Why is gulp-sass not creating output

Question

I'm trying to get gulp to compile some sass, however I can't seem to get it to write out anything. I'm not seeing an error and using sass to compile myself is successful. Here is what my config looks like:

var sass = require('gulp-sass');
var baseOutputDir = '../publish/homepage';
var cssOutputDir = baseOutputDir + '/css';
gulp.task('sass', function () {
    gulp.src(['./css-lib/Bootstrap/3.3.1/assets/stylesheets/_bootstrap.scss'])
        .pipe(sass())
        .on('error', function (err) { console.log(err.message); })
        .pipe(gulp.dest(cssOutputDir));
})

The bootstrap is a clone of https://github.com/twbs/bootstrap-sass

score 11 · Answer 1 · answered Jan 15 '15 at 22:58

11

I'm use to LESS and not Sass. A file with an underscore in front won't generate a css file.

answered Jan 15 '15 at 22:58

Steven

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Thanks so much! This is absolutely absurd and arbitrary and I can't believe how much time I just wasted on this. – bradwilder31415 Aug 21 '18 at 20:35

score 5 · Answer 2 · answered Jul 20 '15 at 09:47

5

I had the same problem. As @Steven pointed out _filename do not generate css files. Just rename/copy the _bootstrap.scss to something without leading underscore

gulp.src(['./css-lib/Bootstrap/3.3.1/assets/stylesheets/bootstrap.scss'])

answered Jul 20 '15 at 09:47

pravin

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Why is gulp-sass not creating output

2 Answers2