分享

js实现基于Base64的编码及解码

 禁忌石 2022-02-17

js实现基于base64的编码解码

知道了编码解码规则之后,就是编程实现了。代码如下:

var Base64 = {

_keyStr: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=",

encode: function(e) {

var t = "";

var n, r, i, s, o, u, a;

var f = 0;

e = Base64._utf8_encode(e);

while (f < e.length) {

n = e.charCodeAt(f++);

r = e.charCodeAt(f++);

i = e.charCodeAt(f++);

s = n >> 2;

o = (n & 3) << 4 | r >> 4;

u = (r & 15) << 2 | i >> 6;

a = i & 63;

if (isNaN(r)) {

u = a = 64

} else if (isNaN(i)) {

a = 64

}

t = t + this._keyStr.charAt(s) + this._keyStr.charAt(o) + this._keyStr.charAt(u) + this._keyStr.charAt(a)

}

return t

},

decode: function(e) {

var t = "";

var n, r, i;

var s, o, u, a;

var f = 0;

e = e.replace(/[^A-Za-z0-9+/=]/g, "");

while (f < e.length) {

s = this._keyStr.indexOf(e.charAt(f++));

o = this._keyStr.indexOf(e.charAt(f++));

u = this._keyStr.indexOf(e.charAt(f++));

a = this._keyStr.indexOf(e.charAt(f++));

n = s << 2 | o >> 4;

r = (o & 15) << 4 | u >> 2;

i = (u & 3) << 6 | a;

t = t + String.fromCharCode(n);

if (u != 64) {

t = t + String.fromCharCode(r)

}

if (a != 64) {

t = t + String.fromCharCode(i)

}

}

t = Base64._utf8_decode(t);

return t

},

_utf8_encode: function(e) {

e = e.replace(/rn/g, "n");

var t = "";

for (var n = 0; n < e.length; n++) {

var r = e.charCodeAt(n);

if (r < 128) {

t += String.fromCharCode(r)

} else if (r > 127 && r < 2048) {

t += String.fromCharCode(r >> 6 | 192);

t += String.fromCharCode(r & 63 | 128)

} else {

t += String.fromCharCode(r >> 12 | 224);

t += String.fromCharCode(r >> 6 & 63 | 128);

t += String.fromCharCode(r & 63 | 128)

}

}

return t

},

_utf8_decode: function(e) {

var t = "";

var n = 0;

var r = c1 = c2 = 0;

while (n < e.length) {

r = e.charCodeAt(n);

if (r < 128) {

t += String.fromCharCode(r);

n++

} else if (r > 191 && r < 224) {

c2 = e.charCodeAt(n + 1);

t += String.fromCharCode((r & 31) << 6 | c2 & 63);

n += 2

} else {

c2 = e.charCodeAt(n + 1);

c3 = e.charCodeAt(n + 2);

t += String.fromCharCode((r & 15) << 12 | (c2 & 63) << 6 | c3 & 63);

n += 3

}

}

return t

}

}

也可以对对象进行编码,编码前将对象转换成字符串就行:

———————————————

    本站是提供个人知识管理的网络存储空间,所有内容均由用户发布,不代表本站观点。请注意甄别内容中的联系方式、诱导购买等信息,谨防诈骗。如发现有害或侵权内容,请点击一键举报。
    转藏 分享 献花(0

    0条评论

    发表

    请遵守用户 评论公约

    类似文章 更多