Task to update .html/.js on Gulp 4?

Question

gulp.js

There is a structure:
app
---css

• -main.scss
  
• -fonts.scss
  

---js

• -main.js
  

---img

• -bg.png
  
• -1bg.png
  

---fonts

• -1.ttf
  
• -1.otf
  

---index.html
dist
---
---
---
How to write a task to transfer .html/.js/fonts/img to dist?
Preferably with watch.

Answer 1

For your assembly will go like this watcher:

gulp.task('watch', function(){
  gulp.watch('файлы html', gulp.series('html'));
  gulp.watch('файлы css', gulp.series('css'));
  gulp.watch('файлы js', gulp.series('js'));
  gulp.watch('файлы img', gulp.series('img'));
})

Instead 'файлы [html, css, js&img]'write the path to the files.
And buildersomething like this:

gulp.task('build', 
  gulp.series(
    сюда можно таск очитки dist папки засунуть, т.е. его название, типа 'clean',
    gulp.parallel(
      'html',
      'styles',
      'js',
      'img'
)));

If the task is not needed for cleaning, then you can delete the line with gulp.series, and do not forget to remove all the brackets associated with this method.