在onClick函数中调用更新函数时，{{State}}不重绘useState

2024-05-04 • 问答

我花了几天时间浏览类似的帖子，但没有解决方案。由于某些原因，当我使用setPostState(myState.posts);时，它不会重新呈现该组件。

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,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！

just483 回答：在onClick函数中调用更新函数时，{{State}}不重绘useState

在React中，组件仅在state发生更改时重新渲染，即prevState！== currentState无论是基于类的组件还是功能组件。在您的情况下，您正在调用setPosts，但它不会更改状态，因为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检查，而只是比较状态下对象的引用。对于您而言，引用不会随着您的突变而发生变化，并且在调用setPosts之后prevState保持等于newState。

为了避免此问题，在使用对象/数组反应设置状态时，需要确保分配新的引用。因此，比较prevState和currState会返回false。有关更多详细信息，请参见相等性检查示例

正确的访问方式和设置状态：

// 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)

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在onClick函数中调用更新函数时，{{State}}不重绘useState

just483 回答：在onClick函数中调用更新函数时，{{State}}不重绘useState

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