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.
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!
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');
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