Michał Sadowski
Michał Sadowski

Reputation: 2149

Gulp sass failing silently

I have a very simple gulp task for transforming scss into css files:

function css() {
  return gulp.src('src/scss/_index.scss')
    .pipe(sass().on('error', sass.logError))
    .pipe(gulp.dest('public/css'))
}

As you probably imagine, _index.scss mostly imports other files.

Now, if I comment out line number 3, the code does what it's supposed to: it copies _index.scss to public/css. But with line 3 it produces no output whatsoever, without printing any errors to the console. All I get is:

[0] [17:32:46] Starting 'css'...
[0] [17:32:46] Finished 'css' after 79 ms

How can I even inspect what might be an issue here? The sass file is correct, as before that I would compile it without using gulp.

Upvotes: 1

Views: 241

Answers (1)

Sean
Sean

Reputation: 8026

Rename _index.scss to index.scss. Sass interprets underscore-prefixed files as partials, and won't output them directly. https://sass-lang.com/guide

Upvotes: 2

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