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
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");
其他回答
它只有2行代码：
fs=require('fs')
fs.readdir("./img/", (err,filename)=>console.log(filename))
图像：
试试这个，它对我有用
import fs from "fs/promises";
const path = "path/to/folder";
export const readDir = async function readDir(path) {
const files = await fs.readdir(path);
// array of file names
console.log(files);
}
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();
}
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");
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"
];
