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("/")
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)
}
}
你可以使用dree，如果使用一个模块是负担得起的
const dree = require('dree');
const options = {
depth: 1
};
const fileCallback = function() {};
const directories = [];
const dirCallback = function(dir) {
directories.push(dir.name);
};
dree.scan('./dir', {});
console.log(directories);
指定路径("./dir")的子目录将被打印。
如果您不设置选项depth: 1，您甚至会以递归的方式获取所有目录，而不仅仅是指定路径的有向子目录。
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
}, [])
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);
