使用ref和操作CSS

一种方法是使用React的ref和使用浏览器的API操作CSS类。它的好处是避免在React中重新渲染，如果唯一的目的是在单击按钮时隐藏/显示一些DOM元素。

// Parent.jsx
import React, { Component } from 'react'

export default class Parent extends Component {
    constructor () {    
        this.childContainer = React.createRef()
    }

    toggleChild = () => {
        this.childContainer.current.classList.toggle('hidden')
    }

    render () {
        return (
            ...

            <button onClick={this.toggleChild}>Toggle Child</button>
            <div ref={this.childContainer}>
                <SomeChildComponent/>
            </div>

            ...
        );
    }
}


// styles.css
.hidden {
    display: none;
}

PS:如果我说错了请指正。：）

状态和效果的应用程序已经并且必须封装在同一个组件中，因此，没有什么比创建一个自定义组件作为钩子来解决在这种情况下是使特定的块或元素可见还是不可见更好的了。

// hooks/useOnScreen.js

import { useState, useEffect } from "react"

const useOnScreen = (ref, rootMargin = "0px") => {

  const [isVisible, setIsVisible] = useState(false)

  useEffect(() => {
    const observer = new IntersectionObserver(
      ([entry]) => {
        setIsVisible(entry.isIntersecting)
      },
      {
        rootMargin
      }
    );

    const currentElement = ref?.current

    if (currentElement) {
      observer.observe(currentElement)
    }

    return () => {
      observer.unobserve(currentElement)
    }
  }, [])

  return isVisible
}

export default useOnScreen

然后自定义钩子嵌入到组件中

import React, { useRef } from "react";
import useOnScreen from "hooks/useOnScreen";

const MyPage = () => {

  const ref = useRef(null)

  const isVisible = useOnScreen(ref)

  const onClick = () => {
    console.log("isVisible", isVisible)
  }
  
  return (
    <div ref={ref}>
      <p isVisible={isVisible}>
        Something is visible
      </p>
      <a
        href="#"
        onClick={(e) => {
          e.preventDefault();
          onClick(onClick)
        }}
      >
        Review
      </a>
    </div>
  )
}

export default MyPage

由useRef钩子控制的ref变量，允许我们在DOM中捕获我们想要控制的块的位置，然后由useOnScreen钩子控制的isVisible变量，允许我们通过useRef钩子在块内部设置I信号。 我相信useState、useeffect和useRef钩子的实现允许您通过使用自定义钩子将它们分开来避免组件呈现。

希望这些知识对你有用。

已经有几个很好的答案了，但我不认为它们解释得很好，而且给出的一些方法包含一些可能会绊倒人们的陷阱。所以我将回顾三种主要的方法(加上一个跑题的选项)来做到这一点，并解释利弊。我写这篇文章主要是因为选项1被推荐了很多，如果使用不当，这个选项会有很多潜在的问题。

选项1:父元素中的条件呈现。

我不喜欢这个方法，除非你只渲染组件一次，然后把它留在那里。问题是，每次切换可见性时，它都会导致react从头创建组件。 下面是一个例子。LogoutButton或LoginButton在父LoginControl中被有条件地呈现。如果运行此命令，您将注意到每次单击按钮时都会调用构造函数。https://codepen.io/Kelnor/pen/LzPdpN?editors=1111

class LoginControl extends React.Component {
  constructor(props) {
    super(props);
    this.handleLoginClick = this.handleLoginClick.bind(this);
    this.handleLogoutClick = this.handleLogoutClick.bind(this);
    this.state = {isLoggedIn: false};
  }

  handleLoginClick() {
    this.setState({isLoggedIn: true});
  }

  handleLogoutClick() {
    this.setState({isLoggedIn: false});
  }

  render() {
    const isLoggedIn = this.state.isLoggedIn;

    let button = null;
    if (isLoggedIn) {
      button = <LogoutButton onClick={this.handleLogoutClick} />;
    } else {
      button = <LoginButton onClick={this.handleLoginClick} />;
    }

    return (
      <div>
        <Greeting isLoggedIn={isLoggedIn} />
        {button}
      </div>
    );
  }
}

class LogoutButton extends React.Component{
  constructor(props, context){
    super(props, context)
    console.log('created logout button');
  }
  render(){
    return (
      <button onClick={this.props.onClick}>
        Logout
      </button>
    );
  }
}

class LoginButton extends React.Component{
  constructor(props, context){
    super(props, context)
    console.log('created login button');
  }
  render(){
    return (
      <button onClick={this.props.onClick}>
        Login
      </button>
    );
  }
}

function UserGreeting(props) {
  return <h1>Welcome back!</h1>;
}

function GuestGreeting(props) {
  return <h1>Please sign up.</h1>;
}

function Greeting(props) {
  const isLoggedIn = props.isLoggedIn;
  if (isLoggedIn) {
    return <UserGreeting />;
  }
  return <GuestGreeting />;
}

ReactDOM.render(
  <LoginControl />,
  document.getElementById('root')
);

Now React is pretty quick at creating components from scratch. However, it still has to call your code when creating it. So if your constructor, componentDidMount, render, etc code is expensive, then it'll significantly slow down showing the component. It also means you cannot use this with stateful components where you want the state to be preserved when hidden (and restored when displayed.) The one advantage is that the hidden component isn't created at all until it's selected. So hidden components won't delay your initial page load. There may also be cases where you WANT a stateful component to reset when toggled. In which case this is your best option.

选项2:子对象中的条件呈现

这将一次性创建两个组件。然后，如果组件被隐藏，其余的呈现代码就会短路。您还可以使用可见道具在其他方法中短路其他逻辑。注意代码依赖页中的console.log。https://codepen.io/Kelnor/pen/YrKaWZ?editors=0011

class LoginControl extends React.Component {
  constructor(props) {
    super(props);
    this.handleLoginClick = this.handleLoginClick.bind(this);
    this.handleLogoutClick = this.handleLogoutClick.bind(this);
    this.state = {isLoggedIn: false};
  }

  handleLoginClick() {
    this.setState({isLoggedIn: true});
  }

  handleLogoutClick() {
    this.setState({isLoggedIn: false});
  }

  render() {
    const isLoggedIn = this.state.isLoggedIn;
    return (
      <div>
        <Greeting isLoggedIn={isLoggedIn} />
        <LoginButton isLoggedIn={isLoggedIn} onClick={this.handleLoginClick}/>
        <LogoutButton isLoggedIn={isLoggedIn} onClick={this.handleLogoutClick}/>
      </div>
    );
  }
}

class LogoutButton extends React.Component{
  constructor(props, context){
    super(props, context)
    console.log('created logout button');
  }
  render(){
    if(!this.props.isLoggedIn){
      return null;
    }
    return (
      <button onClick={this.props.onClick}>
        Logout
      </button>
    );
  }
}

class LoginButton extends React.Component{
  constructor(props, context){
    super(props, context)
    console.log('created login button');
  }
  render(){
    if(this.props.isLoggedIn){
      return null;
    }
    return (
      <button onClick={this.props.onClick}>
        Login
      </button>
    );
  }
}

function UserGreeting(props) {
  return <h1>Welcome back!</h1>;
}

function GuestGreeting(props) {
  return <h1>Please sign up.</h1>;
}

function Greeting(props) {
  const isLoggedIn = props.isLoggedIn;
  if (isLoggedIn) {
    return <UserGreeting />;
  }
  return <GuestGreeting />;
}

ReactDOM.render(
  <LoginControl />,
  document.getElementById('root')
);

Now, if the initialization logic is quick and the children are stateless, then you won't see a difference in performance or functionality. However, why make React create a brand new component every toggle anyway? If the initialization is expensive however, Option 1 will run it every time you toggle a component which will slow the page down when switching. Option 2 will run all of the component's inits on first page load. Slowing down that first load. Should note again. If you're just showing the component one time based on a condition and not toggling it, or you want it to reset when toggledm, then Option 1 is fine and probably the best option.

If slow page load is a problem however, it means you've got expensive code in a lifecycle method and that's generally not a good idea. You can, and probably should, solve the slow page load by moving the expensive code out of the lifecycle methods. Move it to an async function that's kicked off by ComponentDidMount and have the callback put it in a state variable with setState(). If the state variable is null and the component is visible then have the render function return a placeholder. Otherwise render the data. That way the page will load quickly and populate the tabs as they load. You can also move the logic into the parent and push the results to the children as props. That way you can prioritize which tabs get loaded first. Or cache the results and only run the logic the first time a component is shown.

选项3:类隐藏

Class hiding is probably the easiest to implement. As mentioned you just create a CSS class with display: none and assign the class based on prop. The downside is the entire code of every hidden component is called and all hidden components are attached to the DOM. (Option 1 doesn't create the hidden components at all. And Option 2 short circuits unnecessary code when the component is hidden and removes the component from the DOM completely.) It appears this is faster at toggling visibility according some tests done by commenters on other answers but I can't speak to that.

选项4:一个组件，但改变道具。或者根本没有组件，缓存HTML。

This one won't work for every application and it's off topic because it's not about hiding components, but it might be a better solution for some use cases than hiding. Let's say you have tabs. It might be possible to write one React Component and just use the props to change what's displayed in the tab. You could also save the JSX to state variables and use a prop to decide which JSX to return in the render function. If the JSX has to be generated then do it and cache it in the parent and send the correct one as a prop. Or generate in the child and cache it in the child's state and use props to select the active one.

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>
     );
  }
}

//use ternary condition

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

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

{ this.state.yourState == 'string' ? <MyComponent /> : ''}

{ this.state.yourState == 'string' && <MyComponent /> }

//Normal condition

if(this.state.yourState){
 return <MyComponent />
}else{
  return null;
}


<button onClick={()=>this.setState({yourState: !this.props.yourState}>Toggle View</button>

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！