外国字符后得到乱码。

我相信问题来自 jCryption 将 char 字符串转换为字节的方式。它将 char 仅移动 8 位,并将 & 的 2 char 一起移动。外国字符肯定占据> 8位。

任何想法,将不胜感激!
德米特里

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2 回答 2

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您可以先尝试对您的消息进行 base64 编码,然后再进行加密。这将确保输入只有 ascii 字符。

因此,您的算法可能类似于此伪代码。

Encryption
input = "text";
input = base64Encode(input); //there's lots of implementations of this online
cipher = encodeWithJcrypt(input);

Decryption
text = decodeWithJcrypt(cipher);
text = base64Decode(text);
于 2012-02-24T22:53:04.857 回答
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这是 16 位 Unicode 字符(javascript)的 base64 编码:

var Base64 = {};  // Base64 namespace

Base64.code = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";

Base64.encode = function(str) { 
    var o1, o2, bits, h1, h2, h3, h4, h5, d6, e=[], pad = '', c, plain, coded;
    var b64 = Base64.code;

    plain = str;

    c = plain.length % 2;  // pad string to length of multiple of 3
    if (c > 0) { while (c++ < 2) { pad += '==='; plain += '\0'; } }

    for (c=0; c<plain.length; c+=2) {  // pack 2 hexadecets into 6 hexets
        o1 = plain.charCodeAt(c);
        o2 = plain.charCodeAt(c+1);

        bits = o1<<16 | o2;

        h1 = bits>>26 & 0x3f;
        h2 = bits>>20 & 0x3f;
        h3 = bits>>14 & 0x3f;
        h4 = bits>>8 & 0x3f;
        h5 = bits>>2 & 0x3f;
        d6 = bits & 0x3;

        // use hextets to index into code string
        e[c/2] = b64.charAt(h1) + b64.charAt(h2) + b64.charAt(h3) + b64.charAt(h4)
            + b64.charAt(h5) + b64.charAt(d6);
    }
    coded = e.join('');  // join() is far faster than repeated string concatenation in IE

    // replace 'A's from padded nulls with '='s
    coded = coded.slice(0, coded.length-pad.length) + pad;

    return coded;
}

这是 16 位 unicode 字符 (java) 的 base64 解码:

private static String BASE64_CODE = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";
private String decodeBase64(String b64Str) {
    try {
        String decodedBase64 = "";

        for (int c = 0, b=0; c < b64Str.length(); c+=6, b+=4) {
            // unpack 5 hexets and 2-bit into 4 octets
            int h1 = BASE64_CODE.indexOf(b64Str.charAt(c));
            int h2 = BASE64_CODE.indexOf(b64Str.charAt(c + 1));
            int h3 = BASE64_CODE.indexOf(b64Str.charAt(c + 2));
            int h4 = BASE64_CODE.indexOf(b64Str.charAt(c + 3));
            int h5 = BASE64_CODE.indexOf(b64Str.charAt(c + 4));
            int d6 = BASE64_CODE.indexOf(b64Str.charAt(c + 5));

            int bits =  h1<<26 |  h2<<20 |  h3<<14 | h4<<8 | h5<<2 | (d6 & 0x3);
            byte o1 = (byte) (bits>>>24 & 0xff);
            byte o2 = (byte) (bits>>>16 & 0xff);
            byte o3 = (byte) (bits>>>8 & 0xff);
            byte o4 = (byte) (bits & 0xff);

            String partialDecodedStr = new String(new byte[] {o1, o2}, "UTF-16");

            //take care of padding
            // - if BASE64_CODE.indexOf(b64Str.charAt(c + 3)) == 64 OR b64Str.charAt(c + 3) == '='
            if (h4 == 0x40) {
               //This 16-bit char was padding: get rid of it
            }
            else
                partialDecodedStr = partialDecodedStr.concat(new String(new byte[] {o3, o4}, "UTF-16"));

            decodedBase64 = decodedBase64.concat( partialDecodedStr);
        }

    return decodedBase64;
}
catch (UnsupportedEncodingException e) {
    //TODO log4j
    e.printStackTrace();  //To change body of catch statement use File | Settings | File Templates.
    throw new RuntimeException("Unable to decode base64.");
}
}
于 2012-02-26T01:36:23.467 回答