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Gulp watch only detects the first change

I am working on gulp and implementing watch functionality. But the gulp watch detects the changes only for the first time.
I want to write the code so that it detects the changes in CSS and JS files and performs minification and concatenation on the development environment.
I am using the following code:
var gulp = require('gulp');
var concat = require('gulp-concat');
var clean_css = require('gulp-clean-css');
var uglify = require('gulp-uglify');
gulp.task('style', function(){
gulp.src(['assets/css/style.css', 'assets/css/custom.css'])
.pipe(concat('build.min.css'))
.pipe(clean_css())
.pipe(gulp.dest('assets/build'));
});
gulp.task('script', function(){
gulp.src(['assets/js/jquery.js', 'assets/js/custom.js'])
.pipe(concat('build.min.js'))
.pipe(uglify())
.pipe(gulp.dest('assets/build'));
});
gulp.task('watch', function(){
gulp.watch('assets/css/*.css', gulp.series('style') );
gulp.watch('assets/js/*.js', gulp.series('script'));
});

This is probably because gulp does not know the task has finished the first time so it will not re-start the task again when you modify a file next. This can be solved just by adding return statements to your tasks:
gulp.task('style', function(){
return gulp.src(['assets/css/style.css', 'assets/css/custom.css'])
.pipe(concat('build.min.css'))
.pipe(clean_css())
.pipe(gulp.dest('assets/build'));
});
gulp.task('script', function(){
return gulp.src(['assets/js/jquery.js', 'assets/js/custom.js'])
.pipe(concat('build.min.js'))
.pipe(uglify())
.pipe(gulp.dest('assets/build'));
});

AssertionError [ERR_ASSERTION]: Task function must be specified

I am attempting to refactor some legacy code I wrote 2 years ago. A gulpfile.js file to be precise.
It seems like the problem is here:
// gulp.task('default', ['browserify', 'copy'], function() {
// return gulp.watch('src/**/*.*', ['browserify', 'copy']);
// });
I commented it out and replaced it with this:
gulp.task('default', gulp.series('browserify', 'copy'), function() {
return gulp.watch('src/**/*.*', ['browserify', 'copy']);
});
Not good enough. The whole file looks like this:
var gulp = require('gulp');
var browserify = require('browserify');
var reactify = require('reactify'); // Converts jsx to js
var source = require('vinyl-source-stream'); // Converts string to a stream
gulp.task('browserify', function() {
browserify('./src/js/main.js')
.transform('reactify')
.bundle()
.pipe(source('main.js'))
.pipe(gulp.dest('dist/js'));
});
gulp.task('copy', function() {
gulp.src('src/index.html').pipe(gulp.dest('dist'));
gulp.src('src/css/*.*').pipe(gulp.dest('dist/css'));
gulp.src('src/images/*.*').pipe(gulp.dest('dist/images'));
gulp.src('src/js/vendors/*.*').pipe(gulp.dest('dist/js'));
});
// gulp.task('default', ['browserify', 'copy'], function() {
// return gulp.watch('src/**/*.*', ['browserify', 'copy']);
// });
gulp.task('default', gulp.series('browserify', 'copy'), function() {
return gulp.watch('src/**/*.*', ['browserify', 'copy']);
});
I have read through some of the getting started documentation, but what I have read thus far has not helped me refactor this.

With Gulp 4.0 the way you run tasks in series has got changed. You can read and get what you want using below link
https://github.com/gulpjs/gulp/blob/master/docs/recipes/running-tasks-in-series.md.

This issue faced me because of the version of gulp I installed using npm i gulp
to solve this quickly, downgrade to that gulp version you used before 2 years and everything will work fine.

Gulp error: watch task has to be a function

Here is my gulpfile:
// Modules & Plugins
var gulp = require('gulp');
var concat = require('gulp-concat');
var myth = require('gulp-myth');
var uglify = require('gulp-uglify');
var jshint = require('gulp-jshint');
var imagemin = require('gulp-imagemin');
// Styles Task
gulp.task('styles', function() {
return gulp.src('app/css/*.css')
.pipe(concat('all.css'))
.pipe(myth())
.pipe(gulp.dest('dist'));
});
// Scripts Task
gulp.task('scripts', function() {
return gulp.src('app/js/*.js')
.pipe(jshint())
.pipe(jshint.reporter('default'))
.pipe(concat('all.js'))
.pipe(uglify())
.pipe(gulp.dest('dist'));
});
// Images Task
gulp.task('images', function() {
return gulp.src('app/img/*')
.pipe(imagemin())
.pipe(gulp.dest('dist/img'));
});
// Watch Task
gulp.task('watch', function() {
gulp.watch('app/css/*.css', 'styles');
gulp.watch('app/js/*.js', 'scripts');
gulp.watch('app/img/*', 'images');
});
// Default Task
gulp.task('default', gulp.parallel('styles', 'scripts', 'images', 'watch'));
If I run the images, scripts or css task alone it works. I had to add the return in the tasks - this wasn't in the book but googling showed me this was required.
The problem I have is that the default task errors:
[18:41:59] Error: watching app/css/*.css: watch task has to be a function (optionally generated by using gulp.parallel or gulp.series)
at Gulp.watch (/media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/node_modules/gulp/index.js:28:11)
at /media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/gulpfile.js:36:10
at taskWrapper (/media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/node_modules/undertaker/lib/set-task.js:13:15)
at bound (domain.js:287:14)
at runBound (domain.js:300:12)
at asyncRunner (/media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/node_modules/async-done/index.js:36:18)
at nextTickCallbackWith0Args (node.js:419:9)
at process._tickCallback (node.js:348:13)
at Function.Module.runMain (module.js:444:11)
at startup (node.js:136:18)
I think it is because there is also no return in the watch task. Also the error message isn't clear - at least to me. I tried adding a return after the last gulp.watch() but that didn't work either.

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 and KeepWatching 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

It worked for me in Gulp 4.0
gulp.task('watch', function() {
gulp.watch('src/images/*.png', gulp.series('images'));
gulp.watch('src/js/*.js', gulp.series('js'));
gulp.watch('src/scss/*.scss', gulp.series('css'));
gulp.watch('src/html/*.html', gulp.series('html'));
});

//Check what worked for me
gulp.task('watch', function(){
gulp.watch('css/shop.css', gulp.series(['shop']));
});

In my case, work for me this:
(in gulpfile.js)
(install: gulp, gulp sass)
var gulp = require('gulp');
var sass = require('gulp-sass')(require('sass'));
var cssDest = 'style';
var cssInputFile = 'source/style.scss';
var watchedFiles = 'source/**/*.scss';
gulp.task('buildcss', function(){
return gulp.src(cssInputFile)
.pipe(sass({
outputStyle: 'compressed'
}))
.pipe(gulp.dest(cssDest));
});
gulp.task('watch', function(){
gulp.watch(watchedFiles, gulp.series(['buildcss']));
});
Commend: gulp watch
(v 4.0)

On my side, I also had to add "{usePolling: true}" this to get it working:
gulp.watch(paths.js_src + '/*.js', {usePolling: true}, gulp.series(projectScripts, ondemandProjectScripts))
I think it's because my code runs into a docker container.

Gulp 4 watch runs only 1 time

As I mentioned in the title there is a problem running gulp.watch. It runs watch only after first change in the file, when I changing second, third and etc it doesn't run task.
Below is my gulpfile.js:
var gulp = require('gulp');
var babel = require('gulp-babel');
var rename = require("gulp-rename");
var del = require('del');
var less = require('gulp-less');
gulp.task('es6', function () {
return gulp.src('./test.js')
.pipe(rename(function (path) {
path.basename += "-es6";
return path;
}))
.pipe(babel())
.pipe(gulp.dest('./'))
});
gulp.task('clean', function () {
return del('./test-es6.js');
});
gulp.task('watch', function () {
gulp.watch( './test.js', gulp.series('es6') );
console.log('Running watch...');
});
gulp.task('default', gulp.series('clean', 'es6', gulp.parallel('watch') ));
And some logs :
$: gulp
[14:22:40] Using gulpfile /var/www/html/es2015/gulpfile.js
[14:22:40] Starting 'default'...
[14:22:40] Starting 'clean'...
[14:22:40] Finished 'clean' after 11 ms
[14:22:40] Starting 'es6'...
[14:22:43] Finished 'es6' after 2.11 s
[14:22:43] Starting '<parallel>'...
[14:22:43] Starting 'watch'...
Running watch...
[14:22:55] Starting '<series>'...
[14:22:55] Starting 'es6'...
[14:22:55] Finished 'es6' after 42 ms
[14:22:55] Finished '<series>' after 43 ms << first change, but no second third and etc.
I used similar configuration in couple of projects and it was fine, everything worked.
I don't know if this information is important, but I'm using Ubuntu 14.04

After long trying I found a solution for this problem.
I don't know why but this configuration works on Windows, but as appeared on Ubuntu I had to add ** to path of the watched file this part of code:
gulp.task('watch', function () {
gulp.watch( './**/test.js', gulp.series('es6') );
console.log('Running watch...');
});

Get Gulp watch to perform function only on changed file

I am new to Gulp and have the following Gulpfile
var gulp = require('gulp');
var jshint = require('gulp-jshint');
var concat = require('gulp-concat');
var rename = require('gulp-rename');
var uglify = require('gulp-uglify');
gulp.task('compress', function () {
return gulp.src('js/*.js') // read all of the files that are in js with a .js extension
.pipe(uglify()) // run uglify (for minification)
.pipe(gulp.dest('dist/js')); // write to the dist/js file
});
// default gulp task
gulp.task('default', function () {
// watch for JS changes
gulp.watch('js/*.js', function () {
gulp.run('compress');
});
});
I would like to configure this to rename, minify and save only my changed file to the dist folder. What is the best way to do this?

This is how:
// Watch for file updates
gulp.task('watch', function () {
livereload.listen();
// Javascript change + prints log in console
gulp.watch('js/*.js').on('change', function(file) {
livereload.changed(file.path);
gutil.log(gutil.colors.yellow('JS changed' + ' (' + file.path + ')'));
});
// SASS/CSS change + prints log in console
// On SASS change, call and run task 'sass'
gulp.watch('sass/*.scss', ['sass']).on('change', function(file) {
livereload.changed(file.path);
gutil.log(gutil.colors.yellow('CSS changed' + ' (' + file.path + ')'));
});
});
Also great to use gulp-livereload with it, you need to install the Chrome plugin for it to work btw.

See incremental builds on the Gulp docs.
You can filter out unchanged files between runs of a task using the gulp.src function's since option and gulp.lastRun