什么是“更便宜"的性能明智的 $broadcast 或 $watch [英] What is 'cheaper' performance-wise $broadcast or $watch

问题描述

我的应用程序中遇到一种情况，每次用户角色发生变化时我都需要重新加载菜单(一个用户可以在多个公司担任角色).

I have a situation in my application where I need to reload the menu each time the role of the user changes(One user can have roles in several companies).

我想知道解决这个问题的最佳方法是什么.

I was wondering what is the best way to approach this issue.

目前我正在做以下事情:

currently I am doing the following:

app.controller('menuLoadingCtrl', function($location, $scope, authService){
    $scope.model.initialRole = authService.getRole();
    $scope.$watch(function(){return authService.getRole()}, function(val){
        if(val && val != $scope.model.initialRole){
                $scope.layout.menuSrc = 'partials/menu.html';
        }
    });
})

简单地将用户重定向到菜单加载视图，并在角色加载完成后从那里返回菜单视图.我把它包装在一个函数中:

Simple redirecting the user to the menu loading view, and from there, back to the menu view once the role is done loading. I have this wrapped in a function:

 $scope.layout.reloadMenu = function(){
        $scope.layout.menuSrc = 'partials/menuLoading.html';
    }

我在任何想要重新加载菜单的情况下调用它.

which I call at any scenario at which I would like to reload the menu.

我想知道是否可以通过从 $rootScope 上的服务广播此事件，然后在控制器中侦听它来使此过程更加自动化.

I was wondering if I can make this process more automatic by broadcasting this event from the service on the $rootScope, and then listening to it in the controller.

对此的任何想法\建议将不胜感激.

Any thoughts\advice on this will be greatly appreciated.

推荐答案

$watch() 正在进行污垢检查:该函数在每个摘要周期进行比较.另一方面，$broadcast() 仅在有事件时传播事件.自然，$broadcast() 比 $watch() 便宜.

$watch() is doing dirt-checking: the function makes a comparison each digest cycle. On the other hand, $broadcast() propagates an event only when there is one. Naturally, $broadcast() is cheaper than $watch().

但是你真的需要担心这里的性能吗?一种按周期进行的原始比较算不了什么.但是，从概念上讲，$watch() 显然是您所需要的:您希望每次变量更改时执行一个操作.我无法想象在这里使用 $broadcast().

But did you really have to worry about performance here? One primitive comparison by cycle is nothing. However, conceptually, $watch() is clearly what you need: you want to do an action each time a variable changes. I can't imagine using $broadcast() here.

这篇关于什么是“更便宜"的性能明智的 $broadcast 或 $watch的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！