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
承诺
import { readdir } from 'fs/promises'
const getDirectories = async source =>
(await readdir(source, { withFileTypes: true }))
.filter(dirent => dirent.isDirectory())
.map(dirent => dirent.name)
回调
import { readdir } from 'fs'
const getDirectories = (source, callback) =>
readdir(source, { withFileTypes: true }, (err, files) => {
if (err) {
callback(err)
} else {
callback(
files
.filter(dirent => dirent.isDirectory())
.map(dirent => dirent.name)
)
}
})
Syncronous
import { readdirSync } from 'fs'
const getDirectories = source =>
readdirSync(source, { withFileTypes: true })
.filter(dirent => dirent.isDirectory())
.map(dirent => dirent.name)
其他回答
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);
以防其他人从网络搜索到这里，并且已经在他们的依赖列表中有Grunt，这个问题的答案变得微不足道。以下是我的解决方案:
/**
* Return all the subfolders of this path
* @param {String} parentFolderPath - valid folder path
* @param {String} glob ['/*'] - optional glob so you can do recursive if you want
* @returns {String[]} subfolder paths
*/
getSubfolders = (parentFolderPath, glob = '/*') => {
return grunt.file.expand({filter: 'isDirectory'}, parentFolderPath + glob);
}
承诺
import { readdir } from 'fs/promises'
const getDirectories = async source =>
(await readdir(source, { withFileTypes: true }))
.filter(dirent => dirent.isDirectory())
.map(dirent => dirent.name)
回调
import { readdir } from 'fs'
const getDirectories = (source, callback) =>
readdir(source, { withFileTypes: true }, (err, files) => {
if (err) {
callback(err)
} else {
callback(
files
.filter(dirent => dirent.isDirectory())
.map(dirent => dirent.name)
)
}
})
Syncronous
import { readdirSync } from 'fs'
const getDirectories = source =>
readdirSync(source, { withFileTypes: true })
.filter(dirent => dirent.isDirectory())
.map(dirent => dirent.name)
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))];
}
这个答案的CoffeeScript版本，有适当的错误处理:
fs = require "fs"
{join} = require "path"
async = require "async"
get_subdirs = (root, callback)->
fs.readdir root, (err, files)->
return callback err if err
subdirs = []
async.each files,
(file, callback)->
fs.stat join(root, file), (err, stats)->
return callback err if err
subdirs.push file if stats.isDirectory()
callback null
(err)->
return callback err if err
callback null, subdirs
取决于async
或者，使用一个模块! (所有东西都有模块。[引文需要])
