javascript - Angular.js : ng-show element when $state. current.name 是某个值

Question

我希望仅当 $state.current.name 等于 about.list 时才显示 HTML block 。到目前为止，我有以下代码，但它似乎没有根据状态切换元素。

index.html

<nav class="global-navigation">
    <ul class="list">
    <li class="list-item">
      <a class="list-item-link" ui-sref="home">
        Home
      </a>
    </li>
    <li class="list-item">
      <a class="list-item-link" ui-sref="about">
        About
      </a>
    </li>
     <li class="list-item" ng-show="$state.current.name == 'about.list'">
      <a class="list-item-link" ui-sref="about.list">
        List
      </a>
    </li>
    </ul>
  </nav>

app.js

var myApp = angular.module('myApp', ['ui.router'])

  .config(['$urlRouterProvider', '$stateProvider', 

    function($urlRouterProvider, $stateProvider) {

    $urlRouterProvider.otherwise('/404.html');

    $stateProvider.

      // Home
      state('home', {
        url: '/',
        templateUrl: 'partials/_home.html',
        controller: 'homeCtrl'
      }).

      // About
      state('about', {
        url: '/about',
        templateUrl: 'partials/_about.html',
        controller: 'aboutCtrl'
      }).

      // About List
      state('about.list', {
        url: '/list',
        controller: 'aboutCtrl',
        templateUrl: 'partials/_about.list.html',
        views: {
          'list': { templateUrl: 'partials/_about.list.html' }
        }
      });

  }]

);

Answer 1

您可以使用诸如isState或includedByState之类的过滤器。

ng-if="'about.list' | isState" // exactly 'about.list'

ng-if="'about' | includedByState" // works for about and its children about.*