如何在React中从启动切换菜单中制作移动按钮？

npx create-react-app my-app
cd my-app

import React, { useState } from 'react';

const MenuButton = () => {
  const [isOpen, setIsOpen] = useState(false);

  const toggleMenu = () => {
    setIsOpen(!isOpen);
  };

  return (
    <button onClick={toggleMenu}>
      {isOpen ? '关闭菜单' : '打开菜单'}
    </button>
  );
};

export default MenuButton;

import React from 'react';
import MenuButton from './MenuButton';

const App = () => {
  return (
    <div>
      <h1>我的应用</h1>
      <MenuButton />
      {/* 其他应用内容 */}
    </div>
  );
};

export default App;

npm start