base64和Xxtea的加密和解密

base64和Xxtea的加密和解密

数据加密是web数据安全的一种方式，前几天拿到一个base64+xxtea加密的数据，现在在这里整理一下使用的过程。首先当然是全网站找解密方法，但是最后的结果不是很理想，都不是自己想要的，这里只针对前端js的加密和解密

看代码：

``` function long2str(v, w) { var vl = v.length; var sl = v[vl - 1] & 0xffffffff; for (var i = 0; i < vl; i++) { v[i] = String.fromCharCode(v[i] & 0xff, v[i] >>> 8 & 0xff, v[i] >>> 16 & 0xff, v[i] >>> 24 & 0xff); } if (w) { return v.join('').substring(0, sl); } else { return v.join(''); } }

function str2long(s, w) {
var len = s.length;
var v = [];
for (var i = 0; i < len; i += 4)
{
v[i >> 2] = s.charCodeAt(i)
| s.charCodeAt(i + 1) << 8
| s.charCodeAt(i + 2) << 16
| s.charCodeAt(i + 3) << 24;
}
if (w) {
v[v.length] = len;
}
return v;
}

function xxtea_encrypt(str, key) {
if (str == "") {
return "";
}
var v = str2long(str, true);
var k = str2long(key, false);
var n = v.length - 1;

var z = v[n], y = v[0], delta = 0x9E3779B9;
var mx, e, q = Math.floor(6 + 52 / (n + 1)), sum = 0;
while (q-- > 0) {
    sum = sum + delta & 0xffffffff;
    e = sum >>> 2 & 3;
    for (var p = 0; p < n; p++) {
        y = v[p + 1];
        mx = (z >>> 5 ^ y << 2) + (y >>> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z);
        z = v[p] = v[p] + mx & 0xffffffff;
    }
    y = v[0];
    mx = (z >>> 5 ^ y << 2) + (y >>> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z);
    z = v[n] = v[n] + mx & 0xffffffff;
}

return long2str(v, false);

}

function xxtea_decrypt(str, key) {
if (str == "") {
return "";
}
var v = str2long(str, false);
var k = str2long(key, false);
var n = v.length - 1;

var z = v[n - 1], y = v[0], delta = 0x9E3779B9;
var mx, e, q = Math.floor(6 + 52 / (n + 1)), sum = q * delta & 0xffffffff;
while (sum != 0) {
    e = sum >>> 2 & 3;
    for (var p = n; p > 0; p--) {
        z = v[p - 1];
        mx = (z >>> 5 ^ y << 2) + (y >>> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z);
        y = v[p] = v[p] - mx & 0xffffffff;
    }
    z = v[n];
    mx = (z >>> 5 ^ y << 2) + (y >>> 3 ^ z << 4) ^ (sum ^ y) + (k[p & 3 ^ e] ^ z);
    y = v[0] = v[0] - mx & 0xffffffff;
    sum = sum - delta & 0xffffffff;
}

return long2str(v, true);

}

function utf16to8(str) {
var out, i, len, c;

out = [];
len = str.length;
for(i = 0; i < len; i++) {
    c = str.charCodeAt(i);
    if ((c >= 0x0001) && (c <= 0x007F)) {
        out[i] = str.charAt(i);
    } else if (c > 0x07FF) {
        out[i] = String.fromCharCode(0xE0 | ((c >> 12) & 0x0F),
                                     0x80 | ((c >>  6) & 0x3F),
                                     0x80 | ((c >>  0) & 0x3F));
    } else {
        out[i] = String.fromCharCode(0xC0 | ((c >>  6) & 0x1F),
                                     0x80 | ((c >>  0) & 0x3F));
    }
}
return out.join('');

}

function utf8to16(str) {
var out, i, len, c;
var char2, char3;

out = [];
len = str.length;
i = 0;
while(i < len) {
    c = str.charCodeAt(i++);
    switch(c >> 4)
    { 
        case 0: case 1: case 2: case 3: case 4: case 5: case 6: case 7:
        // 0xxxxxxx
        out[out.length] = str.charAt(i-1);
        break;
        case 12: case 13:
        // 110x xxxx   10xx xxxx
        char2 = str.charCodeAt(i++);
        out[out.length] = String.fromCharCode(((c & 0x1F) << 6) | (char2 & 0x3F));
        break;
        case 14:
        // 1110 xxxx  10xx xxxx  10xx xxxx
        char2 = str.charCodeAt(i++);
        char3 = str.charCodeAt(i++);
        out[out.length] = String.fromCharCode(((c & 0x0F) << 12) |
            ((char2 & 0x3F) << 6) | ((char3 & 0x3F) << 0));
        break;
    }
}

return out.join('');

}

//base64解码
function decode_base64(input) {
var keyStr = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";
var output = "";
var chr1, chr2, chr3 = "";
var enc1, enc2, enc3, enc4 = "";
var i = 0;

// remove all characters that are not A-Z, a-z, 0-9, +, /, or =   
var base64test = /[^A-Za-z0-9\+\/\=\n]/g;   
if (base64test.exec(input)) {   
}   
input = input.replace(/[^A-Za-z0-9\+\/\=]/g, "");  

do {   
    enc1 = keyStr.indexOf(input.charAt(i++));   
    enc2 = keyStr.indexOf(input.charAt(i++));   
    enc3 = keyStr.indexOf(input.charAt(i++));   
    enc4 = keyStr.indexOf(input.charAt(i++));  

    chr1 = (enc1 << 2) | (enc2 >> 4);   
    chr2 = ((enc2 & 15) << 4) | (enc3 >> 2);   
    chr3 = ((enc3 & 3) << 6) | enc4;  

    output = output + String.fromCharCode(chr1);  

    if (enc3 != 64) {   
        output = output + String.fromCharCode(chr2);   
    }   
    if (enc4 != 64) {   
        output = output + String.fromCharCode(chr3);   
    }  

    chr1 = chr2 = chr3 = "";   
    enc1 = enc2 = enc3 = enc4 = "";  

} while (i < input.length);  

return output;   

}

<p>然后就是使用：</p>

```

xxtea解密函数的key参数，是数据加密的关键字，这个得着后端的程序员给你提供