最新下载
热门教程
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
使用JavaScript实现node.js中的path.join方法
时间:2022-06-25 15:49:24 编辑:袖梨 来源:一聚教程网
Node.JS中的 path.join 非常方便,能直接按相对或绝对合并路径,使用: path.join([path1], [path2], [...]),有时侯前端也需要这种方法,如何实现呢?
其实直接从 node.js 的 path.js 拿到源码加工一下就可以了:
1. 将 const 等 es6 属性改为 var,以便前端浏览器兼容
2. 添加一个判断路戏分隔符的变量 sep,即左斜杠还是右斜杠,以第一个路戏分隔符为准
3. 将引用的变量和函数放到一个文件里就可以了:
Path 的源码: https://github.com/nodejs/node/blob/master/lib/path.js
var CHAR_FORWARD_SLASH = 47 var CHAR_BACKWARD_SLASH = 92 var CHAR_DOT = 46 function isPathSeparator(code) { return code === CHAR_FORWARD_SLASH || code === CHAR_BACKWARD_SLASH; } function isPosixPathSeparator(code) { return code === CHAR_FORWARD_SLASH; } function normalize(path) { if (path.length === 0) return '.'; var isAbsolute = path.charCodeAt(0) === CHAR_FORWARD_SLASH; var trailingSeparator = path.charCodeAt(path.length - 1) === CHAR_FORWARD_SLASH; // Normalize the path path = normalizeString(path, !isAbsolute, '/', isPosixPathSeparator); if (path.length === 0 && !isAbsolute) path = '.'; if (path.length > 0 && trailingSeparator) path += '/'; if (isAbsolute) return '/' + path; return path; } function normalizeString(path, allowAboveRoot, separator, isPathSeparator) { var res = ''; var lastSegmentLength = 0; var lastSlash = -1; var dots = 0; var code; for (var i = 0; i <= path.length; ++i) { if (i < path.length) code = path.charCodeAt(i); else if (isPathSeparator(code)) break; else code = CHAR_FORWARD_SLASH; if (isPathSeparator(code)) { if (lastSlash === i - 1 || dots === 1) { // NOOP } else if (lastSlash !== i - 1 && dots === 2) { if (res.length < 2 || lastSegmentLength !== 2 || res.charCodeAt(res.length - 1) !== CHAR_DOT || res.charCodeAt(res.length - 2) !== CHAR_DOT) { if (res.length > 2) { const lastSlashIndex = res.lastIndexOf(separator); if (lastSlashIndex !== res.length - 1) { if (lastSlashIndex === -1) { res = ''; lastSegmentLength = 0; } else { res = res.slice(0, lastSlashIndex); lastSegmentLength = res.length - 1 - res.lastIndexOf(separator); } lastSlash = i; dots = 0; continue; } } else if (res.length === 2 || res.length === 1) { res = ''; lastSegmentLength = 0; lastSlash = i; dots = 0; continue; } } if (allowAboveRoot) { if (res.length > 0) res += `${separator}..`; else res = '..'; lastSegmentLength = 2; } } else { if (res.length > 0) res += separator + path.slice(lastSlash + 1, i); else res = path.slice(lastSlash + 1, i); lastSegmentLength = i - lastSlash - 1; } lastSlash = i; dots = 0; } else if (code === CHAR_DOT && dots !== -1) { ++dots; } else { dots = -1; } } return res; } function join() { if (arguments.length === 0) return '.'; var sep = arguments[0].indexOf('/') > -1 ? '/' : '' var joined; var firstPart; for (var i = 0; i < arguments.length; ++i) { var arg = arguments[i]; if (arg.length > 0) { if (joined === undefined) joined = firstPart = arg; else joined += sep + arg; } } if (joined === undefined) return '.'; var needsReplace = true; var slashCount = 0; if (isPathSeparator(firstPart.charCodeAt(0))) { ++slashCount; var firstLen = firstPart.length; if (firstLen > 1) { if (isPathSeparator(firstPart.charCodeAt(1))) { ++slashCount; if (firstLen > 2) { if (isPathSeparator(firstPart.charCodeAt(2))) ++slashCount; else { // We matched a UNC path in the first part needsReplace = false; } } } } } if (needsReplace) { // Find any more consecutive slashes we need to replace for (; slashCount < joined.length; ++slashCount) { if (!isPathSeparator(joined.charCodeAt(slashCount))) break; } // Replace the slashes if needed if (slashCount >= 2) joined = sep + joined.slice(slashCount); } return normalize(joined); }
使用:
join('../var/www', '../abc') > "../var/abc" join('../var/www', 'abc') ../var/www/abc
相关文章
- 王者荣耀侦探能力大测试攻略 王者荣耀侦探能力大测试怎么过 11-22
- 无期迷途主线前瞻兑换码是什么 11-22
- 原神欧洛伦怎么培养 11-22
- 炉石传说网易云音乐联动怎么玩 11-22
- 永劫无间手游确幸转盘怎么样 11-22
- 无期迷途主线前瞻兑换码是什么 无期迷途主线前瞻直播兑换码介绍 11-22