Syntax syrup for gulpfile
In your gulpfile, gulp-syrup works like the following.
const {task, src, dest, $} = require('gulp-syrup')(require('gulp'))
task('default', () =>
src('src/**/*.js')
.pipe($.somePlugin())
.pipe(dest('dist/')))gulp-syrup returns the static versions of the gulp APIs and $ as gulp-load-plugins.
It's useful if it works but be careful.
require('gulp-syrup')(require('gulp')).globals()
task('default', () =>
src('src/**/*.js')
.pipe($.somePlugin())
.pipe(dest('dist/')))This exposes the gulp APIs and gulp-load-plugins to the global namespace of gulpfile i.e. exposes src, dest, task, watch and $.
npm install --save-dev gulp-syrup
MIT