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

如果上面的许多选项看起来太复杂,或者不是您想要的,这里是另一种使用node-dir的方法https://github.com/fshost/node-dir

npm install node-dir

下面是一个somple函数,用于列出在子目录中搜索的所有.xml文件

import * as nDir from 'node-dir' ;

listXMLs(rootFolderPath) {
    let xmlFiles ;

    nDir.files(rootFolderPath, function(err, items) {
        xmlFiles = items.filter(i => {
            return path.extname(i) === '.xml' ;
        }) ;
        console.log(xmlFiles) ;       
    });
}

其他回答

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。

它只有2行代码:

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

图像:

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();
}

这是一个TypeScript,可选递归,可选错误日志和异步解决方案。可以为要查找的文件名指定正则表达式。

我使用了fs extra,因为这是对fs的一个简单的超集改进。

import * as FsExtra from 'fs-extra'

/**
 * Finds files in the folder that match filePattern, optionally passing back errors .
 * If folderDepth isn't specified, only the first level is searched. Otherwise anything up
 * to Infinity is supported.
 *
 * @static
 * @param {string} folder The folder to start in.
 * @param {string} [filePattern='.*'] A regular expression of the files you want to find.
 * @param {(Error[] | undefined)} [errors=undefined]
 * @param {number} [folderDepth=0]
 * @returns {Promise<string[]>}
 * @memberof FileHelper
 */
public static async findFiles(
    folder: string,
    filePattern: string = '.*',
    errors: Error[] | undefined = undefined,
    folderDepth: number = 0
): Promise<string[]> {
    const results: string[] = []

    // Get all files from the folder
    let items = await FsExtra.readdir(folder).catch(error => {
        if (errors) {
            errors.push(error) // Save errors if we wish (e.g. folder perms issues)
        }

        return results
    })

    // Go through to the required depth and no further
    folderDepth = folderDepth - 1

    // Loop through the results, possibly recurse
    for (const item of items) {
        try {
            const fullPath = Path.join(folder, item)

            if (
                FsExtra.statSync(fullPath).isDirectory() &&
                folderDepth > -1)
            ) {
                // Its a folder, recursively get the child folders' files
                results.push(
                    ...(await FileHelper.findFiles(fullPath, filePattern, errors, folderDepth))
                )
            } else {
                // Filter by the file name pattern, if there is one
                if (filePattern === '.*' || item.search(new RegExp(filePattern, 'i')) > -1) {
                    results.push(fullPath)
                }
            }
        } catch (error) {
            if (errors) {
                errors.push(error) // Save errors if we wish
            }
        }
    }

    return results
}

如果有人:

只想列出项目本地子文件夹中的文件名(不包括目录)

✅ 无其他依赖项✅ 1功能✅ 规格化路径(Unix与Windows)

const fs = require("fs");
const path = require("path");

/**
 * @param {string} relativeName "resources/foo/goo"
 * @return {string[]}
 */
const listFileNames = (relativeName) => {
  try {
    const folderPath = path.join(process.cwd(), ...relativeName.split("/"));
    return fs
      .readdirSync(folderPath, { withFileTypes: true })
      .filter((dirent) => dirent.isFile())
      .map((dirent) => dirent.name.split(".")[0]);
  } catch (err) {
    // ...
  }
};

README.md
package.json
resources
 |-- countries
    |-- usa.yaml
    |-- japan.yaml
    |-- gb.yaml
    |-- provinces
       |-- .........


listFileNames("resources/countries") #=> ["usa", "japan", "gb"]