javascript - Convert gulp watch in gulp@3.9.1 to gulp@4

Question

We are switching from gulp@3.9.1 to gulp@4 and are having trouble switching over. When we run gulp watch, we are getting the following errors and trying to figure out how to resolve it.

What is the proper way to convert the gulp watch task to work with gulp@4?

Error message

AssertionError [ERR_ASSERTION]: Task never defined: minify-css

Command: gulp watch

• This should run minify-js then minify-css in order
• minify-js should run after clean-scripts has completed successfully
• minify-css should run after clean-css has completed successfully

Current tasks.

var gulp = require('gulp'),
    cssmin = require('gulp-clean-css'),
    clean = require('gulp-clean'),
    uglify = require('gulp-uglify');
    
var src = {
  js: 'js/some-dir/**/*.js',
  css: 'css/some-dir/**/*.css'
};

var dest = {
  js: 'js/dest/some-dir/**/*.js',
  css: 'css/dest/some-dir/**/*.css'
};

gulp.task('clean-css', function() {
  return gulp.src(dest.css)
             .pipe(clean({read:false, force: true});
});

gulp.task('minify-css', ['clean-css'], function() {
  gulp.src(src.css)
    .pipe(cssmin())
    .pipe(gulp.dest(dest.css));
});

gulp.task('clean-scripts', function() {
  return gulp.src(dest.js)
             .pipe(clean({read:false, force: true});
});

gulp.task('minify-js', ['clean-scripts'], function() {
   gulp.src(src.js)
       .pipe(uglify())
       .pipe(gulp.dest(dest.js));
});

gulp.task('watch', ['minify-js', 'minify-css'], function() {
  gulp.watch(src.js, ['minify-js']);
  gulp.watch(src.css, ['minify-css']);
});

Answer 1

gulp.task('minify-css', gulp.series('clean-css', function() {
  return gulp.src(src.css)
    .pipe(cssmin())
    .pipe(gulp.dest(dest.css));
}));

gulp.task('minify-js', gulp.series('clean-scripts', function() {
   return gulp.src(src.js)
       .pipe(uglify())
       .pipe(gulp.dest(dest.js));
}));

gulp.task('watch', gulp.series('minify-js', 'minify-css', function() {
  gulp.watch(src.js, gulp.series('minify-js'));
  gulp.watch(src.css, gulp.series('minify-css'));
}));

As @Abdaylan suggested, I also advocate switching to functions. Nevertheless, so you can see where your code was wrong, I have fixed it here. Gulp 4 does not use this syntax:

gulp.task('someTask', ['task1', 'task2'], function () {});  // gulp 3

Gulp 4:

gulp.task('someTask', gulp.series('task1', 'task2', function () {}));  // gulp 4 with string tasks

or gulp.parallel. So you can use your gulp.task syntax (rather than named functions) if you modify them to use the signatures that gulp 4 supports as I did in your modified code at the top of this answer.

Gulp 4 with named functions:

gulp.task(someTask, gulp.series(task1, task2, function () {}));  // gulp 4 with named functions

So with named functions, the tasks are not referred to as strings.

See also task never defined for other potential problems when migrating from gulp3 to gulp4 with the same error message.