有很多MD5 JavaScript实现。 有人知道哪一个是最先进的,修复最多的,最快的吗?

我需要它来做这个工具。


当前回答

Node.js有内置的支持

const crypto = require('crypto')
crypto.createHash('md5').update('hello world').digest('hex')

上面的代码段为字符串hello world计算MD5十六进制字符串

这种解决方案的优点是不需要安装额外的库。

我认为内置解决方案应该是最快的。如果不是,我们应该为Node.js项目创建issue/PR。

其他回答

也许这个包是有用的 https://www.npmjs.com/package/pure-md5

console.time('latin'); 常量拉丁语 = MD5('你好'); console.timeEnd('latin'); console.log('Привет: ', md5('Привет')); console.log('嘿: ', md5('嘿')); <script src=“https://unpkg.com/pure-md5@latest/lib/index.js”></script>

我听说Joseph的Myers实现非常快。此外,他还有一篇关于Javascript优化的长篇文章,描述了他在编写实现时学到的东西。对于任何对性能javascript感兴趣的人来说,这是一本很好的读物。

http://www.webreference.com/programming/javascript/jkm3/

他的MD5实现可以在这里找到

如果应用程序的性能受到MD5的Javascript实现的限制,那么您确实做错了什么。考虑架构变化(提示:减少使用MD5)

我真的不记得我从哪里得到这个。我在网上到处找,唯一能找到的参考文献就是这篇文章。奇怪。 如果任何人有关于这个实现起源的信息,请告诉我。

或者你可以试试这个新的,速度快25%:https://stackoverflow.com/a/74186696/236062:D

MD5 = function(e) { function h(a, b) { var c, d, e, f, g; e = a & 2147483648; f = b & 2147483648; c = a & 1073741824; d = b & 1073741824; g = (a & 1073741823) + (b & 1073741823); return c & d ? g ^ 2147483648 ^ e ^ f : c | d ? g & 1073741824 ? g ^ 3221225472 ^ e ^ f : g ^ 1073741824 ^ e ^ f : g ^ e ^ f } function k(a, b, c, d, e, f, g) { a = h(a, h(h(b & c | ~b & d, e), g)); return h(a << f | a >>> 32 - f, b) } function l(a, b, c, d, e, f, g) { a = h(a, h(h(b & d | c & ~d, e), g)); return h(a << f | a >>> 32 - f, b) } function m(a, b, d, c, e, f, g) { a = h(a, h(h(b ^ d ^ c, e), g)); return h(a << f | a >>> 32 - f, b) } function n(a, b, d, c, e, f, g) { a = h(a, h(h(d ^ (b | ~c), e), g)); return h(a << f | a >>> 32 - f, b) } function p(a) { var b = "", d = "", c; for (c = 0; 3 >= c; c++) d = a >>> 8 * c & 255, d = "0" + d.toString(16), b += d.substr(d.length - 2, 2); return b } var f = [], q, r, s, t, a, b, c, d; e = function(a) { a = a.replace(/\r\n/g, "\n"); for (var b = "", d = 0; d < a.length; d++) { var c = a.charCodeAt(d); 128 > c ? b += String.fromCharCode(c) : (127 < c && 2048 > c ? b += String.fromCharCode(c >> 6 | 192) : (b += String.fromCharCode(c >> 12 | 224), b += String.fromCharCode(c >> 6 & 63 | 128)), b += String.fromCharCode(c & 63 | 128)) } return b }(e); f = function(b) { var a, c = b.length; a = c + 8; for (var d = 16 * ((a - a % 64) / 64 + 1), e = Array(d - 1), f = 0, g = 0; g < c;) a = (g - g % 4) / 4, f = g % 4 * 8, e[a] |= b.charCodeAt(g) << f, g++; a = (g - g % 4) / 4; e[a] |= 128 << g % 4 * 8; e[d - 2] = c << 3; e[d - 1] = c >>> 29; return e }(e); a = 1732584193; b = 4023233417; c = 2562383102; d = 271733878; for (e = 0; e < f.length; e += 16) q = a, r = b, s = c, t = d, a = k(a, b, c, d, f[e + 0], 7, 3614090360), d = k(d, a, b, c, f[e + 1], 12, 3905402710), c = k(c, d, a, b, f[e + 2], 17, 606105819), b = k(b, c, d, a, f[e + 3], 22, 3250441966), a = k(a, b, c, d, f[e + 4], 7, 4118548399), d = k(d, a, b, c, f[e + 5], 12, 1200080426), c = k(c, d, a, b, f[e + 6], 17, 2821735955), b = k(b, c, d, a, f[e + 7], 22, 4249261313), a = k(a, b, c, d, f[e + 8], 7, 1770035416), d = k(d, a, b, c, f[e + 9], 12, 2336552879), c = k(c, d, a, b, f[e + 10], 17, 4294925233), b = k(b, c, d, a, f[e + 11], 22, 2304563134), a = k(a, b, c, d, f[e + 12], 7, 1804603682), d = k(d, a, b, c, f[e + 13], 12, 4254626195), c = k(c, d, a, b, f[e + 14], 17, 2792965006), b = k(b, c, d, a, f[e + 15], 22, 1236535329), a = l(a, b, c, d, f[e + 1], 5, 4129170786), d = l(d, a, b, c, f[e + 6], 9, 3225465664), c = l(c, d, a, b, f[e + 11], 14, 643717713), b = l(b, c, d, a, f[e + 0], 20, 3921069994), a = l(a, b, c, d, f[e + 5], 5, 3593408605), d = l(d, a, b, c, f[e + 10], 9, 38016083), c = l(c, d, a, b, f[e + 15], 14, 3634488961), b = l(b, c, d, a, f[e + 4], 20, 3889429448), a = l(a, b, c, d, f[e + 9], 5, 568446438), d = l(d, a, b, c, f[e + 14], 9, 3275163606), c = l(c, d, a, b, f[e + 3], 14, 4107603335), b = l(b, c, d, a, f[e + 8], 20, 1163531501), a = l(a, b, c, d, f[e + 13], 5, 2850285829), d = l(d, a, b, c, f[e + 2], 9, 4243563512), c = l(c, d, a, b, f[e + 7], 14, 1735328473), b = l(b, c, d, a, f[e + 12], 20, 2368359562), a = m(a, b, c, d, f[e + 5], 4, 4294588738), d = m(d, a, b, c, f[e + 8], 11, 2272392833), c = m(c, d, a, b, f[e + 11], 16, 1839030562), b = m(b, c, d, a, f[e + 14], 23, 4259657740), a = m(a, b, c, d, f[e + 1], 4, 2763975236), d = m(d, a, b, c, f[e + 4], 11, 1272893353), c = m(c, d, a, b, f[e + 7], 16, 4139469664), b = m(b, c, d, a, f[e + 10], 23, 3200236656), a = m(a, b, c, d, f[e + 13], 4, 681279174), d = m(d, a, b, c, f[e + 0], 11, 3936430074), c = m(c, d, a, b, f[e + 3], 16, 3572445317), b = m(b, c, d, a, f[e + 6], 23, 76029189), a = m(a, b, c, d, f[e + 9], 4, 3654602809), d = m(d, a, b, c, f[e + 12], 11, 3873151461), c = m(c, d, a, b, f[e + 15], 16, 530742520), b = m(b, c, d, a, f[e + 2], 23, 3299628645), a = n(a, b, c, d, f[e + 0], 6, 4096336452), d = n(d, a, b, c, f[e + 7], 10, 1126891415), c = n(c, d, a, b, f[e + 14], 15, 2878612391), b = n(b, c, d, a, f[e + 5], 21, 4237533241), a = n(a, b, c, d, f[e + 12], 6, 1700485571), d = n(d, a, b, c, f[e + 3], 10, 2399980690), c = n(c, d, a, b, f[e + 10], 15, 4293915773), b = n(b, c, d, a, f[e + 1], 21, 2240044497), a = n(a, b, c, d, f[e + 8], 6, 1873313359), d = n(d, a, b, c, f[e + 15], 10, 4264355552), c = n(c, d, a, b, f[e + 6], 15, 2734768916), b = n(b, c, d, a, f[e + 13], 21, 1309151649), a = n(a, b, c, d, f[e + 4], 6, 4149444226), d = n(d, a, b, c, f[e + 11], 10, 3174756917), c = n(c, d, a, b, f[e + 2], 15, 718787259), b = n(b, c, d, a, f[e + 9], 21, 3951481745), a = h(a, q), b = h(b, r), c = h(c, s), d = h(d, t); return (p(a) + p(b) + p(c) + p(d)).toLowerCase() }; <!DOCTYPE html> <html> <body onload="md5.value=MD5(a.value);"> <form oninput="md5.value=MD5(a.value)">Enter String: <input type="string" id="a" name="a" value="https://www.zibri.org"></br></br>MD5:<output id="md5" name="md5" for="a"></output> </form> </body> </html>

令我烦恼的是,我找不到一个既快速又支持Unicode字符串的实现。

所以我做了一个支持Unicode字符串的实现,并且仍然比目前最快的ascii-only-strings实现更快(在编写时):

https://github.com/gorhill/yamd5.js

基于Joseph Myers的代码,但使用了TypedArrays,并进行了其他改进。