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reactjs - 在 onClick 函数内调用更新函数时 useState 不会重新渲染

转载 作者:行者123 更新时间:2023-12-03 14:11:06 25 4
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我花了几天时间查看类似的帖子,但没有解决方案。由于某种原因,当我使用 setPostState(myState.posts); 时,它不会重新渲染组件。

我正在使用 react ^16.10.2

下面是我的代码:

import React, {useState, useCallback} from 'react';
import {withStyles, makeStyles} from '@material-ui/core/styles';
import {Paper, TableRow, TableHead, TableCell, TableBody, Table, Badge, Fab} from '@material-ui/core'
import {myState} from '../../PubSub/pub-sub'

import ThumbUpIcon from '@material-ui/icons/ThumbUp';
import ThumbDownIcon from '@material-ui/icons/ThumbDown';

const StyledTableCell = withStyles(...))(TableCell);

const StyledTableRow = withStyles(...))(TableRow);

const useStyles = makeStyles(theme => (...));

export default props => {
console.log("++++++++++++++++Render Body+++++++++++++++++++++");
const classes = useStyles();
let [postState, setPostState] = useState(myState.posts);// why does setPostState not update badge count???? or re-render component???

let upVote = (id) => {
let objIndex = myState.posts.findIndex((obj => obj.id == id));
return (
<Fab key={"upVote4309lk" +id} color="primary" aria-label="add" className={classes.fab}
onClick={() => {
myState.posts[objIndex].up_vote++;
setPostState(myState.posts);//why does this not update badge count???? or re-render component???
}}>
<Badge key={"Ubadge"+objIndex} className={classes.margin} badgeContent={postState[objIndex].up_vote} color="primary"><
ThumbUpIcon> </ThumbUpIcon>
</Badge>
</Fab>
)
};

let downVote = (id) => {
let objIndex = myState.posts.findIndex((obj => obj.id == id));
return (
<Fab key={"downVote0940v" + id} color="primary" aria-label="add" className={classes.fab}
onClick={() => {
myState.posts[objIndex].down_vote++;
setPostState(myState.posts);//why does this not update badge count???? or re-render component???
}}>
<Badge className={classes.margin} badgeContent={myState.posts[objIndex].down_vote} color="primary"><
ThumbDownIcon> </ThumbDownIcon>
</Badge>
</Fab>
)
};

function filter(name) {
return name.toLowerCase().includes(props.searchData.title.toLowerCase());
}

function createData(title, description, user, up_votes, down_votes, id) {
if (filter(title, description, user, up_votes, down_votes)) {
return (
<StyledTableRow key={id + "tableKey"}>
<StyledTableCell>{title}</StyledTableCell>
< StyledTableCell>{description}</StyledTableCell>
<StyledTableCell>{user}</StyledTableCell>
<StyledTableCell>{upVote(id)}</StyledTableCell>
<StyledTableCell>{downVote(id)}</StyledTableCell>
</StyledTableRow>
)
}
}

const rows = myState.posts.map(
obj => createData(obj.title, obj.description, obj.user, obj.up_votes, obj.down_votes, obj.id)
);

return (
<Paper className={classes.root}>
<Table className={classes.table} aria-label="customized table">
<TableHead>
<TableRow>
<StyledTableCell>Title</StyledTableCell>
<StyledTableCell>Description</StyledTableCell>
<StyledTableCell>User</StyledTableCell>
<StyledTableCell>Up Votes</StyledTableCell>
<StyledTableCell>Down Votes</StyledTableCell>
</TableRow>
</TableHead>
<TableBody>
{rows.map(row => (row))}
</TableBody>
</Table>
</Paper>
);
}

任何帮助都会很棒,谢谢!

最佳答案

在 React 中,组件仅在 state 发生变化时重新渲染,即 prevState !== currentState 无论是基于类的还是函数式的成分。在您的情况下,您正在调用 setPosts 但它不会更改状态,因为您在设置状态时分配了相同的对象 myState.posts 。 React 不会对对象执行深度相等性检查,而只是比较处于某种状态的对象的引用。在您的情况下,当您发生变异时,引用永远不会改变,并且在调用 setPosts 后 prevState 保持等于 newState

为了避免这个问题,在 React 中使用对象/数组设置状态时,您需要确保分配一个新的引用。因此,比较 prevStatecurrState 返回 false。有关更多详细信息,请参阅相等性检查示例

访问和设置状态的正确方法:

// Set the initial value using myState.posts and then use the variable
// postState to access the posts and not myState.posts
const [postState, setPostState] = useState(myState.posts)

const makeUpVote = (objIndex) => {
// Make local variable posts to change and set posts while
// using spread operator to make sure we get a new array created instead
// of pointing to the same array in memory
const posts = [...postState]
posts[objIndex].up_vote++
setPostState(posts)
}


let upVote = id => {
// use postState to access instead of myState.posts
let objIndex = postState.findIndex(obj => obj.id == id)
return (
<Fab
// I would recommend creating separate functions to handle this
// instead of writing them inline.
onClick={() => makeUpVote(objIndex)}
></Fab>
)
}

平等检查示例:

// This snippet is just to give you an idea of mutation
const posts = [{id: 1, upvote: 0}, {id: 2, upvote: 0}]
const posts2 = posts

// using spread operator from ES6 to assign a new array with similar values to posts3
const posts3 = [...posts]
posts[0].upvote++
posts3[0].upvote++

// This statement will return true because posts and posts2 have
// the same address in memory (reference) even though we just
// changed posts variable.
// If we set posts2 in state and initial state was posts
// component will NOT re-render
console.log(posts === posts2)

// This will return false because we assigned a new object
// to posts3 using spread operator even though values are same
// If we set posts3 in state an initial state was posts
// component will re-render
console.log(posts === posts3)

// Now another thing to notice is that spread operator does not
// perform deep cloning and therefore the object at index 0 in
// posts has the same reference to object at index 0 in posts 3
// therefore we get upvote = 2
console.log("Posts: ", posts)
console.log("Posts3: ", posts3)

关于reactjs - 在 onClick 函数内调用更新函数时 useState 不会重新渲染,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58784464/

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