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

使用npm列表内容模块。它读取给定目录的内容和子内容,并返回文件和文件夹路径列表。

const list = require('list-contents');

list("./dist",(o)=>{
  if(o.error) throw o.error;
   console.log('Folders: ', o.dirs);
   console.log('Files: ', o.files);
});

其他回答

下面是一个仅使用本机fs和路径模块的简单解决方案:

// sync version
function walkSync(currentDirPath, callback) {
    var fs = require('fs'),
        path = require('path');
    fs.readdirSync(currentDirPath).forEach(function (name) {
        var filePath = path.join(currentDirPath, name);
        var stat = fs.statSync(filePath);
        if (stat.isFile()) {
            callback(filePath, stat);
        } else if (stat.isDirectory()) {
            walkSync(filePath, callback);
        }
    });
}

或异步版本(改用fs.readder):

// async version with basic error handling
function walk(currentDirPath, callback) {
    var fs = require('fs'),
        path = require('path');
    fs.readdir(currentDirPath, function (err, files) {
        if (err) {
            throw new Error(err);
        }
        files.forEach(function (name) {
            var filePath = path.join(currentDirPath, name);
            var stat = fs.statSync(filePath);
            if (stat.isFile()) {
                callback(filePath, stat);
            } else if (stat.isDirectory()) {
                walk(filePath, callback);
            }
        });
    });
}

然后您只需调用(同步版本):

walkSync('path/to/root/dir', function(filePath, stat) {
    // do something with "filePath"...
});

或异步版本:

walk('path/to/root/dir', function(filePath, stat) {
    // do something with "filePath"...
});

不同之处在于节点在执行IO时如何阻塞。考虑到上面的API是相同的,您可以只使用异步版本来确保最大性能。

然而,使用同步版本有一个优点。在遍历完成后立即执行一些代码更容易,就像在遍历后的下一条语句中一样。对于异步版本,您需要一些额外的方法来知道何时完成。也许首先创建所有路径的映射,然后枚举它们。对于简单的build/util脚本(与高性能web服务器相比),您可以使用同步版本而不会造成任何损坏。

获取排序的文件名。您可以基于特定扩展名(如“.txt”、“.jpg”等)过滤结果。

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

function getFilenames(path, extension) {
    return fs
        .readdirSync(path)
        .filter(
            item =>
                fs.statSync(Path.join(path, item)).isFile() &&
                (extension === undefined || Path.extname(item) === extension)
        )
        .sort();
}

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"
];

这将起作用,并将结果存储在test.txt文件中,该文件将位于同一目录中

  fs.readdirSync(__dirname).forEach(file => {
    fs.appendFileSync("test.txt", file+"\n", function(err){
    })
})

开箱即用

如果您想要一个具有开箱即用的目录结构的对象,我强烈建议您检查目录树。

假设你有这样的结构:

photos
│   june
│   └── windsurf.jpg
└── january
    ├── ski.png
    └── snowboard.jpg
const dirTree = require("directory-tree");
const tree = dirTree("/path/to/photos");

将返回:

{
  path: "photos",
  name: "photos",
  size: 600,
  type: "directory",
  children: [
    {
      path: "photos/june",
      name: "june",
      size: 400,
      type: "directory",
      children: [
        {
          path: "photos/june/windsurf.jpg",
          name: "windsurf.jpg",
          size: 400,
          type: "file",
          extension: ".jpg"
        }
      ]
    },
    {
      path: "photos/january",
      name: "january",
      size: 200,
      type: "directory",
      children: [
        {
          path: "photos/january/ski.png",
          name: "ski.png",
          size: 100,
          type: "file",
          extension: ".png"
        },
        {
          path: "photos/january/snowboard.jpg",
          name: "snowboard.jpg",
          size: 100,
          type: "file",
          extension: ".jpg"
        }
      ]
    }
  ]
}

自定义对象

否则,如果要使用自定义设置创建目录树对象,请查看以下代码段。在这个代码沙盒上可以看到一个活生生的例子。

// my-script.js
const fs = require("fs");
const path = require("path");

const isDirectory = filePath => fs.statSync(filePath).isDirectory();
const isFile = filePath => fs.statSync(filePath).isFile();

const getDirectoryDetails = filePath => {
  const dirs = fs.readdirSync(filePath);
  return {
    dirs: dirs.filter(name => isDirectory(path.join(filePath, name))),
    files: dirs.filter(name => isFile(path.join(filePath, name)))
  };
};

const getFilesRecursively = (parentPath, currentFolder) => {
  const currentFolderPath = path.join(parentPath, currentFolder);
  let currentDirectoryDetails = getDirectoryDetails(currentFolderPath);

  const final = {
    current_dir: currentFolder,
    dirs: currentDirectoryDetails.dirs.map(dir =>
      getFilesRecursively(currentFolderPath, dir)
    ),
    files: currentDirectoryDetails.files
  };

  return final;
};

const getAllFiles = relativePath => {
  const fullPath = path.join(__dirname, relativePath);
  const parentDirectoryPath = path.dirname(fullPath);
  const leafDirectory = path.basename(fullPath);

  const allFiles = getFilesRecursively(parentDirectoryPath, leafDirectory);
  return allFiles;
};

module.exports = { getAllFiles };

然后,您可以简单地执行以下操作:

// another-file.js 

const { getAllFiles } = require("path/to/my-script");

const allFiles = getAllFiles("/path/to/my-directory");