How can I use a glob to ignore files that start with an underscore?

Question

I am using this file matching in gulp:

'content/css/*.css'

But I would like to only include the files that do not start with an underscore. Can someone tell me how I can do this?

Here's the code:

gulp.task('less', function () {
    gulp.src('content/less/*.less')
        .pipe(less())
        .pipe(gulp.dest('content/css'));
});

That doesn't look like a regexp, it looks like a filename wildcard. — Barmar, Dec 29 '14 at 12:57
I think it is referred to as a glob in gulp but I am a bit confused by it all. — Alan2, Dec 29 '14 at 13:08

Barmar · Accepted Answer · 2014-12-29T13:15:01.013

44

If Gulp wildcards are like shell wildcards:

content/css/[^_]*.css

The ^ at the beginning of a character set means to match any characters not in the set.

edited Dec 29 '14 at 13:15

answered Dec 29 '14 at 12:58

Barmar

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score 8 · Answer 2 · answered Dec 06 '15 at 16:02

8

You can add a second glob whitch will filter files or folders with an underscore.

For ignore folders I use:

gulp.src(['./src/**/*.html', '!**/_*/**'])

'!**/_*/**' will filter folders that start with an underscore.

answered Dec 06 '15 at 16:02

Denis535

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Pratap Koritala · Answer 3 · 2014-12-29T13:22:14.530

2

Use this plugin https://github.com/robrich/gulp-ignore

var gulpIgnore = require('gulp-ignore');
var condition = '_*.css'; //exclude condition

gulp.task('less', function() {
  gulp.src('content/less/*.less')
    .pipe(gulpIgnore.exclude(condition))
    .pipe(less())
    .pipe(gulp.dest('/dist/'));
});

I think gulp supports file glob not full-regex, But You can give a try to content/css/[^_]*.less

edited Dec 29 '14 at 13:22

answered Dec 29 '14 at 13:10

Pratap Koritala

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I'm very open to your new suggestiong but can you check my updated question where I show the code that's being used. I think my version might be a bit simpler as it does not need gulp-ignore. I welcome your comments on this. Thanks – Alan2 Dec 29 '14 at 13:12
Check the updated post, But your code is including all the files ending with .less I am not sure if gulp by default supports full regex in filenames!! – Pratap Koritala Dec 29 '14 at 13:17

score 1 · Answer 4 · answered Dec 29 '14 at 13:00

1

You could use this regex.

content/css/[^_].*\.css

Add $ at the last if necessary.

answered Dec 29 '14 at 13:00

Avinash Raj

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Can you tell me what the ".*\" in the middle does here? Also I am looking for files that do not start with an underscore. – Alan2 Dec 29 '14 at 13:10
`[^_]` matches the first letter of the filename, only if it's not an underscore. So the following `.*` matches any character non-greedily, it matches underscores also. And the final `.css` asserts that the filenames must end with .css` – Avinash Raj Dec 29 '14 at 13:15

score 0 · Answer 5 · answered Oct 12 '21 at 20:59

0

For those using Python (coming from Google like me), the correct would be:

import glob
glob.glob('content/css/[!_]*.css')

answered Oct 12 '21 at 20:59

Fernando Wittmann

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How can I use a glob to ignore files that start with an underscore?

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