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
这应该做到:
CoffeeScript(同步)
fs = require 'fs'
getDirs = (rootDir) ->
files = fs.readdirSync(rootDir)
dirs = []
for file in files
if file[0] != '.'
filePath = "#{rootDir}/#{file}"
stat = fs.statSync(filePath)
if stat.isDirectory()
dirs.push(file)
return dirs
CoffeeScript(异步)
fs = require 'fs'
getDirs = (rootDir, cb) ->
fs.readdir rootDir, (err, files) ->
dirs = []
for file, index in files
if file[0] != '.'
filePath = "#{rootDir}/#{file}"
fs.stat filePath, (err, stat) ->
if stat.isDirectory()
dirs.push(file)
if files.length == (index + 1)
cb(dirs)
JavaScript(异步)
var fs = require('fs');
var getDirs = function(rootDir, cb) {
fs.readdir(rootDir, function(err, files) {
var dirs = [];
for (var index = 0; index < files.length; ++index) {
var file = files[index];
if (file[0] !== '.') {
var filePath = rootDir + '/' + file;
fs.stat(filePath, function(err, stat) {
if (stat.isDirectory()) {
dirs.push(this.file);
}
if (files.length === (this.index + 1)) {
return cb(dirs);
}
}.bind({index: index, file: file}));
}
}
});
}
使用路径列出目录。
function getDirectories(path) {
return fs.readdirSync(path).filter(function (file) {
return fs.statSync(path+'/'+file).isDirectory();
});
}
var getDirectories = (rootdir , cb) => {
fs.readdir(rootdir, (err, files) => {
if(err) throw err ;
var dirs = files.map(filename => path.join(rootdir,filename)).filter( pathname => fs.statSync(pathname).isDirectory());
return cb(dirs);
})
}
getDirectories( myDirectories => console.log(myDirectories));``
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。
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])
