List all files in array with gulp.src()

Question

I'm trying to create an index of all file(path)s within a given folder. So far I worked with gulp.src(filePath) to achieve that. According to this blog post, it should work:

gulp.src(files) is a string or array containing the file(s) / file paths.

My current code:

gulp.task("createFileIndex", function(){
    var index = gulp.src(['./content/**/*.*']);
    console.log("INDEX:", index[0]);
});

By outputing the returned values of gulp.src() with index[0] I get undefined and the whole index outputs only a large dictionary without any filepaths.

score 14 · Answer 1 · answered Nov 28 '17 at 12:42

14

The current solution is:

var gulp = require('gulp');
var debug = require('gulp-debug');

gulp.src(sources)
  .pipe(debug());

answered Nov 28 '17 at 12:42

Blitz

249
3
16

This is the best answer: Avoids all unnecessary extra code and only shows the minimal code needed to test this functionality. – Capy Dec 23 '19 at 09:05

score 9 · Answer 2 · answered Jul 29 '16 at 19:00

9

As the OP stated in a comment, a simple solution to this problem would be to use fs.readdirSync instead of gulp.src:

fs = require("fs");
fs.readdirSync(directoryPath); // ["file1", "file2"]

answered Jul 29 '16 at 19:00

Chase Sandmann

4,795
3
30
42

score 7 · Answer 3 · answered Apr 20 '18 at 08:33

var through = require('through2');
gulp.task('getFileList', function () {
    var fileList = [];
    gulp.src(['./someFolder/**/*.ext', '!./someFolder/unwantedFolder/**/*'])
        .pipe(through.obj(function (file, enc, cb) {
            fileList.push(file.path);
            cb(null);
        }))
        .pipe(gulp.dest('./temp/'))
        .on ('end', function () {
            console.log(fileList);
        });
});

score 4 · Accepted Answer · edited Jun 20 '20 at 09:12

According to the gulp documentation on gulp.src (https://github.com/gulpjs/gulp/blob/master/docs/API.md#gulpsrcglobs-options)

gulp.src(globs[, options])

Emits files matching provided glob or an array of globs. Returns a stream of Vinyl files that can be piped to plugins.
gulp.src('client/templates/*.jade')
  .pipe(jade())
  .pipe(minify())
  .pipe(gulp.dest('build/minified_templates'));
glob refers to node-glob syntax or it can be a direct file path.

globs

Type: String or Array

Glob or array of globs to read.

options

Type: Object

Options to pass to node-glob through glob-stream.

gulp adds some additional options in addition to the options supported by node-glob and glob-stream

So it seems you need to look in further on this. Otherwise this maybe helpful Get the current file name in gulp.src()

I created a workaround with nodejs' internal `js` library. `fs.readdirSync()` lists all files of a defined folder. — user3147268, Aug 02 '15 at 16:53

score 4 · Answer 5 · edited Dec 17 '18 at 17:15

4

If all you want is the array of filenames from a glob (like gulp.src) use:

const glob = require('glob');

const fileArray = glob.sync('./content/**/*.*');

edited Dec 17 '18 at 17:15

Community

1
1

answered Apr 20 '18 at 14:36

Mark

143,421
24
428
436

score 0 · Answer 6 · answered Jan 17 '20 at 14:04

0

Or you can use native Node.js function readdirSync

const fs = require('fs');

const syncImgFolder = (done) => {
    const files = fs.readdirSync('/path/');
}

answered Jan 17 '20 at 14:04

Tibor

73
1
1
8

score -2 · Answer 7 · answered Oct 15 '19 at 15:26

-2

You can do the following in gulp 4 :

gulp.src(['./lib/file3.js', './lib/file1.js', './lib/file2.js'])

answered Oct 15 '19 at 15:26

Wildhammer

2,017
1
27
33

List all files in array with gulp.src()

7 Answers7