本文介绍了如何在 React.memo 中使用带有泛型的 Props的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在尝试将以下内容转换为使用 React.memo:

I am trying to convert the following to use React.memo:

interface Props<TRowData> {
  // props...
}

export function Table<TRowData>({
  propA,
  propB
}: Props<TRowData>) {

}

像这样(不正确):

interface Props<TRowData> {
  // props...
}



export const Table = memo<Props<TRowData>>(
({
  propA,
  propB
}) => {

})

如何更正此语法?目前它有这个错误:

How can I correct this syntax? Currently it has this error:

// Cannot find name 'TRowData'.
export const Table = memo<Props<TRowData>>(
                                ~~~~~~~~

推荐答案

使用当前的 React 类型声明,不可能从 React.memo 创建通用组件.没有类型断言的解决方案是添加一个额外的 memo 函数重载以利用 TS 3.4 高阶函数类型推断:

With current React type declarations, it is not possible to create a generic component out of React.memo. A solution without type assertions is to add an additional memo function overload to leverage TS 3.4 higher order function type inference:

import React, { memo } from "react"

declare module "react" { // augment React types
  function memo<A, B>(Component: (props: A) => B): (props: A) => ReactElement | null
  // return type is same as ReturnType<ExoticComponent<any>>
}

然后您将能够使 Table 组件通用.只需确保将通用函数传递给 memo:

You then will be able to make Table component generic. Just make sure to pass a generic function to memo:

interface Props<T> {
  a: T
}

const TableWrapped = <T extends {}>(props: Props<T>) => <div>{props.a}</div>

const Table = memo(TableWrapped)

const App = () => (
  <>
    <Table a="foo" /> {/* (props: Props<string>) => ... */}
    <Table a={3} /> {/* (props: Props<number>) => ... */}
  </>
)

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09-17 21:22