通过<a> link

Question

In the example below, I have two functional components that give the idea of being separate pages, though, of course, they only display different colored <div> boxes.

const Home = () =>
    <div style={{background:'red', height: 100, width: 100}}></div>;

const About = () =>
    <div style={{background:'blue', height: 100, width: 100}}></div>;

Now, if I wanted to create a navigation bar that links to certain parts of my site -- ignoring the possible wrong use of the <a> links, but it is to show that I want those to render other parts of the site as a response to clicking on a Navbar option.

const Navbar = () =>
    <div>
        <a href='#'>Home</a>
        <a href='#'>About</a>
    </div>;

If I have a component named App that renders different pages for me:

const App = () =>
    <div>
        <Navbar />
        // Render current page here; default is the `Home` component.
    </div>;

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

How could I actively update my App component to render different pages based off the clicking of links, or elements, in my Navbar?

EDIT: Soviut和finalfreq渲染其他组件，谢谢您的回答！我来自Codecademy的ReactJS课程，所以过渡到Javascript和React的“真实世界”有点困难，因为我必须学习ES6，并且不了解大多数JS库。无论如何，我感谢你们指出了我react-router的方向，并给了我一个如何在没有它的情况下实现我的目标的例子。欢呼声。

Answer 1

这就是react-router的目的。

虽然不推荐手动操作，但它可以通过发出点击事件来调用从包含你的＆＃34;页面＆＃34;的父组件传入的回调函数。该回调将告诉父容器单击了哪个导航链接，它将决定显示哪个页面以及停止渲染哪个页面。

Answer 2

每个页面都没有路由器/控制器，你可以这样做：

const pages = ['array', 'of', 'pages'];

class App extends React.Component {
  constructor(props) {
    super();
    this.state = {
      activePage: pages[0]
    }
  }

  switchPage = (page) => {
    this.setState({
    activePage: page
    })
  }

  renderActivePage = () => {
    switch(this.state.activePage) {
      case 'home': 
        return <HomeComponent />
    }
  }
  render() {

    return (
      <div>
       <Navbar switchPageFunction={this.switchPage}/>
       {renderActivePage()}
       </div>;)
  }

}

然后只需将onclick设置为您的标签，实际上是switchPageFunction并将页面名称作为参数传递。然后，父组件将接收更新，更新其自己的状态，并呈现正确的组件。

我不建议使用这条路线，而是选择仅针对此用例构建的react-router