Get the most recent file in a directory, Node.js

Question

I am trying to find the most recently created file in a directory using Node.js and cannot seem to find a solution. The following code seemed to be doing the trick on one machine but on another it was just pulling a random file from the directory - as I figured it might. Basically, I need to find the newest file and ONLY that file.

var fs = require('fs'); //File System
var audioFilePath = 'C:/scanner/audio/'; //Location of recorded audio files
    var audioFile = fs.readdirSync(audioFilePath)
        .slice(-1)[0]
        .replace('.wav', '.mp3');

Many thanks!

Answer 1

A more functional version might look like:

import { readdirSync, lstatSync } from "fs";

const orderReccentFiles = (dir: string) =>
  readdirSync(dir)
    .filter(f => lstatSync(f).isFile())
    .map(file => ({ file, mtime: lstatSync(file).mtime }))
    .sort((a, b) => b.mtime.getTime() - a.mtime.getTime());

const getMostRecentFile = (dir: string) => {
  const files = orderReccentFiles(dir);
  return files.length ? files[0] : undefined;
};

Answer 2

Surprisingly, there is no example in this questions that explicitly uses Array functions, functional programming.

Here is my take on getting the latest file of a directory in nodejs. By default, it will get the latest file by any extension. When passing the extension property, the function will return the latest file for that extension.

The advantage of this code is that its declarative and modular and easy to understand as oppose to using "logic branching/control flows", of course given you understand how these array functions work

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Answer 3

Assuming availability of underscore (http://underscorejs.org/) and taking synchronous approach (which doesn't utilize the node.js strengths, but is easier to grasp):

var fs = require('fs'),
    path = require('path'),
    _ = require('underscore');

// Return only base file name without dir
function getMostRecentFileName(dir) {
    var files = fs.readdirSync(dir);

    // use underscore for max()
    return _.max(files, function (f) {
        var fullpath = path.join(dir, f);

        // ctime = creation time is used
        // replace with mtime for modification time
        return fs.statSync(fullpath).ctime;
    });
}

Answer 4

[Extended umair's answer to correct a bug with current working directory]

function getNewestFile(dir, regexp) {
    var fs = require("fs"),
     path = require('path'),
    newest = null,
    files = fs.readdirSync(dir),
    one_matched = 0,
    i

    for (i = 0; i < files.length; i++) {

        if (regexp.test(files[i]) == false)
            continue
        else if (one_matched == 0) {
            newest = files[i];
            one_matched = 1;
            continue
        }

        f1_time = fs.statSync(path.join(dir, files[i])).mtime.getTime()
        f2_time = fs.statSync(path.join(dir, newest)).mtime.getTime()
        if (f1_time > f2_time)
            newest[i] = files[i]  
    }

    if (newest != null)
        return (path.join(dir, newest))
    return null
}

Answer 5

Using pure JavaScript and easy to understand structure :

function getLatestFile(dirpath) {

  // Check if dirpath exist or not right here

  let latest;

  const files = fs.readdirSync(dirpath);
  files.forEach(filename => {
    // Get the stat
    const stat = fs.lstatSync(path.join(dirpath, filename));
    // Pass if it is a directory
    if (stat.isDirectory())
      return;

    // latest default to first file
    if (!latest) {
      latest = {filename, mtime: stat.mtime};
      return;
    }
    // update latest if mtime is greater than the current latest
    if (stat.mtime > latest.mtime) {
      latest.filename = filename;
      latest.mtime = stat.mtime;
    }
  });

  return latest.filename;
}