// Try this way

class App extends Component{

  state = {
     isActive:false
  }

  showHandler = ()=>{
      this.setState({
          isActive: true
      })
  }

  hideHandler = () =>{
      this.setState({
          isActive: false
      })
  }

   render(){
       return(
           <div>
           {this.state.isActive ? <h1>Hello React jS</h1> : null }
             <button onClick={this.showHandler}>Show</button>
             <button onClick={this.hideHandler}>Hide</button>
           </div>
       )
   }
}

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)

class Toggle extends React.Component {
  state = {
    show: true,
  }

  render() {
    const {show} = this.state;
    return (
      <div>
        <button onClick={()=> this.setState({show: !show })}>
          toggle: {show ? 'show' : 'hide'}
        </button>    
        {show && <div>Hi there</div>}
      </div>
     );
  }
}

下面是三元操作符的另一种语法:

{ this.state.showMyComponent ? <MyComponent /> : null }

等价于:

{ this.state.showMyComponent && <MyComponent /> }

学习的原因

还有显示的替代语法:'none';

<MyComponent style={this.state.showMyComponent ? {} : { display: 'none' }} />

然而，如果你过度使用display: 'none'，这将导致DOM污染，并最终降低应用程序的速度。

在react中隐藏和显示元素是非常简单的。

有很多种方法，但我只展示两种。

方式1:

const [isVisible, setVisible] = useState(false)

let onHideShowClick = () =>{
    setVisible(!isVisible)
}

return (<div> 
        <Button onClick={onHideShowClick} >Hide/Show</Button>
         {(isVisible) ? <p>Hello World</p> : ""}
</div>)

方式2:

const [isVisible, setVisible] = useState(false)

let onHideShowClick = () =>{
    setVisible(!isVisible)
}

return (<div> 
        <Button onClick={onHideShowClick} >Hide/Show</Button>
        <p style={{display: (isVisible) ? 'block' : 'none'}}>Hello World</p>
</div>)

它就像if和else一样工作。

在方法一中，它将删除并重新渲染Dom中的元素。 在第二种方式中，你只是将元素显示为false或true。

谢谢你！

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>