find file with wild card matching

Question

In node.js, can I list files with wild card matching like

fs.readdirSync(\'C:/tmp/*.csv\')?

I did not find the information on wild card ma

Answer 1

If glob is not quite what you want, or a little confusing, there is also glob-fs. The documentation covers many usage scenarios with examples.

// sync 
var files = glob.readdirSync('*.js', {});

// async 
glob.readdir('*.js', function(err, files) {
  console.log(files);
});

// stream 
glob.readdirStream('*.js', {})
  .on('data', function(file) {
    console.log(file);
  });

// promise 
glob.readdirPromise('*.js')
  .then(function(files) {
    console.log(file);
  });

Answer 2

If you don't want to add a new dependency to your project (like glob), you can use plain js/node functions, like:

var files = fs.readdirSync('C:/tmp').filter(fn => fn.endsWith('.csv'));

Regex may help in more complex comparisons

Answer 3

This is not covered by Node core. You can check out this module for what you are after.

Setup

npm i glob

Usage

var glob = require("glob")

// options is optional
glob("**/*.js", options, function (er, files) {
  // files is an array of filenames.
  // If the `nonull` option is set, and nothing
  // was found, then files is ["**/*.js"]
  // er is an error object or null.
})

Answer 4

Just in case you want to search files by regex (for complex matches), then consider using file-regex, which supports recursive search and concurrency control (for faster results).

Sample usage

import FindFiles from 'file-regex'

// This will find all the files with extension .js
// in the given directory
const result = await FindFiles(__dirname, /\.js$/);
console.log(result)

Answer 5

dont reinvent the wheel, if you are on *nix the ls tool can easily do this (node api docs)

var options = {
  cwd: process.cwd(),
}
require('child_process')
.exec('ls -1 *.csv', options, function(err, stdout, stderr){
  if(err){ console.log(stderr); throw err };
  // remove any trailing newline, otherwise last element will be "":
  stdout = stdout.replace(/\n$/, '');
  var files = stdout.split('\n');
});