首页 > 解决方案 > 如何根据我在 ReactJS 中悬停的内容来渲染组件?

问题描述

我可以渲染 1 个组件,但是否有另一种方法可以根据我悬停的内容来渲染其他组件?

constructor(props) {
        super(props)
        
        this.handleMouseHover = this.handleMouseHover.bind(this)
        this.state={
            isHovering: false
        }
    }

    handleMouseHover(){
        this.setState(this.toggleHoverState)
    }

    toggleHoverState(state) {
        return {
            isHovering:!state.isHovering
        }
    }
<a href="#" onMouseEnter={this.handleMouseHover} onMouseLeave={this.handleMouseHover}>Hover me1</a>
<a href="#" onMouseEnter={this.handleMouseHover} onMouseLeave={this.handleMouseHover}>Hover me2</a>
{isHovering && (<Component/>)}

标签: javascriptreactjs

解决方案


switch如果基于一个元素的悬停有多个元素要显示,则可以使用。

在下面的示例中,我考虑了 3 个anchor元素并添加了相应mouse overmouse out事件处理程序,以便在它们下方显示相应的悬停元素。

const {useState} = React;

const App = () => {
  const [hoveredOn, setHoveredOn] = useState(-1);
  
  const getElement = () => {
    switch(hoveredOn) {
      case 0: return <p>Hovered Element 0</p>
      case 1: return <p>Hovered Element 1</p>
      case 2: return <p>Hovered Element 2</p>
    }
  }
  
  const onMouseOver = (index) => {
    setHoveredOn(index);
  }
  
  const onMouseOut = () => {
    setHoveredOn(-1);
  }
  
  return (
    <div>
      <a href="#" onMouseOver={() => {onMouseOver(0)}} onMouseOut={onMouseOut}>Element 0</a>
      <a href="#" onMouseOver={() => {onMouseOver(1)}} onMouseOut={onMouseOut}>Element 1</a>
      <a href="#" onMouseOver={() => {onMouseOver(2)}} onMouseOut={onMouseOut}>Element 2</a>
      {getElement()}
    </div>
  )
}

ReactDOM.render(<App />, document.getElementById("react"));
a {
  display: block;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.8.0/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.8.0/umd/react-dom.production.min.js"></script>

<div id="react"></div>

您还可以将默认情况添加到开关情况,以便根据您的要求显示默认组件或任何消息。


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