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
以下解决方案使用ES6并遵循绑定以及通过方法设置ref的最佳实践。
要将其付诸行动:
挂钩实施反应16.3后的类实现反应16.3之前的类实现
挂钩实施:
import React, { useRef, useEffect } from "react";
/**
* Hook that alerts clicks outside of the passed ref
*/
function useOutsideAlerter(ref) {
useEffect(() => {
/**
* Alert if clicked on outside of element
*/
function handleClickOutside(event) {
if (ref.current && !ref.current.contains(event.target)) {
alert("You clicked outside of me!");
}
}
// 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 OutsideAlerter(props) {
const wrapperRef = useRef(null);
useOutsideAlerter(wrapperRef);
return <div ref={wrapperRef}>{props.children}</div>;
}
类实现:
16.3之后
import React, { Component } from "react";
/**
* Component that alerts if you click outside of it
*/
export default class OutsideAlerter extends Component {
constructor(props) {
super(props);
this.wrapperRef = React.createRef();
this.handleClickOutside = this.handleClickOutside.bind(this);
}
componentDidMount() {
document.addEventListener("mousedown", this.handleClickOutside);
}
componentWillUnmount() {
document.removeEventListener("mousedown", this.handleClickOutside);
}
/**
* Alert if clicked on outside of element
*/
handleClickOutside(event) {
if (this.wrapperRef && !this.wrapperRef.current.contains(event.target)) {
alert("You clicked outside of me!");
}
}
render() {
return <div ref={this.wrapperRef}>{this.props.children}</div>;
}
}
16.3之前
import React, { Component } from "react";
/**
* Component that alerts if you click outside of it
*/
export default class OutsideAlerter extends Component {
constructor(props) {
super(props);
this.setWrapperRef = this.setWrapperRef.bind(this);
this.handleClickOutside = this.handleClickOutside.bind(this);
}
componentDidMount() {
document.addEventListener("mousedown", this.handleClickOutside);
}
componentWillUnmount() {
document.removeEventListener("mousedown", this.handleClickOutside);
}
/**
* Set the wrapper ref
*/
setWrapperRef(node) {
this.wrapperRef = node;
}
/**
* Alert if clicked on outside of element
*/
handleClickOutside(event) {
if (this.wrapperRef && !this.wrapperRef.contains(event.target)) {
alert("You clicked outside of me!");
}
}
render() {
return <div ref={this.setWrapperRef}>{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>
);
}
我有一个需要有条件地将孩子插入模态的例子。像这样,贝娄。
const [view, setView] = useState(VIEWS.SomeView)
return (
<Modal onClose={onClose}>
{VIEWS.Result === view ? (
<Result onDeny={() => setView(VIEWS.Details)} />
) : VIEWS.Details === view ? (
<Details onDeny={() => setView(VIEWS.Result) /> />
) : null}
</Modal>
)
所以parent.contains(event.target)在这里不起作用,因为一旦分离了子级,parent(modal)就不再包含event.targe。
我的解决方案(到目前为止有效,没有任何问题)是这样写:
const listener = (event: MouseEvent) => {
if (parentNodeRef && !event.path.includes(parentNodeRef)) callback()
}
若父级包含已经分离的树中的元素,那个么它不会触发回调。
编辑:event.path是新的,尚未在所有浏览器中退出。请改用composedPath。
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>;
}
}