样式化组件 - 基于 props 有条件地渲染整个 css 块

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 DRY - 我如何（是否可能）基于 props渲染整个css 样式块？

就像是：

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一个css块：

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)}
`;

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'));