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


当前回答

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]);

其他回答

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>
  );
}

只需使用mui(material ui)中的ClickAwayListener:

<ClickAwayListener onClickAway={handleClickAway}>
    {children}
<ClickAwayListener >

有关更多信息,请查看:https://mui.com/base/react-click-away-listener/

我有一个类似的用例,我必须开发一个自定义下拉菜单。当用户在外面单击时,它应该自动关闭。以下是最近的React Hooks实现-

从“react”导入{useEffect,useRef,useState};导出常量应用程序=()=>{const-ref=useRef();const[isMenuOpen,setIsMenuOpen]=useState(false);使用效果(()=>{常量checkIfClickedOutside=(e)=>{//如果菜单是打开的并且点击的目标不在菜单内,//然后关闭菜单if(isMenuOpen&&ref.current&&!ref.current.contents(e.target)){setIsMenuOpen(false);}};document.addEventListener(“mousedown”,checkIfClickedOutside);返回()=>{//清理事件侦听器document.removeEventListener(“mousedown”,checkIfClickedOutside);};},[isMenuOpen]);返回(<div className=“wrapper”ref={ref}><按钮className=“button”onClick={()=>setIsMenuOpen((oldState)=>!oldState)}>单击我</按钮>{isMenuOpen&&(<ul className=“list”><li className=“list item”>下拉选项1</li><li className=“list item”>下拉选项2</li><li className=“list item”>下拉选项3</li><li className=“list item”>下拉选项4</li></ul>)}</div>);}