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
因为对我来说!ref.current.contains(e.target)无法工作,因为ref中包含的DOM元素正在更改,我提出了一个稍微不同的解决方案:
function useClickOutside<T extends HTMLElement>(
element: T | null,
onClickOutside: () => void,
) {
useEffect(() => {
function handleClickOutside(event: MouseEvent) {
const xCoord = event.clientX;
const yCoord = event.clientY;
if (element) {
const { right, x, bottom, y } = element.getBoundingClientRect();
if (xCoord < right && xCoord > x && yCoord < bottom && yCoord > y) {
return;
}
onClickOutside();
}
}
document.addEventListener('click', handleClickOutside);
return () => {
document.removeEventListener('click', handleClickOutside);
};
}, [element, onClickOutside]);
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。
这是我的方法(演示-https://jsfiddle.net/agymay93/4/):
我创建了一个名为WatchClickOutside的特殊组件,它可以像这样使用(我假设JSX语法):
<WatchClickOutside onClickOutside={this.handleClose}>
<SomeDropdownEtc>
</WatchClickOutside>
以下是WatchClickOutside组件的代码:
import React, { Component } from 'react';
export default class WatchClickOutside extends Component {
constructor(props) {
super(props);
this.handleClick = this.handleClick.bind(this);
}
componentWillMount() {
document.body.addEventListener('click', this.handleClick);
}
componentWillUnmount() {
// remember to remove all events to avoid memory leaks
document.body.removeEventListener('click', this.handleClick);
}
handleClick(event) {
const {container} = this.refs; // get container that we'll wait to be clicked outside
const {onClickOutside} = this.props; // get click outside callback
const {target} = event; // get direct click event target
// if there is no proper callback - no point of checking
if (typeof onClickOutside !== 'function') {
return;
}
// if target is container - container was not clicked outside
// if container contains clicked target - click was not outside of it
if (target !== container && !container.contains(target)) {
onClickOutside(event); // clicked outside - fire callback
}
}
render() {
return (
<div ref="container">
{this.props.children}
</div>
);
}
}
在我的DROPDOWN案例中,Ben Bud的解决方案工作得很好,但我有一个单独的切换按钮和一个onClick处理程序。因此,外部单击逻辑与单击切换按钮冲突。下面是我如何通过传递按钮的ref来解决这个问题:
import React, { useRef, useEffect, useState } from "react";
/**
* Hook that triggers onClose when clicked outside of ref and buttonRef elements
*/
function useOutsideClicker(ref, buttonRef, onOutsideClick) {
useEffect(() => {
function handleClickOutside(event) {
/* clicked on the element itself */
if (ref.current && !ref.current.contains(event.target)) {
return;
}
/* clicked on the toggle button */
if (buttonRef.current && !buttonRef.current.contains(event.target)) {
return;
}
/* If it's something else, trigger onClose */
onOutsideClick();
}
// Bind the event listener
document.addEventListener("mousedown", handleClickOutside);
return () => {
// Unbind the event listener on clean up
document.removeEventListener("mousedown", handleClickOutside);
};
}, [ref]);
}
/**
* Component that alerts if you click outside of it
*/
export default function DropdownMenu(props) {
const wrapperRef = useRef(null);
const buttonRef = useRef(null);
const [dropdownVisible, setDropdownVisible] = useState(false);
useOutsideClicker(wrapperRef, buttonRef, closeDropdown);
const toggleDropdown = () => setDropdownVisible(visible => !visible);
const closeDropdown = () => setDropdownVisible(false);
return (
<div>
<button onClick={toggleDropdown} ref={buttonRef}>Dropdown Toggler</button>
{dropdownVisible && <div ref={wrapperRef}>{props.children}</div>}
</div>
);
}