vuejs2 - vuex-router-sync 有什么用？

Question

据我所知vuex-router-sync仅用于同步 route与 vuex store并且开发者可以访问route如下:

store.state.route.path
store.state.route.params

不过，我也可以处理 route来自 this.$route哪个更简洁。

什么时候需要使用store中的路由，需要vuex-router-sync的场景是什么？

Answer 1

这是我的两分钱。您不需要导入 vuex-router-sync如果您无法在您的项目中找出它的用例，但是当您尝试使用 route 时可能需要它。您的 vuex 中的对象的方法( this.$route 在 vuex 的领域中效果不佳)。

我想在这里举一个例子。
假设您想在一个组件中显示一条消息。您想显示类似 Have a nice day, Jack 的消息在几乎每个页面中，除了 Welcome back, Jack应在用户浏览首页时显示。

您可以在 vuex-router-sync 的帮助下轻松实现它.

const Top = {
  template: '<div>{{message}}</div>',
  computed: {
    message() {
      return this.$store.getters.getMessage;
    }
  },
};
const Bar = {
  template: '<div>{{message}}</div>',
  computed: {
    message() {
      return this.$store.getters.getMessage;
    }
  }
};

const routes = [{
    path: '/top',
    component: Top,
    name: 'top'
  },
  {
    path: '/bar',
    component: Bar,
    name: 'bar'
  },
];

const router = new VueRouter({
  routes
});

const store = new Vuex.Store({
  state: {
    username: 'Jack',
    phrases: ['Welcome back', 'Have a nice day'],
  },
  getters: {
    getMessage(state) {
      return state.route.name === 'top' ?
        `${state.phrases[0]}, ${state.username}` :
        `${state.phrases[1]}, ${state.username}`;
    },
  },
});

// sync store and router by using `vuex-router-sync`
sync(store, router);

const app = new Vue({
  router,
  store,
}).$mount('#app');












// vuex-router-sync source code pasted here because no proper cdn service found
function sync(store, router, options) {
  var moduleName = (options || {}).moduleName || 'route'

  store.registerModule(moduleName, {
    namespaced: true,
    state: cloneRoute(router.currentRoute),
    mutations: {
      'ROUTE_CHANGED': function(state, transition) {
        store.state[moduleName] = cloneRoute(transition.to, transition.from)
      }
    }
  })

  var isTimeTraveling = false
  var currentPath

  // sync router on store change
  store.watch(
    function(state) {
      return state[moduleName]
    },
    function(route) {
      if (route.fullPath === currentPath) {
        return
      }
      isTimeTraveling = true
      var methodToUse = currentPath == null ?
        'replace' :
        'push'
      currentPath = route.fullPath
      router[methodToUse](route)
    }, {
      sync: true
    }
  )

  // sync store on router navigation
  router.afterEach(function(to, from) {
    if (isTimeTraveling) {
      isTimeTraveling = false
      return
    }
    currentPath = to.fullPath
    store.commit(moduleName + '/ROUTE_CHANGED', {
      to: to,
      from: from
    })
  })
}

function cloneRoute(to, from) {
  var clone = {
    name: to.name,
    path: to.path,
    hash: to.hash,
    query: to.query,
    params: to.params,
    fullPath: to.fullPath,
    meta: to.meta
  }
  if (from) {
    clone.from = cloneRoute(from)
  }
  return Object.freeze(clone)
}

.router-link-active {
  color: red;
}

<script src="https://unpkg.com/vue/dist/vue.js"></script>
<script src="https://unpkg.com/vue-router/dist/vue-router.js"></script>
<script src="https://unpkg.com/vuex/dist/vuex.js"></script>

<div id="app">
  <p>
    <router-link to="/top">Go to Top</router-link>
    <router-link to="/bar">Go to Bar</router-link>
  </p>
  <router-view></router-view>
</div>

fiddle here

如您所见，组件与 vuex 很好地解耦。和 vue-router的逻辑。
对于您不关心当前路由与 vuex 的 getter 返回值之间的关系的情况，这种模式有时非常有效。