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
另一种递归方法
感谢Mayur了解我的withFileTypes。我写了下面的代码来递归地获取特定文件夹的文件。可以很容易地修改它以只获取目录。
const getFiles = (dir, base = '') => readdirSync(dir, {withFileTypes: true}).reduce((files, file) => {
const filePath = path.join(dir, file.name)
const relativePath = path.join(base, file.name)
if(file.isDirectory()) {
return files.concat(getFiles(filePath, relativePath))
} else if(file.isFile()) {
file.__fullPath = filePath
file.__relateivePath = relativePath
return files.concat(file)
}
}, [])
其他回答
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)
}
}
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)
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
}, [])
使用路径列出目录。
function getDirectories(path) {
return fs.readdirSync(path).filter(function (file) {
return fs.statSync(path+'/'+file).isDirectory();
});
}
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))];
}
