Question

状态突变的Vuex方法如下：

const store = new Vuex.Store({
  state: {
    fetching: false,
    board: null
  },
  mutations: {
    setFething (state) {
      state.fetching = true
    },
    setCurrentBoard (state, board) {
       state.board = board
       state.fetching = false
    }
  }
})

但是我担心它会独立触发board和fetching的两次更改而不是一次，我的视图会针对每个属性更新两次。这只是一个简单的例子，我有更复杂的属性突变，最好通过一个突变进行突变。是否可以在vuex？

我喜欢用redux方法返回状态对象只变一次：

initialState = { board: null, fetching: false };
export default function reducer(state = initialState, action = {}) {
  switch (action.type) {
    case Constants.SET_FETCHING:
      return { ...state, fetching: true };

    case Constants.SET_CURRENT_BOARD:
      return { ...state, ...action.board, fetching: false };
 }

Answer 1

那么，你在找这样的东西吗？

＆＃13;

<div id="app">
  {{ $store.state.id }}
  <br>
  {{ $store.state.name }}
  <br>
  {{ $store.state.last }}
</div>

<script src="https://unpkg.com/vue"></script>
<script src="https://unpkg.com/vuex"></script>

＆＃13;

a-sky

＆＃13;

vuex如何在redux中改变商店中的多个属性？

1 个答案: