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

我的单行代码:

const fs = require("fs")
const path = 'somePath/'

const filesArray = fs.readdirSync(path).filter(file => fs.lstatSync(path+file).isFile())

其他回答

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
}

但是,上面的答案不会对目录执行递归搜索。以下是我对递归搜索所做的操作(使用node walk:npm install walk)

var walk    = require('walk');
var files   = [];

// Walker options
var walker  = walk.walk('./test', { followLinks: false });

walker.on('file', function(root, stat, next) {
    // Add this file to the list of files
    files.push(root + '/' + stat.name);
    next();
});

walker.on('end', function() {
    console.log(files);
});

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'))

如果有人还在搜索这个,我会这样做:

从“fs”导入fs;从“path”导入路径;const getAllFiles=目录=>fs.readdirSync(dir).reduce((files,file)=>{常量名称=路径.连接(目录,文件);const isDirectory=fs.statSync(名称).isDirectory();return isDirectory?[…file,…getAllFiles(名称)]:[…files,名称];}, []);

它的工作对我很好

从Node v10.10.0开始,可以将fs.readdir和fs.readderSync的新withFileTypes选项与dirent.isDirectory()函数结合使用,以过滤目录中的文件名。看起来像这样:

fs.readdirSync('./dirpath', {withFileTypes: true})
.filter(item => !item.isDirectory())
.map(item => item.name)

返回的数组的格式为:

['file1.txt', 'file2.txt', 'file3.txt']