避免使用oclazyload和webpack代码拆分进行角度路由中的代码重复

时间:2017-02-24 09:31:55

标签: angularjs webpack oclazyload

我目前在项目中使用angular 1.5,我的路由如下:

const routes = ($stateProvider) => {

  $stateProvider
    .state('update', {
      url: '/update',
      template: '<update></update>',
      resolve: {
        lazyload: [ '$q', '$ocLazyLoad', function ($q, $ocLazyLoad) {
             console.log('start lazy');
             let deferred = $q.defer();
             require.ensure([], function () {
             let module = require('../components/update/update.module');
               let module = require('../components/update/update.module');
               $ocLazyLoad.load({
                 name: module.default.name
               });
               deferred.resolve(module);
             });
             return deferred.promise;
           }
         ]
      }
    })
    .state('login', {
      url: '/login',
      template: '<login></login>',
      resolve: {
        lazyload: [ '$q', '$ocLazyLoad', function ($q, $ocLazyLoad) {
            let deferred = $q.defer();
            require.ensure([], function () {
              let module = require('../components/login/login.module');
              $ocLazyLoad.load({
                name: module.default.name
              });
              deferred.resolve(module)
            });
            return deferred.promise;
          }
        ]
      }
    });

};

当我尝试在像这样的单独函数中提取重复代码时:

function load (modulePath) {
    return function ($q, $ocLazyLoad) {
      let deferred = $q.defer();
      require.ensure([], function () {
      let module = require(modulePath);
        $ocLazyLoad.load({
          name: module.default.name
        });
        deferred.resolve(module);
      });
      return deferred.promise;
    }
  }

并以这种方式使用load函数:

$stateProvider
    .state('update', {
      url: '/update',
      template: '<update></update>',
      resolve: {
        lazyload: [ '$q', '$ocLazyLoad', load('../components/update/update.module')]
      }
    })

路由不再起作用,我收到此错误&#34;找不到模块&#34;。&#34;在webpackMissingModule&#34;

1 个答案:

答案 0 :(得分:0)

使用动态需求时,请不要在参数中包含路径和扩展名。

将所有懒惰模块需求放在单独的文件夹lazy

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function load (state) {
  return function ($q, $ocLazyLoad) {
    let deferred = $q.defer();
    require.ensure([], function () {
      let module = require('../components/lazy/' +state+ '.module');
      $ocLazyLoad.load({
        name: module.default.name
      });
      deferred.resolve(module);
    });
    return deferred.promise;
  }
}

$stateProvider
.state('update', {
  url: '/update',
  template: '<update></update>',
  resolve: {
    lazyload: [ '$q', '$ocLazyLoad', load('update')]
  }
})
&#13;
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您可以使用与建议的结构不同的结构。请参阅https://webpack.github.io/docs/context.html#dynamic-requires