当在 ReactJs React-Redux 中仅创建或更新列表中的一个项目时,如何停止重新渲染整个项目列表?
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【中文标题】当在 ReactJs React-Redux 中仅创建或更新列表中的一个项目时,如何停止重新渲染整个项目列表?【英文标题】:How to stop re-rendering a whole list of items when only one item of the list is created or updated in ReactJs React-Redux? 【发布时间】:2021-03-20 19:36:47 【问题描述】:我正在制作这个 Web 应用程序,它有帖子,用户可以在这些帖子中给出答案。我使用 React-Redux 来管理应用程序的状态。每次我创建或更新特定帖子的答案时,属于该帖子的整个答案列表都会重新呈现,我想停止它并仅呈现新创建或更新的答案。我对 post cmets 使用了完全相同的方式,并且效果很好。评论不会重新呈现,但答案会。我只是无法弄清楚这里有什么问题。请参考下面的代码。
我也尝试使用React.memo()
,但它也不起作用!
应答渲染组件,
export function Answer()
const classes = useStyles();
const dispatch = useDispatch();
const postId = useParams();
const postAnswers = useSelector(state => state.Answers);
const [answers, setAnswers] = React.useState(postAnswers.answers);
React.useEffect(() =>
if(postAnswers.status === 'idle') dispatch(fetchAnswers(postId));
, [dispatch]);
React.useEffect(() =>
if(postAnswers.answers) handleAnswers(postAnswers.answers);
, [postAnswers]);
const handleAnswers = (answers) =>
setAnswers(answers);
;
const AnswersList = answers ? answers.map(item =>
const displayContent = item.answerContent;
return(
<Grid item key=item.id>
<Grid container direction="column">
<Grid item>
<Paper component="form" className=classes.root elevation=0 variant="outlined" >
<div className=classes.input>
<Typography>displayContent</Typography>
</div>
</Paper>
</Grid>
</Grid>
</Grid>
);
): undefined;
return(
<Grid container direction="column" spacing=2>
<Grid item>
<Divider/>
</Grid>
<Grid item>
<Grid container direction="column" alignItems="flex-start" justify="center" spacing=2>
AnswersList
</Grid>
</Grid>
<Grid item>
<Divider/>
</Grid>
</Grid>
);
获取答案 redux 应用,
export const fetchAnswers = (postId) => (dispatch) =>
dispatch(answersLoading());
axios.get(baseUrl + `/answer_api/?postBelong=$postId`)
.then(answers =>
dispatch(addAnswers(answers.data))
)
.catch(error =>
console.log(error);
dispatch(answersFailed(error));
);
发布答案,
export const postAnswer = (data) => (dispatch) =>
axios.post(baseUrl + `/answer_api/answer/create/`,
data
)
.then(response =>
console.log(response);
dispatch(fetchAnswers(postBelong)); //This is the way that I update answers state every time a new answer is created or updated
)
.catch(error =>
console.log(error);
);
任何帮助都会很棒。谢谢!
【问题讨论】:
【参考方案1】:添加项目后,您会从 api 获取所有项目,以便在状态下重新创建所有项目。如果您为容器组件提供项目的 id 并让选择器将项目作为 JSON 获取,然后解析回对象,您可以记住它并防止重新渲染,但我认为重新渲染可能会更好。
以下是该项目的记忆 JSON 示例:
const Provider, useDispatch, useSelector = ReactRedux;
const createStore, applyMiddleware, compose = Redux;
const createSelector = Reselect;
const fakeApi = (() =>
const id = ((num) => () => ++num)(1);
const items = [ id: 1 ];
const addItem = () =>
Promise.resolve().then(() =>
items.push(
id: id(),
)
);
const updateFirst = () =>
Promise.resolve().then(() =>
items[0] = ...items[0], updated: id() ;
);
const getItems = () =>
//this is what getting all the items from api
// would do, it re creates all the items
Promise.resolve(JSON.parse(JSON.stringify(items)));
return
addItem,
getItems,
updateFirst,
;
)();
const initialState =
items: [],
;
//action types
const GET_ITEMS_SUCCESS = 'GET_ITEMS_SUCCESS';
//action creators
const getItemsSuccess = (items) => (
type: GET_ITEMS_SUCCESS,
payload: items,
);
const getItems = () => (dispatch) =>
fakeApi
.getItems()
.then((items) => dispatch(getItemsSuccess(items)));
const update = () => (dispatch) =>
fakeApi.updateFirst().then(() => getItems()(dispatch));
const addItem = () => (dispatch) =>
fakeApi.addItem().then(() => getItems()(dispatch));
const reducer = (state, type, payload ) =>
if (type === GET_ITEMS_SUCCESS)
return ...state, items: payload ;
return state;
;
//selectors
const selectItems = (state) => state.items;
const selectItemById = createSelector(
[selectItems, (_, id) => id],
(items, id) => items.find((item) => item.id === id)
);
const createSelectItemAsJSON = (id) =>
createSelector(
[(state) => selectItemById(state, id)],
//return the item as primitive (string)
(item) => JSON.stringify(item)
);
const createSelectItemById = (id) =>
createSelector(
[createSelectItemAsJSON(id)],
//return the json item as object
(item) => JSON.parse(item)
);
//creating store with redux dev tools
const composeEnhancers =
window.__REDUX_DEVTOOLS_EXTENSION_COMPOSE__ || compose;
const store = createStore(
reducer,
initialState,
composeEnhancers(
applyMiddleware(
( dispatch, getState ) => (next) => (action) =>
//simple thunk implementation
typeof action === 'function'
? action(dispatch, getState)
: next(action)
)
)
);
const Item = React.memo(function Item( item )
const rendered = React.useRef(0);
rendered.current++;
return (
<li>
rendered:rendered.current times, item:' '
JSON.stringify(item)
</li>
);
);
const ItemContainer = ( id ) =>
const selectItem = React.useMemo(
() => createSelectItemById(id),
[id]
);
const item = useSelector(selectItem);
return <Item item=item />;
;
const ItemList = () =>
const items = useSelector(selectItems);
return (
<ul>
items.map(( id ) => (
<ItemContainer key=id id=id />
))
</ul>
);
;
const App = () =>
const dispatch = useDispatch();
React.useEffect(() => dispatch(getItems()), [dispatch]);
return (
<div>
<button onClick=() => dispatch(addItem())>
add item
</button>
<button onClick=() => dispatch(update())>
update first item
</button>
<ItemList />
</div>
);
;
ReactDOM.render(
<Provider store=store>
<App />
</Provider>,
document.getElementById('root')
);
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.8.4/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.8.4/umd/react-dom.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/redux/4.0.5/redux.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-redux/7.2.0/react-redux.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/reselect/4.0.0/reselect.min.js"></script>
<div id="root"></div>
【讨论】:
非常感谢您的回答,为什么您认为每次创建或更新一个答案时重新呈现答案列表会更好?会不会影响申请质量?还是有其他方法可以做到这一点? @janithahn 如果您不进行大量渲染,那么它会使代码更简单。但是,如果您发现性能问题,您可以尝试我的答案中使用的记忆选择器。 啊,好吧。知道了!非常感谢!【参考方案2】:我刚刚发现导致上述问题的问题出在哪里。
在我的状态管理系统中,有一个名为 answers
的操作来处理帖子答案的状态,如下所示。
import * as ActionTypes from '../ActionTypes';
export const Answers = (state =
status: 'idle',
errMess: null,
answers: []
, action) =>
switch(action.type)
case ActionTypes.ADD_ANSWER_LIST:
return ...state, status: 'succeeded', errMess: null, answers: action.payload
case ActionTypes.ANSWER_LIST_LOADING:
return ...state, status: 'loading', errMess: null, answers: []
case ActionTypes.ANSWER_LIST_FAILED:
return ...state, status: 'failed', errMess: action.payload, answers: []
default:
return state;
这里的问题是我放在ANSWER_LIST_LOADING
和ANSWER_LIST_FAILED
案例中的空数组。每次动作创建者获取新数据时,它都会通过loading
状态并在那里获得一个空数组,该数组导致整个答案列表被重新渲染和重新创建,这是不必要的。所以我改变了实现如下,它解决了问题。
export const Answers = (state =
status: 'idle',
errMess: null,
answers: []
, action) =>
switch(action.type)
case ActionTypes.ADD_ANSWER_LIST:
return ...state, status: 'succeeded', errMess: null, answers: action.payload
case ActionTypes.ANSWER_LIST_LOADING:
return ...state, status: 'loading', errMess: null, answers: [...state.answers]
case ActionTypes.ANSWER_LIST_FAILED:
return ...state, status: 'failed', errMess: action.payload, answers: [...state.answers]
default:
return state;
问题一直出在我从未想过的地方。在我的问题中,我什至没有提到这个action
。但是你去吧。
【讨论】:
天啊...你帮我省了很多痛苦... :D 很高兴为您提供帮助!以上是关于当在 ReactJs React-Redux 中仅创建或更新列表中的一个项目时,如何停止重新渲染整个项目列表?的主要内容,如果未能解决你的问题,请参考以下文章
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