Doesn't compile when importing scss file with underscores

This does not compile:

gulp.task('sass', function () {
    return gulp.src([
        'bower_components/bootstrap-sass/assets/stylesheets/_bootstrap.scss'
    ])
        .pipe(sass())
        .pipe(gulp.dest('css'));
});

I have to rename _bootstrap.scss to bootstrap.scss to be able to compile it, like this:

gulp.task('sass', function () {
    return gulp.src([
        'bower_components/bootstrap-sass/assets/stylesheets/bootstrap.scss'
    ])
        .pipe(sass())
        .pipe(gulp.dest('css'));
});

[Snugug]

Files starting with underscores are partials in Sass and, by the language definition, are not compiled.

On Sep 2, 2015, at 6:46 AM, George Geladaris [e-mail] wrote:

This does not compile:

gulp.task(‘sass’, function () { return gulp.src([ ‘bower_components/bootstrap-sass/assets/stylesheets/_bootstrap.scss’ ]) .pipe(sass()) .pipe(gulp.dest(‘css’)); }); I have to rename _bootstrap.scss to bootstrap.scss to be able to compile it, like this:

gulp.task(‘sass’, function () { return gulp.src([ ‘bower_components/bootstrap-sass/assets/stylesheets/bootstrap.scss’ ]) .pipe(sass()) .pipe(gulp.dest(‘css’)); }); — Reply to this email directly or view it on GitHub.

[LoveAndHappiness]

Yes, you (@Keats, @OEvgeny) are right. I misunderstood how to work properly with bootstrap. I’ll close the ticket, because I understand it now.