javascript - 样式化组件 - 根据 props 有条件地渲染整个 css block

Question

我知道样式可以有条件地呈现，例如:

const HelloWorldLabel= styled("div")<{ centered?: boolean }>`
  display: ${({ centered }) => (centered ? "block" : "flex")};;
  margin: ${({ centered }) => (centered ? "auto 0" : "unset")};
  padding: ${({ centered }) => (centered ? "0 15px" : "unset")};
`;

这看起来并不枯燥 - 我如何(是否可能)基于 props 渲染整个 block css 样式？

类似于:

const HelloWorldLabel= styled("div")<{ centered?: boolean }>`
   if (centered) {
    display: "block" ;
    margin: $"auto 0";
     padding: "0 15px" ;
   } else {
     ......
   }
`;

Answer 1

使用 styled-component 或任何 CSS-in-JS，您可以有条件地渲染 css block :

import styled, { css } from 'styled-components';

const light = css`
  background-color: white;
  color: black;
`;

const dark = css`
  background-color: black;
  color: white;
`;

const Box = styled.div`
  ${({ isDark }) => (isDark ? light : dark)}
`;

<小时/>

完整示例:

import React, { useState } from 'react';
import ReactDOM from 'react-dom';
import styled, { css } from 'styled-components';

const light = css`
  background-color: white;
  border: 2px solid black;
  color: black;
`;

const dark = css`
  background-color: black;
  color: white;
`;

const FlexBox = styled.div`
  margin: 20px;
  padding: 20px;
  ${({ isDark }) => (isDark ? light : dark)}
`;

const App = () => {
  const [isDark, setIsDark] = useState(false);

  const toggle = () => setIsDark(b => !b);

  return (
    <FlexBox isDark={isDark}>
      <div>Some Text</div>
      <button onClick={toggle}>Change Block</button>
    </FlexBox>
  );
};

ReactDOM.render(<App />, document.getElementById('root'));