我希望这是一件简单的事情，但我找不到任何东西在那里这样做。
我只想获得给定文件夹/目录内的所有文件夹/目录。
例如:
<MyFolder>
|- SomeFolder
|- SomeOtherFolder
|- SomeFile.txt
|- SomeOtherFile.txt
|- x-directory
我期望得到一个数组:
["SomeFolder", "SomeOtherFolder", "x-directory"]
或者上面的路径，如果它是这样提供的……
那么，有什么东西已经存在了吗?
当前回答
函数式编程
const fs = require('fs')
const path = require('path')
const R = require('ramda')
const getDirectories = pathName => {
const isDirectory = pathName => fs.lstatSync(pathName).isDirectory()
const mapDirectories = pathName => R.map(name => path.join(pathName, name), fs.readdirSync(pathName))
const filterDirectories = listPaths => R.filter(isDirectory, listPaths)
return {
paths:R.pipe(mapDirectories)(pathName),
pathsFiltered: R.pipe(mapDirectories, filterDirectories)(pathName)
}
}
其他回答
使用fs-extra，承诺async fs调用，以及新的await async语法:
const fs = require("fs-extra");
async function getDirectories(path){
let filesAndDirectories = await fs.readdir(path);
let directories = [];
await Promise.all(
filesAndDirectories.map(name =>{
return fs.stat(path + name)
.then(stat =>{
if(stat.isDirectory()) directories.push(name)
})
})
);
return directories;
}
let directories = await getDirectories("/")
使用node.js版本>= v10.13.0, fs. js。readdirSync将返回一个fs数组。如果withFileTypes选项设置为true，则直接对象。
所以你可以用，
const fs = require('fs')
const directories = source => fs.readdirSync(source, {
withFileTypes: true
}).reduce((a, c) => {
c.isDirectory() && a.push(c.name)
return a
}, [])
完全异步的版本与ES6，只有本机包，fs。Promises和async/await，并行执行文件操作:
const fs = require('fs');
const path = require('path');
async function listDirectories(rootPath) {
const fileNames = await fs.promises.readdir(rootPath);
const filePaths = fileNames.map(fileName => path.join(rootPath, fileName));
const filePathsAndIsDirectoryFlagsPromises = filePaths.map(async filePath => ({path: filePath, isDirectory: (await fs.promises.stat(filePath)).isDirectory()}))
const filePathsAndIsDirectoryFlags = await Promise.all(filePathsAndIsDirectoryFlagsPromises);
return filePathsAndIsDirectoryFlags.filter(filePathAndIsDirectoryFlag => filePathAndIsDirectoryFlag.isDirectory)
.map(filePathAndIsDirectoryFlag => filePathAndIsDirectoryFlag.path);
}
经过测试，它工作得很好。
你可以使用graph-fs
const {Node} = require("graph-fs");
const directory = new Node("/path/to/directory");
const subDirectories = directory.children.filter(child => child.is.directory);
递归解决方案
我来这里是为了寻找一种方法来获取所有子目录，以及它们的所有子目录，等等。在这个公认的答案的基础上，我写道:
const fs = require('fs');
const path = require('path');
function flatten(lists) {
return lists.reduce((a, b) => a.concat(b), []);
}
function getDirectories(srcpath) {
return fs.readdirSync(srcpath)
.map(file => path.join(srcpath, file))
.filter(path => fs.statSync(path).isDirectory());
}
function getDirectoriesRecursive(srcpath) {
return [srcpath, ...flatten(getDirectories(srcpath).map(getDirectoriesRecursive))];
}
