避免使用event.persist()——你想让React回收合成事件。我认为无论你使用类还是钩子，最干净的方法是将回调函数分成两部分:

没有deboundation的回调 只使用您需要的事件片段调用已撤销的函数(这样合成的事件就可以循环使用)

类

handleMouseOver = throttle(target => {
  console.log(target);
}, 1000);

onMouseOver = e => {
  this.handleMouseOver(e.target);
};

<div onMouseOver={this.onMouseOver} />

功能

const handleMouseOver = useRef(throttle(target => {
  console.log(target);
}, 1000));

function onMouseOver(e) {
  handleMouseOver.current(e.target);
}

<div onMouseOver={this.onMouseOver} />

注意，如果你的handleMouseOver函数从组件中使用状态，你应该使用usemo而不是useRef，并将它们作为依赖项传递，否则你将使用过时的数据(当然不适用于类)。

I was searching for a solution to the same problem and came across this thread as well as some others but they had the same problem: if you are trying to do a handleOnChange function and you need the value from an event target, you will get cannot read property value of null or some such error. In my case, I also needed to preserve the context of this inside the debounced function since I'm executing a fluxible action. Here's my solution, it works well for my use case so I'm leaving it here in case anyone comes across this thread:

// at top of file:
var myAction = require('../actions/someAction');

// inside React.createClass({...});

handleOnChange: function (event) {
    var value = event.target.value;
    var doAction = _.curry(this.context.executeAction, 2);

    // only one parameter gets passed into the curried function,
    // so the function passed as the first parameter to _.curry()
    // will not be executed until the second parameter is passed
    // which happens in the next function that is wrapped in _.debounce()
    debouncedOnChange(doAction(myAction), value);
},

debouncedOnChange: _.debounce(function(action, value) {
    action(value);
}, 300)

与其在debounce()中包装handleOnChange，不如在debounce()中包装回调函数中的ajax调用，从而不破坏事件对象。就像这样:

handleOnChange: function (event) {
   debounce(
     $.ajax({})
  , 250);
}

你也可以使用自己编写的mixin，就像这样:

var DebounceMixin = {
  debounce: function(func, time, immediate) {
    var timeout = this.debouncedTimeout;
    if (!timeout) {
      if (immediate) func();
      this.debouncedTimeout = setTimeout(function() {
        if (!immediate) func();
        this.debouncedTimeout = void 0;
      }.bind(this), time);
    }
  }
};

然后像这样在你的组件中使用它:

var MyComponent = React.createClass({
  mixins: [DebounceMixin],
  handleClick: function(e) {
    this.debounce(function() {
      this.setState({
        buttonClicked: true
      });
    }.bind(this), 500, true);
  },
  render: function() {
    return (
      <button onClick={this.handleClick}></button>
    );
  }
});

现在，React和React Native在2019年底有了另一个解决方案:

react-debounce-component

<input>
<Debounce ms={500}>
  <List/>
</Debounce>

它是一个组件，易于使用，体积小，支持广泛

例子:

import React from 'react';
import Debounce from 'react-debounce-component';

class App extends React.Component {
  constructor (props) {
    super(props);
    this.state = {value: 'Hello'}
  }
  render () {
    return (
      <div>
        <input value={this.state.value} onChange={(event) => {this.setState({value: event.target.value})}}/>
        <Debounce ms={1000}>
          <div>{this.state.value}</div>
        </Debounce>
      </div>
    );
  }
}

export default App;

我是这个组件的创建者

有一个使用react钩子的简单方法。

步骤1:定义一个状态来维护搜索的文本

const [searchTerm, setSearchTerm] = useState('')

步骤2:使用useEffect捕获搜索Term中的任何变化

useEffect(() => {
  const delayDebounceFn = setTimeout(() => {
    if (searchTerm) {
      // write your logic here
    }
  }, 400)

  return () => clearTimeout(delayDebounceFn)
}, [searchTerm])

步骤3:编写一个函数来处理输入更改

function handleInputChange(value) {
  if (value) {
    setSearchTerm(value)
  }
}

就这些!在需要时调用此方法