Question

在我的食谱应用中，当用户单击特定食谱时，它将在组件 recipeById 中呈现所单击食谱的所有详细信息。当我导航回到登录页面并在我的UI中选择另一个食谱时；它首先呈现先前选择的配方的数据，然后重新呈现新选择的配方的数据，我该怎么做以防止出现这种情况？

recipeReducer.js

import { GET_RECIPES, GET_RECIPE_BY_ID} from "../actions/types.js"; 
const initialState = { 
      recipes: [],
      recipe:{}
};

export default function (state = initialState, action) { 
    switch(action.type) {
        case GET_RECIPES:
             return {
                     ...state,  
                     recipes:action.payload
                    }; 
        case GET_RECIPE_BY_ID:
             return {
                     ...state,  
                     recipe:action.payload
                    }; 
        default:
             return state; 
}

recipeActions.js

import { GET_RECIPES, GET_RECIPE_BY_ID} from "./types.js"; 
import axios from 'axios'; 

export const getRecipes = () =>async dispatch => { ... }
export const getRecipeById = (id) =>async dispatch => { 
     const res = await axios.get(`/api/recipe/${id})
     dispatch({
           type:GET_RECIPE_BY_ID,
           payload: res.data
     }); 
}

recipeById.js

import React, {component} from 'react'; 
import {connect} from 'react-redux'; 
import {getRecipeById} from '../../actions/recipeActions.js'; 
import RecipeCard from './RecipeCard'; 

class RecipeById extends Component {
    constructor(props) {
       super(props); 
     } 

  componentDidMount = async() => {
    this.props.getRecipeById(this.props.match.params.id);    
  }

  render() {
     return(
         <RecipeCard 
           title={recipe.title}
           description={recipe.description}
           image= {recipe.image}
          />
     )
  }
}
const mapStateToProps = state => ({
       recipes: state.recipe.recipes,
       recipe: state.recipe.recipe
}); 
export default connect(mapStateToProps, {getRecipeById})(RecipeById);

Answer 1

您需要先清除数据，然后再卸载该组件。

为此，请创建另一个操作（例如：clearData）

然后，在您的RecipeDetail组件中，添加具有声明的操作的componentWillUnmount()：

componentWillUnmount() { 
  this.props.clearData();
}

在减速器中：

case CLEAR_DATA: 
  return {
    ...state,
    recipe: {}
  }

因此，在导航回列表页面之前，将清除详细信息页面中的数据。

Answer 2

什么解决了我的问题：不传递整个状态（即... state），而不传递组件所需的条件。...

import { GET_RECIPES, GET_RECIPE_BY_ID} from "../actions/types.js"; 
const initialState = { 
      recipes: [],
      recipe:{}
};

export default function (state = initialState, action) { 
    switch(action.type) {
        case GET_RECIPES:
             return {
                      //...state
                     recipes:action.payload
                    }; 
        case GET_RECIPE_BY_ID:
             return {
                     //...state
                     recipe:action.payload
                    }; 
        default:
             return state; 
}

如何清除状态，以使先前状态不会在当前组件中呈现？

2 个答案: