我试图使用Node.js获取目录中所有文件的名称列表。我希望输出文件名数组。我该怎么做?


当前回答

采用@湖南罗斯托米扬的一般方法,使其更加简洁,并添加了excludeDirs论点。使用includeDirs进行扩展很简单,只需遵循相同的模式:

import * as fs from 'fs';
import * as path from 'path';

function fileList(dir, excludeDirs?) {
    return fs.readdirSync(dir).reduce(function (list, file) {
        const name = path.join(dir, file);
        if (fs.statSync(name).isDirectory()) {
            if (excludeDirs && excludeDirs.length) {
                excludeDirs = excludeDirs.map(d => path.normalize(d));
                const idx = name.indexOf(path.sep);
                const directory = name.slice(0, idx === -1 ? name.length : idx);
                if (excludeDirs.indexOf(directory) !== -1)
                    return list;
            }
            return list.concat(fileList(name, excludeDirs));
        }
        return list.concat([name]);
    }, []);
}

示例用法:

console.log(fileList('.', ['node_modules', 'typings', 'bower_components']));

其他回答

function getFilesRecursiveSync(dir, fileList, optionalFilterFunction) {
    if (!fileList) {
        grunt.log.error("Variable 'fileList' is undefined or NULL.");
        return;
    }
    var files = fs.readdirSync(dir);
    for (var i in files) {
        if (!files.hasOwnProperty(i)) continue;
        var name = dir + '/' + files[i];
        if (fs.statSync(name).isDirectory()) {
            getFilesRecursiveSync(name, fileList, optionalFilterFunction);
        } else {
            if (optionalFilterFunction && optionalFilterFunction(name) !== true)
                continue;
            fileList.push(name);
        }
    }
}

它只有2行代码:

fs=require('fs')
fs.readdir("./img/", (err,filename)=>console.log(filename))

图像:

我从你的问题中假设你不需要目录名,只需要文件。

目录结构示例

animals
├── all.jpg
├── mammals
│   └── cat.jpg
│   └── dog.jpg
└── insects
    └── bee.jpg

步行功能

根据这一要点,Justin Maier将获得积分

如果只需要一个文件路径数组,请使用return_object:false:

const fs = require('fs').promises;
const path = require('path');

async function walk(dir) {
    let files = await fs.readdir(dir);
    files = await Promise.all(files.map(async file => {
        const filePath = path.join(dir, file);
        const stats = await fs.stat(filePath);
        if (stats.isDirectory()) return walk(filePath);
        else if(stats.isFile()) return filePath;
    }));

    return files.reduce((all, folderContents) => all.concat(folderContents), []);
}

用法

async function main() {
   console.log(await walk('animals'))
}

输出

[
  "/animals/all.jpg",
  "/animals/mammals/cat.jpg",
  "/animals/mammals/dog.jpg",
  "/animals/insects/bee.jpg"
];

依赖关系。

var fs = require('fs');
var path = require('path');

释义

// String -> [String]
function fileList(dir) {
  return fs.readdirSync(dir).reduce(function(list, file) {
    var name = path.join(dir, file);
    var isDir = fs.statSync(name).isDirectory();
    return list.concat(isDir ? fileList(name) : [name]);
  }, []);
}

用法

var DIR = '/usr/local/bin';

// 1. List all files in DIR
fileList(DIR);
// => ['/usr/local/bin/babel', '/usr/local/bin/bower', ...]

// 2. List all file names in DIR
fileList(DIR).map((file) => file.split(path.sep).slice(-1)[0]);
// => ['babel', 'bower', ...]

请注意,fileList过于乐观。对于任何严重的问题,请添加一些错误处理。

获取所有分区中的文件

const fs=require('fs');

function getFiles (dir, files_){
    files_ = files_ || [];
    var files = fs.readdirSync(dir);
    for (var i in files){
        var name = dir + '/' + files[i];
        if (fs.statSync(name).isDirectory()){
            getFiles(name, files_);
        } else {
            files_.push(name);
        }
    }
    return files_;
}

console.log(getFiles('path/to/dir'))