Gulp is beautifully installed both globally and locally. Locally installed sass in the project folder.

I want to process .scss file and save it in another folder. When you call on the command line, everything ends successfully, but the build folder does not appear.

If I delete the line with the sass processing call, the folder appears, but the .scss file is of course copied in its original form.

What could be the problem?

var gulp = require('gulp'); var sass = require('gulp-sass'); gulp.task('publish_css', function() { gulp.src('css/**/*.scss') .pipe(sass().on('error', sass.logError)) .pipe(gulp.dest('build/styles')); });

    1 answer 1

    The code above is working. As it turned out, in the code itself, in vain I wanted to explicitly nest the nesting of folders, it does not rob gulp.src ('css / folder / ** / *. Scss')

    You just have to gulp.src ('css / ** / *. Scss') Features of semantics, you see. The question is closed.