如何为这种类型的脚手架设置gulp？

Question

文件夹结构

Components ------component1 -partials - js - html - scss - component1.css - component1.js ------component2 -partials - js - html - scss - component2.css - component2.js

功能是我的所有 js，html 和 scss 文件转换为一个 css 和 js 很好，但在组件文件夹中。如果我创建一个新组件，我不想每次都将它们单独添加到 gulp 中，它将通过 gulp 自动添加。如何编写 gulp 来实现此类功能？

非常感谢帮助...!!!

Answer 1

您可以使用我为您的案例创建的代码：

项目结构： folder structure tested

gulpfile.js

/**
 * Created by Anderson on 22/04/2016.
 */

'use strict';

var gulp        = require('gulp');
var data        = require('gulp-data');
var concat      = require('gulp-concat');
var sass        = require('gulp-sass');
var gcallback   = require('gulp-callback');
var rename      = require('gulp-rename');
var del         = require('del');
var path        = require('path');
var fs          = require('fs');


gulp.task('main',function(){
    //scan componens folder all files and folders
    gulp.src("./components/**")
        .pipe(data(function(file,cb){
            var filepath = file.path;
            var stats = fs.lstatSync(filepath);

            var base = file.base;
            var fpath = filepath.replace(file.base,"");
            //console.log(fpath);

            var array = fpath.split("\");
            var basename = path.basename(fpath);
            var componentName = array[0];
            //destiny
            var dest = base + fpath.replace(basename,"");
                dest = dest.replace("partials\","");

            //process the scss
            if(stats.isDirectory() && basename == "scss"){
                //console.log(array,componentName,basename);
                //console.log(file.path);
                var scssScan = base + fpath+"\*.scss";
                console.log("scan",scssScan );
                console.log("dest",dest);
                //scan dir for scss
                gulp.src(scssScan)
                    //concat all scss in a temporary file
                    .pipe(concat(componentName+".scss"))
                    .pipe(gulp.dest(dest))
                    .pipe(gcallback(function(){
                        //scan to process the scss
                        var componentPath = base + componentName;
                        var componentScssScan = componentPath + "\*.scss";
                        //console.log(componentPath,componentScssScan);
                        //console.log(componentPath + "\" + componentName+".scss");
                        //scan the component dir for scss
                        gulp.src(componentScssScan)
                            //process the sass
                            .pipe(sass())
                            .pipe(gulp.dest(dest));
                            //delete the temporary scss
                            //.pipe(del(componentPath + "\" + componentName+".scss"));
                    }));

            }

            //process the js
            if(stats.isDirectory() && basename == "js"){
                //console.log(array,componentName,basename);
                //console.log(file.path);
                var scssScan = base + fpath+"\*.js";
                console.log("scan",scssScan );
                console.log("dest",dest);
                //scan dir for js
                gulp.src(scssScan)
                    //concat all js in a temporary file
                    .pipe(concat(componentName+".js"))
                    .pipe(gulp.dest(dest));

            }

            cb(undefined,undefined);
        }));
});

Answer 2

我认为你需要做的是为 css 和 js 文件和运行添加基于更改的特定任务的观察者，这里有一些例子：

var gulp = require('gulp'),
    less = require('gulp-less'),
    cleancss = require('gulp-clean-css'),
    server = require('gulp-server-livereload'),
    rename = require('gulp-rename');

gulp.task('server', ['watch', 'webserver']);

gulp.task('styles', function() {
    return gulp.src('public/assets/less/app.less')
        .pipe(less())
        .pipe(rename('app.css'))
        .pipe(cleancss())
        .pipe(rename({suffix: '.min'}))
        .pipe(gulp.dest('public/dist/css'))
});

// Add watcher for the changes in files
gulp.task('watch', function() {
    gulp.watch('public/assets/less/*.less', ['styles']);
})

gulp.task('webserver', function() {
    gulp.src('public')
    .pipe(server({
        livereload: true,
        port: 8080,
        open: true,
    }))
});

并且您还可以使用 Livereload Chrome Extension 启用浏览器在您的代码发生更改时自动刷新。

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