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javascript - 为什么我的代码返回 React Type Error?

转载 作者:行者123 更新时间:2023-11-28 04:23:07 24 4
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当我将 props 传递给子元素时,出现以下错误:

TS2322: Type '{}' is not assignable to type 'IntrinsicAttributes & FooterRightSideProps & { children?: ReactNode; }'.
Type '{}' is not assignable to type 'FooterRightSideProps'.
Property 'onClickCreate' is missing in type '{}'.

我的代码如下:

import React from "react";

import CSSModules from "react-css-modules";

import styles from "./Footer.module.sass";

import { Icon } from "@components/icon/Icon";
import { Link } from "@components/typography/link";
import { Button } from "@components/button/Button";

export interface FooterProps {
}

export const Footer: React.SFC<FooterProps> =
CSSModules(styles)
(
(props: FooterProps) =>
<div styleName="footer">
<FooterLeftSide />
<FooterRightSide { ...props } /> //an error occurs here
</div>
);

export const FooterLeftSide: React.SFC =
CSSModules(styles)(
() =>
<div styleName="footer-left-side"></div>
);

export interface FooterRightSideProps {
onClickAbandon?: () => void;
onClickCreate: () => void;
}

export const FooterRightSide: React.SFC<FooterRightSideProps> =
CSSModules(styles)
(
(props: FooterRightSideProps) =>
<div styleName="footer-right-side">
<Link
className="option-back"
styleName="header-option"
onClick={props.onClickAbandon}
>
<div styleName="icon-with-label">
<Icon name="left" />
</div>
Abandon
</Link>
<Button
onClick={props.onClickCreate}
theme="primary-white"
>
Create Profile
</Button>
</div>
);

你们中有谁知道如何将这个 onClickCreate 属性传递给嵌套在父元素中的子元素吗?

最佳答案

您传递给 FooterRightSide

props 的类型为 FooterProps,但此组件需要满足 FooterRightSideProps 的 props >.

关于javascript - 为什么我的代码返回 React Type Error?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45288351/

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