在node.js环境下模拟atob/btoa 的Base64算法

废话不多说，直接上源码：

Method 1

_keyStr = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";

function _utf8_encode (string) {
    var string = string.replace(/\r\n/g,"\n");
    var utftext = "";
    for (var n = 0; n < string.length; n++) {
        var c = string.charCodeAt(n);
        if (c < 128) {
            utftext += String.fromCharCode(c);
        } else if((c > 127) && (c < 2048)) {
            utftext += String.fromCharCode((c >> 6) | 192);
            utftext += String.fromCharCode((c & 63) | 128);
        } else {
            utftext += String.fromCharCode((c >> 12) | 224);
            utftext += String.fromCharCode(((c >> 6) & 63) | 128);
            utftext += String.fromCharCode((c & 63) | 128);
        }

    }
    return utftext;
}

function _utf8_decode (utftext) {
    var string = "";
    var i = 0;
    var c = 0;
    var c1 = 0;
    var c2 = 0;
    var c3 = 0;
    while ( i < utftext.length ) {
        c = utftext.charCodeAt(i);
        if (c < 128) {
            string += String.fromCharCode(c);
            i++;
        } else if((c > 191) && (c < 224)) {
            c2 = utftext.charCodeAt(i+1);
            string += String.fromCharCode(((c & 31) << 6) | (c2 & 63));
            i += 2;
        } else {
            c2 = utftext.charCodeAt(i+1);
            c3 = utftext.charCodeAt(i+2);
            string += String.fromCharCode(((c & 15) << 12) | ((c2 & 63) << 6) | (c3 & 63));
            i += 3;
        }
    }
    return string;
}

var xazxBase64 = {
    'decode': function (input){
        output = "";
        var chr1, chr2, chr3;
        var enc1, enc2, enc3, enc4;
        i = 0;
        input = input.replace(/[^A-Za-z0-9+\/=]/g, "");
        while (i < input.length) {
            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);
            }
        }
        output = _utf8_decode(output);
        return output;
    },

    'encode': function (input){
        output = "";
        var chr1, chr2, chr3, enc1, enc2, enc3, enc4;
        i = 0;
        input = _utf8_encode(input);
        while (i < input.length) {
            chr1 = input.charCodeAt(i++);
            chr2 = input.charCodeAt(i++);
            chr3 = input.charCodeAt(i++);
            enc1 = chr1 >> 2;
            enc2 = ((chr1 & 3) << 4) | (chr2 >> 4);
            enc3 = ((chr2 & 15) << 2) | (chr3 >> 6);
            enc4 = chr3 & 63;
            if (isNaN(chr2)) {
                enc3 = enc4 = 64;
            } else if (isNaN(chr3)) {
                enc4 = 64;
            }
            output = output +
                _keyStr.charAt(enc1) + _keyStr.charAt(enc2) +
                _keyStr.charAt(enc3) + _keyStr.charAt(enc4);
        }
        return output;
    }
};

Method 2

global.Buffer = global.Buffer || require('buffer').Buffer;

if (typeof btoa === 'undefined') {
    global.btoa = function (str) {
        return new Buffer.from(str, 'binary').toString('base64');
    };
}

if (typeof atob === 'undefined') {
    global.atob = function (b64Encoded) {
        return new Buffer.from(b64Encoded, 'base64').toString('binary');
    };
}

Method 3

var atob = function(r) {
    e = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";
    var o = String(r).replace(/=+$/, "");
    if (o.length % 4 == 1)
        throw new t("'atob' failed: The string to be decoded is not correctly encoded.");
    for (var n, a, i = 0, c = 0, d = ""; a = o.charAt(c++); ~a && (n = i % 4 ? 64 * n + a : a,
    i++ % 4) ? d += String.fromCharCode(255 & n >> (-2 * i & 6)) : 0)
        a = e.indexOf(a);
    return d
}