Handling exactly the file that has changed ( Gulp )?

Question

Node.js

I use gulp and the ssh module to upload the project to the hosting.
Now, when a file is changed, all files are processed and sent to the hosting. (Example below)
I want the file that was changed to be processed and sent.

gulp = require "gulp"
coffee = require "gulp-coffee"
uglyfly = require "gulp-uglyfly"
sftp = require 'gulp-sftp'



gulp.task "s", [], ->
  gulp.src ["dev/*.coffee", "dev/*/*.coffee"]
  .pipe do coffee
  .pipe sftp 
    host: '......'
    remotePath: "/var/www/..."
    user: '....'
    pass: '.....'
  .pipe gulp.dest "result"

gulp.task "watch", ->
  gulp.watch ["dev/*.coffee", "dev/*/*.coffee"], ["s"]


gulp.task "default", ["s", "watch"]

Answer 1

Feature:
Incremental Builds
This feature works with Gulp 4.
For Gulp 3, try the gulp-watch plugin . I read somewhere that with its help they solved such a problem, but I didn’t do it myself, I use version 4.