考虑下面的钩子示例

   import { useState } from 'react';

   function Example() {
       const [count, setCount] = useState(0);

       return (
           <div>
               <p>You clicked {count} times</p>
               <button onClick={() => setCount(count + 1)}>
                  Click me
               </button>
          </div>
        );
     }

基本上,我们使用this.forceUpdate()方法强制组件立即在React类组件中重新渲染,如下例所示

    class Test extends Component{
        constructor(props){
             super(props);
             this.state = {
                 count:0,
                 count2: 100
             }
             this.setCount = this.setCount.bind(this);//how can I do this with hooks in functional component 
        }
        setCount(){
              let count = this.state.count;
                   count = count+1;
              let count2 = this.state.count2;
                   count2 = count2+1;
              this.setState({count});
              this.forceUpdate();
              //before below setState the component will re-render immediately when this.forceUpdate() is called
              this.setState({count2: count
        }

        render(){
              return (<div>
                   <span>Count: {this.state.count}></span>. 
                   <button onClick={this.setCount}></button>
                 </div>
        }
 }

但我的问题是,我如何才能强制上述功能组件重新渲染立即与挂钩?


当前回答

有点晚了,但我注意到大多数(所有)的答案都错过了可以传递回调到forceUpdate生命周期方法的部分。

根据react源代码,此回调具有与setState方法中的行为相同的行为-它在更新后执行。

因此,最正确的实现应该是这样的:

/** *增加导致重传并执行回调的状态 * @param {function}回调-状态更新后执行的回调 * @返回{函数} * / export const useForceUpdate = (callback) => { const [state, updater] = useReducer((x) => x + 1,0); useEffect(() => { Callback && Callback (); },[状态]); 返回useCallback(() => { 更新(); },[]); };

其他回答

React Hooks FAQ官方解决方案的forceUpdate:

const [_, forceUpdate] = useReducer((x) => x + 1, 0);
// usage
<button onClick={forceUpdate}>Force update</button>

工作示例

const App = () => { const [_, forceUpdate] = useReducer((x) => x + 1, 0); return ( <div> <button onClick={forceUpdate}>Force update</button> <p>Forced update {_} times</p> </div> ); }; ReactDOM.render(<App />, document.getElementById("root")); <script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.10.1/umd/react.production.min.js" integrity="sha256-vMEjoeSlzpWvres5mDlxmSKxx6jAmDNY4zCt712YCI0=" crossorigin="anonymous"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.10.1/umd/react-dom.production.min.js" integrity="sha256-QQt6MpTdAD0DiPLhqhzVyPs1flIdstR4/R7x4GqCvZ4=" crossorigin="anonymous"></script> <script>var useReducer = React.useReducer</script> <div id="root"></div>

const useForceRender = () => {
  const [, forceRender] = useReducer(x => !x, true)
  return forceRender
}

使用

function Component () {
  const forceRender = useForceRender() 
  useEffect(() => {
    // ...
    forceRender()
  }, [])

一句话解决方案:

useState()[1];

useState返回一对值:当前状态和更新状态的函数——state和setter,这里我们只使用setter来强制重新呈现。

潜在的选项是仅在特定组件上使用键强制更新。更新键会触发组件的呈现(之前更新失败)

例如:

const [tableKey, setTableKey] = useState(1);
...

useEffect(() => {
    ...
    setTableKey(tableKey + 1);
}, [tableData]);

...
<DataTable
    key={tableKey}
    data={tableData}/>

你最好只让你的组件依赖于状态和道具,它会像预期的那样工作,但如果你真的需要一个函数来强制组件重新呈现,你可以使用useState钩子,并在需要时调用这个函数。

例子

const { useState, useEffect } = React; 函数 Foo() { const [, forceUpdate] = useState(); useEffect(() => { setTimeout(forceUpdate, 2000); }, []); 返回 <div>{Date.now()}</div>; } ReactDOM.render(<Foo />, document.getElementById(“root”)); <script src=“https://unpkg.com/react@16.7.0-alpha.0/umd/react.production.min.js”></script> <script src=“https://unpkg.com/react-dom@16.7.0-alpha.0/umd/react-dom.production.min.js”></script> <div id=“root”></div>