Проблеммы с gulp
Выдает ошибку Еype Error: cannot read property 'html'
of undefined
let project_folder="dist";
let sourse_folder="#src";
let path = {
buid: {
html: project_folder + "/",
css: project_folder + "/css/",
js: project_folder + "/js/",
img: project_folder + "/img/",
fonts: project_folder + "/fonts/",
},
src: {
html: sourse_folder + "/*.html",
css: sourse_folder + "/scss/style.scss",
js: sourse_folder + "/js/script.js",
img: sourse_folder + "/img/**/*.{jpg,png,svg,gif,ico,webp}",
fonts: sourse_folder + "/fonts/*.ttf",
},
watch: {
html: sourse_folder + "/**/*.html",
css: sourse_folder + "/scss/**/*.scss",
js: sourse_folder + "/js/**/*.js",
img: sourse_folder + "/img/**/*.{jpg,png,svg,gif,ico,webp}",
},
clean: "./" + project_folder + "/"
}
let { src, dest } = require('gulp'),
gulp = require('gulp'),
browsersync = require('browser-sync').create();
function browserSync(params) {
browsersync.init({
server: {
baseDir: "./" + project_folder + "/"
},
port: 3000,
notify: false
})
}
function html() {
return src(path.src.html)
.pipe(dest(path.build.html))
.pipe(browsersync.stream())
}
let build = gulp.series(html);
let watch = gulp.parallel(build,browserSync);
exports.html = html;
exports.build = build;
exports.watch = watch;
exports.default = watch;