Question

我无法弄清楚为什么在调用details方法时以下组件中的fetch()计算属性没有更新：

<template>
  <div>
  {{ haveData }} //remains undefined
  </div>
</template>

<script>
export default {
  props: {
    group: {
      type: Object,
      required: true
    },
  },
  computed: {
    currentGroup() {
      return this.$store.getters['user/navbar_menu_app_current_group'](
        this.group.id
      )

      /*-- which is the following function 
        navbar_menu_app_current_group: state => item => {
        return state.menu.find(m => {
          return m.id == item
        })
        }
      */

      /*-- this function returns an object like so 
        {
          id: 1,
          label: 'kity cats',
        }
        ***details --> IS NOT DEFINED. If I add it to the current group as null, my problem goes away. However, this is a previous API call that does not set the `details` parameter.
      */
    },
    details() {
      let c = this.currentGroup.details
      console.log(c) // returns undefined, which makes sense, but it should be updated after this.fetch() is called
      return c
    },
    haveData() {
      return this.details != null
    }
  },
  methods: {
    async fetch() {
      await this.$store.dispatch(
        'user/navbar_menu_app_details_get',
        this.group.id
      )
      //This is setting the "details" part of the state on menu which is referred to in the computed properties above
      //Previous to this there is no state "this.group.details"
      //If I add a console log to the mutation the action calls, I get what is expected.
    }
  },
  created() {
    if (!this.haveData) {
      this.fetch()
    }
  }
}
</script>

如果我将数组项更改为包含details，它将起作用：

{
  id: 1,
  label: 'kity cats',
  details: null  // <-- added
}

不幸的是，该数组是通过一个大型API调用创建的，添加details似乎是不必要的，因为它可能永远不需要。

如何在不将details:null添加到默认状态的情况下使计算出的属性起作用？

尝试1：

// Vuex mutation
navbar_menu_app_details_set(state, vals) {
  let app = state.menu.find(item => {
    return item.id == vals[0] //-> The group id passing in the dispatch function 
  })

  //option 1 = doesn't work
  app = { app, details: vals[1] } //-> vals[1] = the details fetched from the action (dispatch)

  //option 2 = doesnt work
  app.details = vals[1]

  //option 3 = working but want to avoid using Vue.set()
  import Vue from 'vue' //Done outside the actual function
  Vue.set( app, 'details', vals[1])
},

尝试2：

// Vuex action
navbar_menu_app_details_get(context, id) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      context.commit('navbar_menu_app_details_set', [
        context.getters.navbar_menu_app_current(id), //-> the same as find function in the mutation above
        apps[id]
      ])
      resolve()
    }, 1000)
  })
}

// --> mutation doesn't work
navbar_menu_app_details_set(state, vals) {
  vals[0].details = vals[1]
},

Answer 1

可以通过this._vm通过Vuex突变获得Vue实例，您可以使用vm.$set()（相当于Vue.set()）将details添加到菜单项：< / p>

navbar_menu_app_details_set(state, vals) {
  let app = state.menu.find(item => {
    return item.id == vals[0]
  })

  this._vm.$set(app, 'details', vals[1])
},

Answer 2

Vue 中的所有对象是反应性，并且其设计方式使得仅当重新分配对象时< / strong>，更改将被捕获，并且将进行更改检测。

在您的情况下，跟踪应该很好。

app = { ...app, details: vals[1] }

Vue计算的属性不响应状态更改

2 个答案: