我在我的应用程序中使用样式组件在CodeSandbox中。请参考以下网址 https://lrn6vmq297.sse.codesandbox.io/
每次我进行一些更改时,控制台都会说。
Warning: Prop `className` did not match.
It looks like you've wrapped styled() around your React component (Component), but the className prop is not being passed down to a child. No styles will be rendered unless className is composed within your React component.
并且UI不能按预期呈现。 有人知道我为什么遇到这个问题吗?请查看上面的网址。
谢谢
答案 0 :(得分:6)
基本上,您需要传递this.props.className
生成的props.className
或className
或经过解构的styled-components
并将其手动应用于要设置样式的组件。否则,您不会将className
应用于任何内容,因此不会看到任何样式更改。
工作示例:
components / LinkComponent.js (此function
接受className
和styled()
生成的props
,这些Link
传递给样式化的组件在下面创建的-您需要将它们手动应用于import React from "react";
import PropTypes from "prop-types";
import { Link } from "react-router-dom";
const LinkComponent = ({ className, children, link }) => (
<Link className={className} to={link}>
{children}
</Link>
);
LinkComponent.propTypes = {
className: PropTypes.string.isRequired,
link: PropTypes.string.isRequired,
children: PropTypes.string.isRequired
};
export default LinkComponent;
组件)
function
components / StyledLink.js (导入上方的styled()
并将其传递给styled()
-您也可以创建一个styled themed来更新{{1 }}元素
import styled from "styled-components";
import LinkComponent from "./LinkComponent";
const StyledLink = styled(LinkComponent)`
color: ${props => (!props.primary && !props.danger ? "#03a9f3" : "#ffffff")};
background-color: ${props => {
if (props.primary) return "#03a9f3";
if (props.danger) return "#f56342";
return "transparent";
}};
font-weight: bold;
margin-right: 20px;
padding: 8px 16px;
transition: all 0.2s ease-in-out;
border-radius: 4px;
border: 2px solid
${props => {
if (props.primary) return "#03a9f3";
if (props.danger) return "#f56342";
return "#03a9f3";
}};
&:hover {
color: ${props => (!props.primary && !props.danger ? "#0f7ae5" : "#ffffff")};
background-color: ${props => {
if (props.primary) return "#0f7ae5";
if (props.danger) return "#be391c";
return "transparent";
}};
text-decoration: none;
border: 2px solid ${props => (props.danger ? "#be391c" : "#0f7ae5")}};
}
`;
export default StyledLink;
components / Header.js (导入上面创建的样式化组件StyledLink
并加以利用-传递给该组件的所有其他道具都将自动传递给function
,但是在这种情况下,您需要解构prop
才能使用它)
import React from "react";
import StyledLink from "./StyledLink";
export default () => (
<nav className="container">
<StyledLink primary link="/">Home</StyledLink>
<StyledLink danger link="/about">About</StyledLink>
<StyledLink link="/portfolio">Portfolio</StyledLink>
</nav>
);
答案 1 :(得分:0)
链接并没有真正起作用(或者我不明白您想显示什么),但是从错误消息看来,您应该像这样传递className
styled(<Component className={your source for classnames} />)
答案 2 :(得分:0)
对于共享组件,最好使用forwardRef,也可以只传递props:
import React from 'react'
import styled from 'styled-components'
function MainComponent() {
return (
<LoadingStyled />
)
})
const LoadingStyled = styled(LoadingComponent)`
margin-top: 40px;
`
import React, { forwardRef } from 'react'
export const LoadingComponent = forwardRef((props, ref) => {
return (
<div {...props}>
I got all props and styles, yeeeee!
</div>
)
})
不带有forwardRef的替代项。
import React from 'react'
export const LoadingComponent = (props) => {
return (
<div {...props}>
I got all props and styles, yeeeee!
</div>
)
}
答案 3 :(得分:0)
我有类似的情况,我需要使用由styled-component创建的组件,并将css属性传递给该组件。希望这会有所帮助!
主要组件(在此处定义CSS属性)
import Wrapper from 'components/Wrapper'
const CustomWrapper = styled(Wrapper)`
&:hover {
background-color: blue; // defining css property I want to pass down
}
`;
...
render() {
return (
... <CustomWrapper /> // using my CustomWrapper component made from 'styled-component'
)
}
`;
Wrapper.js-功能组件(在此处使用定义的CSS)
const Wrapper = props => {
const { className } = props; // see how className is destructed and used below
return (
<div className={className}> // 'className' is used here
{YOUR_CONTENT}
</div>
)
}