我正在寻找一种方法来检测单击事件是否发生在组件之外,如本文所述。jQueryclosest()用于查看单击事件的目标是否将dom元素作为其父元素之一。如果存在匹配项,则单击事件属于其中一个子项,因此不被视为在组件之外。
因此,在我的组件中,我想将一个单击处理程序附加到窗口。当处理程序启动时,我需要将目标与组件的dom子级进行比较。
click事件包含类似“path”的财产,它似乎保存了事件经过的dom路径。我不知道该比较什么,或者如何最好地遍历它,我想肯定有人已经把它放在了一个聪明的效用函数中。。。不
我为所有场合制定了解决方案。
你应该使用一个高阶组件来包装你想要监听的组件。
这个组件示例只有一个属性:“onClickedOutside”,它接收函数。
ClickedOutside.js
import React, { Component } from "react";
export default class ClickedOutside extends Component {
componentDidMount() {
document.addEventListener("mousedown", this.handleClickOutside);
}
componentWillUnmount() {
document.removeEventListener("mousedown", this.handleClickOutside);
}
handleClickOutside = event => {
// IF exists the Ref of the wrapped component AND his dom children doesnt have the clicked component
if (this.wrapperRef && !this.wrapperRef.contains(event.target)) {
// A props callback for the ClikedClickedOutside
this.props.onClickedOutside();
}
};
render() {
// In this piece of code I'm trying to get to the first not functional component
// Because it wouldn't work if use a functional component (like <Fade/> from react-reveal)
let firstNotFunctionalComponent = this.props.children;
while (typeof firstNotFunctionalComponent.type === "function") {
firstNotFunctionalComponent = firstNotFunctionalComponent.props.children;
}
// Here I'm cloning the element because I have to pass a new prop, the "reference"
const children = React.cloneElement(firstNotFunctionalComponent, {
ref: node => {
this.wrapperRef = node;
},
// Keeping all the old props with the new element
...firstNotFunctionalComponent.props
});
return <React.Fragment>{children}</React.Fragment>;
}
}
Ez的方式。。。(2022年更新)
创建挂钩:useOutsideClick.ts
export function useOutsideClick(ref: any, onClickOut: () => void){
useEffect(() => {
const onClick = ({target}: any) => !ref.contains(target) && onClickOut?.()
document.addEventListener("click", onClick);
return () => document.removeEventListener("click", onClick);
}, []);
}
将componentRef添加到组件并调用useOutsideClick
export function Example(){
const componentRef = useRef();
useOutsideClick(componentRef.current!, () => {
// do something here
});
return (
<div ref={componentRef as any}> My Component </div>
)
}
在这里尝试了许多方法之后,我决定使用github.com/Pomax/react-onclickoutside,因为它非常完整。
我通过npm安装了模块并将其导入到组件中:
import onClickOutside from 'react-onclickoutside'
然后,在组件类中,我定义了handleClickOutside方法:
handleClickOutside = () => {
console.log('onClickOutside() method called')
}
导出组件时,我将其包装在onClickOutside()中:
export default onClickOutside(NameOfComponent)
就是这样。
这已经有很多答案了,但它们没有解决e.stopPropagation()和阻止单击要关闭的元素之外的react链接的问题。
由于React有自己的人工事件处理程序,您无法将文档用作事件侦听器的基础。在这之前,您需要e.stopPropagation(),因为React使用文档本身。如果改用document.querySelector('body')。您可以防止点击React链接。下面是我如何实现单击外部并关闭的示例。这使用ES6和React 16.3。
import React, { Component } from 'react';
class App extends Component {
constructor(props) {
super(props);
this.state = {
isOpen: false,
};
this.insideContainer = React.createRef();
}
componentWillMount() {
document.querySelector('body').addEventListener("click", this.handleClick, false);
}
componentWillUnmount() {
document.querySelector('body').removeEventListener("click", this.handleClick, false);
}
handleClick(e) {
/* Check that we've clicked outside of the container and that it is open */
if (!this.insideContainer.current.contains(e.target) && this.state.isOpen === true) {
e.preventDefault();
e.stopPropagation();
this.setState({
isOpen: false,
})
}
};
togggleOpenHandler(e) {
e.preventDefault();
this.setState({
isOpen: !this.state.isOpen,
})
}
render(){
return(
<div>
<span ref={this.insideContainer}>
<a href="#open-container" onClick={(e) => this.togggleOpenHandler(e)}>Open me</a>
</span>
<a href="/" onClick({/* clickHandler */})>
Will not trigger a click when inside is open.
</a>
</div>
);
}
}
export default App;