小编Ric*_*nes的帖子

我正在尝试将 gulpfile 任务拆分为不同的文件。我之前做过并将其添加到主 gulpfile 中，只需添加require('./path/taskName') 但这在 gulp 4 中不起作用

这是我的styles.js文件：

var gulp = require('gulp'),
    postcss = require('gulp-postcss'),
    autoprefixer = require('autoprefixer'),
    cssvars = require('postcss-simple-vars'),
    nestingCss = require('postcss-nested'),
    cssImport = require('postcss-import');

function styles(){
    console.log('Changes happened in CSS file');
    return gulp.src('./public/styles/style.css')
    .pipe(postcss([cssImport, cssvars, nestingCss, autoprefixer]))
    .pipe(gulp.dest('./public/temp/styles'));
}

 exports.styles = styles;

这是我的watch.js文件

var gulp = require('gulp'),
browserSync = require('browser-sync').create();

function html(done){
    console.log('Changes happened in html');
    browserSync.reload();
    done();
}

function cssInject(){
    return gulp.src('./public/temp/styles/style.css')
        .pipe(browserSync.stream());
}

function watch(done){
    browserSync.init({
        server: { …