react hooks写法中使用防抖节流

在Class继承写法中，我们的防抖节流都是这样写的

import debounce from 'lodash/debounce';

export default class Search extends Component {
    constructor(props) {
      super(props)
      this.handleSearch = debounce(this.handleSearch, 500);
    }
    handleSubmit = (e) => {
        e.preventDefault();
        this.handleSearch();
    }
    handleSearch = () => {
    ...
    }
    render() {
        return (
            

        )
    }
}

改写成hooks是这样

import React, {useState} from "react";
import debounce from 'lodash/debounce';
const sendQuery = (query) => console.log(`Querying for ${query}`);
const Search = () => {
  const [userQuery, setUserQuery] = useState("");
  const delayedQuery = debounce(q => sendQuery(q), 500);
  const onChange = e => {
    setUserQuery(e.target.value);
    delayedQuery(e.target.value);
  };
  return (
    

      Search:
      
    
  );
}

但这样子做防抖函数并没有发挥作用，仅仅是把反应时间延后了500毫秒。这是因为函数组件每次渲染结束之后，内部的变量都会被释放，重新渲染时所有的变量会被重新初始化，产生的结果就是每一次都注册和执行了setTimeout函数。想要得到正确的运行结果，必须以某种方式存储那些本会被删除的变量和方法的引用。很遗憾，没办法直接使用useState去存储，我们可以使用useRef和useCallback我们可以更简单的去解决这个问题：

const SearchFixed = () => {
  const [userQuery, setUserQuery] = useState("");
  const delayedQuery = useCallback(debounce(q => sendQuery(q), 500), []);
  const onChange = e => {
    setUserQuery(e.target.value);
    delayedQuery(e.target.value);
  };
  return (
    

      Search Fixed:
      
    
  );
};

或者使用useRef

const SearchFixed = () => {
  const [userQuery, setUserQuery] = useState("");
  const delayedQuery = useRef(debounce(q => sendQuery(q), 500)).current;
  const onChange = e => {
    setUserQuery(e.target.value);
    delayedQuery(e.target.value);
  };
  return (
    

      Search Fixed:
      
    
  );
};

也可以自定义hook去解决