Gulp-sass error with notify

Question

I wondered if there is a way to have notify display a message on gulp-sass error. preferably the actual message that is displayed in the console.

my gulp task looks like this:

gulp.task('styles', function() {
  return gulp.src('src/scss/style.scss')
    .pipe(sass({ style: 'compressed', errLogToConsole: true }))
    .pipe(autoprefixer('last 2 version', 'safari 5', 'ie 8', 'ie 9', 'opera 12.1', 'ios 6', 'android 4'))
    .pipe(gulp.dest(''))
    .pipe(livereload(server))
    .pipe(notify({ message: 'Styles task complete' }));
});

I'd like to pipe the notify to some kind of error callback.

Any help appreciated.

Answer 1

After struggling with this myself I found that this worked:

gulp.task('styles', function() {
  return gulp.src('src/scss/style.scss')
    .pipe(sass({
        style: 'compressed',
        errLogToConsole: false,
        onError: function(err) {
            return notify().write(err);
        }
    }))
    .pipe(autoprefixer('last 2 version', 'safari 5', 'ie 8', 'ie 9', 'opera 12.1', 'ios 6', 'android 4'))
    .pipe(gulp.dest(''))
    .pipe(livereload(server))
    .pipe(notify({ message: 'Styles task complete' }));
});

You need to catch the error using the onError option that gulp-sass provides.

Hope that helps!

Answer 2

I'm a little late to the party here but the issue I was having was that the sass would stop compiling if there was an error in the code and I would have to restart gulp. Here is what I ended up doing:

gulp.task('sass', function() {
    return gulp.src('assets/scss/style.scss')
        .pipe(sass({ errLogToConsole: false, }))
        .on('error', function(err) {
            notify().write(err);
            this.emit('end');
        })
        .pipe(gulp.dest('assets/css'))
        .pipe(notify({ message: 'SCSS Compiled' }));
});

In my case I had to add this.emit('end');