Gulp error: watch task has to be a function
In gulp 3.x you could just pass the name of a task to gulp.watch()
like this:
gulp.task('watch', function() { gulp.watch('app/css/*.css', ['styles']); gulp.watch('app/js/*.js', ['scripts']); gulp.watch('app/img/*', ['images']);});
In gulp 4.x this is no longer the case. You have to pass a function. The customary way of doing this in gulp 4.x is to pass a gulp.series()
invocation with only one task name. This returns a function that only executes the specified task:
gulp.task('watch', function() { gulp.watch('app/css/*.css', gulp.series('styles')); gulp.watch('app/js/*.js', gulp.series('scripts')); gulp.watch('app/img/*', gulp.series('images'));});
GULP-V4.0
It is a bit late to answer this right now but still. I was stuck in this problem as well and this is how I got it working.
In detail analysis what I was doing wrong
- I forgot to call the reload function when the
watch
noticed some changes in my html files. - Since
fireUp
andKeepWatching
are blocking. They need to be started in parallel rather than serially. So I used the parallel function in the variable run.
thanks for all
gulp.task('watch', function(){ gulp.watch('app/sass/**/*.sass', gulp.series('sass'));});
for version gulp 4.xx