在与许多方面合作时,我曾试图把我的头.倒。 目前,有两种办法储存多笔回馈。
射线
[useRef(null), useRef(null), ...]
或目前阵列。
useRef([])
我在阅读了这套书,但就这个专题做了一些研究。 两者都奏效,但应当使用什么? 也许后一种选择较好,因为一般规则是避免过度使用?
在与许多方面合作时,我曾试图把我的头.倒。 目前,有两种办法储存多笔回馈。
射线
[useRef(null), useRef(null), ...]
或目前阵列。
useRef([])
我在阅读了这套书,但就这个专题做了一些研究。 两者都奏效,但应当使用什么? 也许后一种选择较好,因为一般规则是避免过度使用?
我处理许多问题的方法,请你参考:
const self = useRef({}).current
const saveRef = key => r => { self[key] = r }
return (
<div ref={saveRef("container")}>
<input ref={saveRef("input")} />
</div>
)
正如你所说的那样,我不认为对此有明确答案。 换文中提到了一种可变的价值,因此,我猜测这取决于你认为 mu价值是什么。 如果您的阵列出现突变,那么在使用Ref中储存阵列是有意义的。 如果您的数值清单有所变动,那么每值就会产生反射。 然而,我认为这确实是个人的偏好。
一种谨慎的词;对 re的 value价值进行更新,不会迫使重新投标。 我假定你知道这一点,因为你已清楚地读到。 The docs recent, but consider I would only calls out.
EDIT:同样,正如你所说的那样,你应当避免过度使用,如果你需要一整座大楼,你可能想考虑重新计算,以减少每部分的回馈数量。 我非常关心你需要这么多回想的用词。
同样,
const refs = useRef(Array.from({ length: numberOfElements }, () => React.createRef()));
// Example of accessing the first ref
const firstRef = refs.current[0];
采用这种办法,所有回馈都储存在<代码>ref. 现行财产内的一个单阵列中。 当你想一提管理所有<代码>refs时,这是有用的。
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