Gulp watch() method is not watching file changes - gulp

Question

I am trying to automate some tasks with Gulp for Wordpress theme development. I have managed to get everything to work, except that gulp.watch doesn't watch for changes on my files automatically.
Gulp Version:
CLI version 2.0.1
Local version 3.9.1
Operating System:
Windows 10
Package.json:
{
"name": "theme-development-boilerplate",
"version": "1.0.0",
"description": "A boilerplate for Theme Development automated with Gulp. Includes SASS, Bootstrap, jQuery, Normalize, Mordernizer, among other stacks",
"main": "gulpfile.js",
"scripts": {
"test": "echo \"Error: no test specified\" && exit 1"
},
"author": "Manuel Abascal",
"license": "MIT",
"devDependencies": {
"gulp": "^3.9.1",
"gulp-concat": "^2.6.1",
"gulp-error-notifier": "^1.1.0",
"gulp-imagemin": "^5.0.3",
"gulp-livereload": "^4.0.1",
"gulp-sass": "^4.0.2",
"gulp-uglify": "^3.0.1"
},
"dependencies": {
"browser-sync": "^2.26.3",
"gulp-watch": "^5.0.1"
}
}
Gulpfile:
var gulp = require('gulp');
var watch = require('gulp-watch');
var imagemin = require('gulp-imagemin');
var uglify = require('gulp-uglify');
var errorNotifier = require('gulp-error-notifier');
var sass = require('gulp-sass');
var concat = require('gulp-concat');
// Logs Message
gulp.task('message', function(){
return console.log('Gulp is running...');
});
// Optimize Images
gulp.task('imageMin', function (){
gulp.src('src/images/*')
.pipe(imagemin())
.pipe(gulp.dest('dist/images'))
});
// Minify JS
gulp.task('minify', function(){
gulp.src('src/js/*.js')
.pipe(uglify())
.pipe(gulp.dest('dist/js'));
});
// Compile Sass into CSS
gulp.task('sass', function(){
gulp.src('src/sass/*.scss')
.pipe(errorNotifier.handleError(sass({outputStyle: 'compressed'})).on('error', sass.logError))
.pipe(gulp.dest('dist/css'));
});
// Scripts Concatenation
gulp.task('scripts', function(){
gulp.src('src/js/*.js')
.pipe(concat('main.js'))
.pipe(uglify())
.pipe(gulp.dest('dist/js'));
});
gulp.task('default', ['message', 'imageMin', 'sass', 'scripts']);
gulp.task('watch', function(){
gulp.watch('src/js/*.js', ['scripts']);
gulp.watch('src/images/*', ['imageMin']);
gulp.watch('src/sass/*.scss', ['sass']);
});
When I use the command gulp it works like a charm, however when I run gulp watch I get this output in the console:
Starting 'watch'...
Finished 'watch' after 12 ms
But it doesn't watch for anything. Any ideas on how to debug this? I read that Windows Operating System might be the issue or the version. I don't know at this point.
Thanks in advance!

Answer 1

There is no problem with windows, I use gulp on windows on a daily basis.
And about the version, I don't think that is the culprit here either, but you should update to the latest version anyway (which is 4.0.0).
Try using a function in your watch instead of a task, like this:
(This is what's done in the official examples here, and also what I do and it works)
// Scripts Concatenation
function transpile_scripts(cb) {
gulp.src('src/js/*.js')
.pipe(concat('main.js'))
.pipe(uglify())
.pipe(gulp.dest('dist/js'))
.on('end', cb);
}
gulp.watch('src/js/*.js', transpile_scripts)
Also note that I passed a callback to my function, and I run it after the pipe is finished. This callback is provided implicitly by gulp, and it sort of keeps track of when a task is finished. In some cases not passing and executing this causes warnings and even errors. Anyway, it won't harm to do it just in case.