首先,我将每个angularjs文件导入到html中,并且工作正常 但是,我想使用gulp将这些文件连接到一个文件中以减少连接量 我的文件夹结构是:
--- public
|- app.js
|- controller.js
|- directives.js
|- factories.js
|- filter.js
|---
|--- events
|- app.js
|- controller.js
....
在我的/public/app.js中,我声明了路由器
$stateProvider
.state('events', {
url: "/",
templateUrl: "/public/partial/events",
controller: "events",
reloadOnSearch: false
}).....
我将每个页面声明为一个文件夹,例如事件,并包含app.js和controller.js。
我的活动/ app.js是
angular.module('events', ['ui.router', 'ngRoute', 'events.controller'])
.config(['$routeProvider', '$locationProvider', '$stateProvider', '$urlRouterProvider',
function($routeProvider, $locationProvider, $stateProvider, $urlRouterProvider) {
}
]);
events / controller.js是
angular.module('events.controller', ['restangular', 'LocalStorageModule', 'oitozero.ngSweetAlert', 'Alertify', 'daterangepicker', 'ngTable', 'slick'])
.controller('events', ['$scope','$compile', ....,
function ($scope, $compile, ....) {
我的gulpfile.js只是简单地连接所有angularjs文件但是出错了!
gulp.src([
"/public/app-v8.js",
"/public/controller-v22.js",
"/public/directives-v6.js",
"/public/factories-v5.js",
"/public/filters-v1.js",
"/public/events/app-v0.js",
"/public/events/controller-v23.js",
.....
])
答案 0 :(得分:0)
在你的gulpfile中,giv
var gulp = require('gulp');
var concat = require('gulp-concat');
var minify = require('gulp-minify');
var cleanCss = require('gulp-clean-css');
gulp.task('pack-js',function(){
return gulp.src(['app.js',
'common/controllers.js',
'common/services.js',
....
])
.pipe(concat('bundle.js'))
.pipe(minify())
.pipe(gulp.dest('myFolder/build/js/'))
});
'bundle.js' contain all your JS files. Similarly you can do for CSS files also.
At last you can run your task like this.
gulp.task('bundle-assets',['pack-js','pack-css']);