var Search = React.createClass({ getInitialState: function() { return { showResults: false }; }, onClick: function() { this.setState({ showResults: true }); }, render: function() { return ( <div> <input type="checkbox" value="Search" onClick={this.onClick} /> { this.state.showResults ? <Results /> : null } </div> ); } }); var Results = React.createClass({ render: function() { return ( <div id="results" className="search-results"> <input type="text" /> </div> ); } }); ReactDOM.render( <Search /> , document.getElementById('container')); <script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.6.2/react.min.js"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/15.6.2/react-dom.min.js"></script> <div id="container"> <!-- This element's contents will be replaced with your component. --> </div>

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:)

我创建了一个小组件来处理这个问题:react-toggle-display

它根据隐藏或显示道具将样式属性设置为display: none !important。

使用示例:

var ToggleDisplay = require('react-toggle-display');

var Search = React.createClass({
    getInitialState: function() {
        return { showResults: false };
    },
    onClick: function() {
        this.setState({ showResults: true });
    },
    render: function() {
        return (
            <div>
                <input type="submit" value="Search" onClick={this.onClick} />
                <ToggleDisplay show={this.state.showResults}>
                    <Results />
                </ToggleDisplay>
            </div>
        );
    }
});

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

React.renderComponent(<Search />, document.body);

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)

在最新的react 0.11版本中，你也可以只返回null，不呈现任何内容。

渲染为空

你必须在代码中做一些小的改变来持续隐藏和显示

const onClick = () => {setShowResults(!showResults}

问题会得到解决

const Search = () => {
  const [showResults, setShowResults] = React.useState(false)
  const onClick = () => setShowResults(true)
  const onClick = () => {setShowResults(!showResults}
  return (
    <div>
      <input type="submit" value="Search" onClick={onClick} />
      { showResults ? <Results /> : null }
    </div>
  )
}

const Results = () => (
  <div id="results" className="search-results">
    Some Results
  </div>
)

ReactDOM.render(<Search />, document.querySelector("#container"))

<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.13.1/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.13.1/umd/react-dom.production.min.js"></script>

<div id="container">
  <!-- This element's contents will be replaced with your component. -->
</div>
```