将eventListener与react组件分离

时间:2019-02-13 15:07:06

标签: reactjs onmouseover react-component

我有一个react组件,当遇到事件(onMouseOver)时,我想调用一个函数。在组件首次出现后,是否可以将eventListener与组件分离? (使用状态是一种解决方案,但我想知道是否还有另一个)

handleOnMouseOver = () => {
    console.log('event has occured');
}

render() {
   return(
     <div onMouseOver={this.handleOnMouseOver}>
   );
}

2 个答案:

答案 0 :(得分:1)

您可以返回闭包而不污染状态,并保留一个标志,以在第一次出现后禁用监听器。

handleOnMouseOver = () => {
  let isActive = true;
  return (evt) => {
    if(isActive) {
       console.log('event has occured');
       isActive = false; // disable listener
    }
}

render() {
  return(
    <div onMouseOver={this.handleOnMouseOver()}>
  );
}

答案 1 :(得分:1)

您可以在状态中存储一个值,指示您的事件是否已经发生。

如果这样做,您将不会在事件处理程序中返回任何回调函数:

class App extends React.Component {

	state = {
		hovered: false
	}

	handleOnMouseOver = () => {
		this.setState({ hovered: true })
		console.log('event has occured');
	}

	render() {
		return (
			<div onMouseOver={!this.state.hovered && this.handleOnMouseOver}>AAAA</div>
		);
	}
}

ReactDOM.render(<App/>, document.getElementById('root'))
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.5.2/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.5.2/umd/react-dom.production.min.js"></script>
<div id='root'>