以下是我的方法。

import React, { useState } from 'react';

function ToggleBox({ title, children }) {
  const [isOpened, setIsOpened] = useState(false);

  function toggle() {
    setIsOpened(wasOpened => !wasOpened);
  }

  return (
    <div className="box">
      <div className="boxTitle" onClick={toggle}>
        {title}
      </div>
      {isOpened && (
        <div className="boxContent">
          {children}
        </div>
      )}
    </div>
  );
}

在上面的代码中，为了实现这一点，我使用了如下代码:

{opened && <SomeElement />}

仅当opened为true时才会呈现SomeElement。它的工作原理在于JavaScript解析逻辑条件的方式:

true && true && 2; // will output 2
true && false && 2; // will output false
true && 'some string'; // will output 'some string'
opened && <SomeElement />; // will output SomeElement if `opened` is true, will output false otherwise (and false will be ignored by react during rendering)
// be careful with 'falsy' values eg
const someValue = [];
someValue.length && <SomeElement /> // will output 0, which will be rednered by react
// it'll be better to:
someValue.length > 0 && <SomeElement /> // will render nothing as we cast the value to boolean

使用这种方法而不是CSS“display: none”的原因;

While it might be 'cheaper' to hide an element with CSS - in such case 'hidden' element is still 'alive' in react world (which might make it actually way more expensive) it means that if props of the parent element (eg. <TabView>) will change - even if you see only one tab, all 5 tabs will get re-rendered the hidden element might still have some lifecycle methods running - eg. it might fetch some data from the server after every update even tho it's not visible the hidden element might crash the app if it'll receive incorrect data. It might happen as you can 'forget' about invisible nodes when updating the state you might by mistake set wrong 'display' style when making element visible - eg. some div is 'display: flex' by default, but you'll set 'display: block' by mistake with display: invisible ? 'block' : 'none' which might break the layout using someBoolean && <SomeNode /> is very simple to understand and reason about, especially if your logic related to displaying something or not gets complex in many cases, you want to 'reset' element state when it re-appears. eg. you might have a slider that you want to set to initial position every time it's shown. (if that's desired behavior to keep previous element state, even if it's hidden, which IMO is rare - I'd indeed consider using CSS if remembering this state in a different way would be complicated)

我从React团队的声明开始:

在React中，你可以创建封装行为的不同组件 你所需要的。然后，您只能渲染其中的一些，这取决于 您的申请状态。 条件渲染在React中的工作方式与条件的工作方式相同 JavaScript。使用JavaScript操作符，如if或条件 运算符来创建表示当前状态的元素，并让 React更新UI以匹配它们。

基本上需要显示组件的按钮被点击时,你可以两种方式,使用纯粹的反应或使用CSS,使用纯的反应方式,你可以做类似下面的代码在你的情况下,所以在第一次运行,结果并不像hideResults显示是正确的,但通过单击按钮,状态会改变和hideResults是假和组件再次呈现新的价值条件下,这是很常见的使用改变组件视图的反应……

var Search = React.createClass({
  getInitialState: function() {
    return { hideResults: true };
  },

  handleClick: function() {
    this.setState({ hideResults: false });
  },

  render: function() {
    return (
      <div>
        <input type="submit" value="Search" onClick={this.handleClick} />
        { !this.state.hideResults && <Results /> }
      </div> );
  }

});

var Results = React.createClass({
  render: function() {
    return (
    <div id="results" className="search-results">
      Some Results
    </div>);
   }
});

ReactDOM.render(<Search />, document.body);

如果你想进一步研究React中的条件渲染，可以看看这里。

下面是一个简单，有效和最佳的解决方案，使用无类React组件来显示/隐藏元素。React- hooks的使用可以在最新的使用React 16的create-react-app项目中使用

import React, {useState} from 'react';
function RenderPara(){
const [showDetail,setShowDetail] = useState(false);

const handleToggle = () => setShowDetail(!showDetail);

return (
<React.Fragment>
    <h3>
        Hiding some stuffs 
    </h3>
    <button onClick={handleToggle}>Toggle View</button>
   {showDetail && <p>
        There are lot of other stuffs too
    </p>}
</React.Fragment>)

}  
export default RenderPara;

快乐编码:)

根据文档，最佳实践如下:

{this.state.showFooter && <Footer />}

只有在状态有效时才呈现元素。

只要找到一种新的、神奇的方法来使用(useReducer)功能组件

const [state, handleChangeState] = useReducer((state) => !state, false); 改变状态

我能够使用css属性“隐藏”。不知道可能的缺点。

export default function App() {
    const [hidden, setHidden] = useState(false);
    return (
      <div>
        <button onClick={() => setHidden(!hidden)}>HIDE</button>
        <div hidden={hidden}>hidden component</div>
      </div>
    );
  }