Glob Pattern String Error- gulp-ruby-sass

Question

I have looked at some answers here, but I am completely new to this so I wasn't able to understand. Could someone take a look at my code and tell me what could have gone wrong. Also if you could explain what this error means. Attached is a screenshot of my terminal. Here is my gulp.js file:

var gulp = require('gulp'),
gutil = require('gulp-util'),
uglify = require('gulp-uglify'),
sass = require('gulp-ruby-sass'),
concat = require('gulp-concat'),
livereload = require('gulp-livereload'),
lr = require('tiny-lr'),
server = lr();

var jsSources = [
   'components/scripts/scriptOne.js',
   'components/scripts/scriptTwo.js'
];


var sassSources = [
    'components/sass/*.scss'
];

gulp.task('js', function () {
  gulp.src(jsSources)
        .pipe(uglify())
        .pipe(concat('scripts.js'))
        .pipe(gulp.dest('js'));
});

gulp.task('sass', function () {
  gulp.src(sassSources)
    .pipe(sass({style: 'expanded', lineNumbers: true}))
    .pipe(concat('style.css'))
    .pipe(gulp.dest('css'))
    .pipe(livereload());
});

gulp.task('watch', function () {
  var server = livereload();
   gulp.watch(jsSources, ['js']);
   gulp.watch(sassSources, ['sass']);
   gulp.watch(['js/scripts.js', '*.html'], function (e) {
    server.changed(e.path);
});
});
gulp.task('default', ['sass', 'js',    'watch']);

enter image description here

Answer 1

As you are using gulp-ruby-sass, you should call it directly instead of using the gulp.src(), like this:

gulp.task('sass', function () {
  sass(sassSources)
    .pipe(concat('style.css'))
    .pipe(gulp.dest('css'))
    .pipe(livereload());
});

Alternatively you can require gulp-sass instead of the gulp-ruby-sass.