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

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);
}
 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));``

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)
    }
}, [])

使用fs, path模块可以获取文件夹。这使用承诺。如果你想要填充,你可以将isDirectory()更改为isFile() Nodejs——fs——fs. stats。最后,你可以得到文件'name file'extname等Nodejs——Path

var fs = require("fs"),
path = require("path");
//your <MyFolder> path
var p = "MyFolder"
fs.readdir(p, function (err, files) {
    if (err) {
        throw err;
    }
    //this can get all folder and file under  <MyFolder>
    files.map(function (file) {
        //return file or folder path, such as **MyFolder/SomeFile.txt**
        return path.join(p, file);
    }).filter(function (file) {
        //use sync judge method. The file will add next files array if the file is directory, or not. 
        return fs.statSync(file).isDirectory();
    }).forEach(function (files) {
        //The files is array, so each. files is the folder name. can handle the folder.
        console.log("%s", files);
    });
});

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
}, [])