vue页面跳转到新页面之后，再由新页面返回到原页面时候若想返回调出原页面的初始位置，怎么来解决这个问题呢？首先我们应该在跳出页面时候记录下跳出的scrollY，返回原页面的时候在设置返回位置为记录下的scrolly即可，scrolly我用的是vuex状态管理器来保存的。整个环境是基于vue-cli搭建的  

//引用vuex
import Vuex from 'vuex'
Vue.use(Vuex)

var store = new Vuex.Store({
  state: {
    recruitScrollY:0
  },
  getters: {
    recruitScrollY:state => state.recruitScrollY
  },
  mutations: {
    changeRecruitScrollY(state,recruitScrollY) { 
      state.recruitScrollY = recruitScrollY
    }
  },
  actions: {
 
  },
  modules: {}
})

beforeRouteLeave(to, from, next) {
    let position = window.scrollY  //记录离开页面的位置
    if (position == null) position = 0
    this.$store.commit('changeRecruitScrollY', position) //离开路由时把位置存起来
    next()
  },
  watch: {
    '$route' (to, from) {
      if (to.name === 'NewRecruit') {//跳转的的页面的名称是"NewRecruit",这里就相当于我们listview页面，或者原始页面
        let recruitScrollY = this.$store.state.recruitScrollY
        window.scroll(0, recruitScrollY)
      }
    }
  }

<keep-alive v-if="$route.meta.keepAlive">
      <router-view></router-view>
</keep-alive>
<router-view v-if="!$route.meta.keepAlive"></router-view>

Vue.use(Router)
 
const routerApp = new Router({
  routes: [{
    {
      path: '/NewRecruit',
      name: 'NewRecruit',
      component: NewRecruit,
      meta: {
        keepAlive: true
      }
    },
    {
      path: '/NewRecruitDesc/:id',
      name: 'NewRecruitDesc',
      component: NewRecruitDesc,
      meta: {
        keepAlive: true
      }
    },
    {
      path: '/SubmitSucess',
      name: 'SubmitSucess',
      component: SubmitSucess,
      meta: {
        keepAlive: false
      }
    }
  ]
})
 
export default routerApp

(3)在回退页面的时候，如果出现有个滚动条条转的过程，会有一下的界面跳动，可以在路由切换时加个极短时间的动画，来完美错过这个跳动的时间。就用到了官网给出的transition，直接用多个transition包裹，具体代码：

<transition :name="transitionName" mode="out-in">
//...
</transition>

链接: https://fly63.com/article/detial/1078