如何在反應狀態鈎子上使用map函數

Question

有人可以解釋一下，為什么我不能在React狀態鈎子上運行JavaScript map函數？

const [sequenceNames, setSequenceNames] = useState(null);

useEffect(() => {
  fetch('/savedFiles')
    .then(response => response.json())
    .then(data => setSequenceNames(data));
}, []);

const table = sequenceNames.map(name => Sequence(name));

這適用於for循環，但是我的linter禁止使用in in。

const table = [];

for (const name in sequenceNames) {
  table.push(Sequence(sequenceNames[name]));
}

當我使用.map時，我收到以下錯誤。

TypeError: Cannot read property 'map' of null
    at main.a21158832f7ed8c55e25.bundle.js:1
    at Bi (main.a21158832f7ed8c55e25.bundle.js:1)
    at main.a21158832f7ed8c55e25.bundle.js:1
    at f (main.a21158832f7ed8c55e25.bundle.js:1)
    at d (main.a21158832f7ed8c55e25.bundle.js:1)
    at main.a21158832f7ed8c55e25.bundle.js:1

即使我的sequenceNames數組不應為null。

Answer 1

您遇到的問題是您將sequenceNames的初始狀態設置為null ，因此您遇到了競爭條件，第一次嘗試在sequenceNames運行循環時，它為null 。 你可以將sequenceNames初始化為一個空數組[] ，但實現你想做的事情的最好方法是擁抱鈎子。

import React, { useMemo, useState } from 'react';
import { Sequence } from '.';

export interface DisplayNamesProps {
    id?: number;
}

export function DisplayNames(props: DisplayNamesProps) {
    const { id } = props;
    const [sequenceNames, setSequenceNames] = useState<Sequence[]>([]);

    // Use useEffect with the 2nd param, to guard against `fetch`
    // executing on each render. In this example I use `id` as
    // a var that should be unique for each `fetch` call.
    useEffect(() => {
        fetch(`/savedFiles/${id}`)
            .then(response => response.json())
            .then(data => setSequenceNames(data));
    }, [id]);

    // Create a memoized var `table` that updates when there
    // is a change to `sequenceNames`.
    const table = useMemo(() => {
        return sequenceNames.map(name => Sequence(name))
    }, [sequenceNames]);

    return <div>{table}</div>;
}

Answer 2

更改

    const table = sequenceNames.map(name => Sequence(name));

至

    const table = Array.isArray(sequenceNames) && sequenceNames.map(name => Sequence(name));