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

MUI有一个小组件来解决这个问题:https://mui.com/base/react-click-away-listener/它的重量低于1kB,支持移动、IE 11和门户网站。

其他回答

战略示例

我喜欢所提供的解决方案,这些解决方案通过围绕组件创建包装器来完成相同的任务。

由于这更多是一种行为,我想到了战略,并提出了以下建议。

我是React的新手,我需要一些帮助来保存用例中的样板

请回顾并告诉我你的想法。

ClickOutside行为

import ReactDOM from 'react-dom';

export default class ClickOutsideBehavior {

  constructor({component, appContainer, onClickOutside}) {

    // Can I extend the passed component's lifecycle events from here?
    this.component = component;
    this.appContainer = appContainer;
    this.onClickOutside = onClickOutside;
  }

  enable() {

    this.appContainer.addEventListener('click', this.handleDocumentClick);
  }

  disable() {

    this.appContainer.removeEventListener('click', this.handleDocumentClick);
  }

  handleDocumentClick = (event) => {

    const area = ReactDOM.findDOMNode(this.component);

    if (!area.contains(event.target)) {
        this.onClickOutside(event)
    }
  }
}

示例用法

import React, {Component} from 'react';
import {APP_CONTAINER} from '../const';
import ClickOutsideBehavior from '../ClickOutsideBehavior';

export default class AddCardControl extends Component {

  constructor() {
    super();

    this.state = {
      toggledOn: false,
      text: ''
    };

    this.clickOutsideStrategy = new ClickOutsideBehavior({
      component: this,
      appContainer: APP_CONTAINER,
      onClickOutside: () => this.toggleState(false)
    });
  }

  componentDidMount () {

    this.setState({toggledOn: !!this.props.toggledOn});
    this.clickOutsideStrategy.enable();
  }

  componentWillUnmount () {
    this.clickOutsideStrategy.disable();
  }

  toggleState(isOn) {

    this.setState({toggledOn: isOn});
  }

  render() {...}
}

笔记

我想到了存储传递的组件生命周期挂钩,并用类似的方法覆盖它们:

const baseDidMount = component.componentDidMount;

component.componentDidMount = () => {
  this.enable();
  baseDidMount.call(component)
}

component是传递给ClickOutsideBehavior构造函数的组件。这将从该行为的用户中删除启用/禁用样板,但看起来不太好

import React, { useState, useEffect, useRef } from "react";

const YourComponent: React.FC<ComponentProps> = (props) => {
  const ref = useRef<HTMLDivElement | null>(null);
  const [myState, setMyState] = useState(false);
  useEffect(() => {
    const listener = (event: MouseEvent) => {
      // we have to add some logic to decide whether or not a click event is inside of this editor
      // if user clicks on inside the div we dont want to setState
      // we add ref to div to figure out whether or not a user is clicking inside this div to determine whether or not event.target is inside the div
      if (
        ref.current &&
        event.target &&
        // contains is expect other: Node | null
        ref.current.contains(event.target as Node)
      ) {
        return;
      }
      // if we are outside
      setMyState(false);
    };
    // anytime user clics anywhere on the dom, that click event will bubble up into our body element
    // without { capture: true } it might not work
    document.addEventListener("click", listener, { capture: true });
    return () => {
      document.removeEventListener("click", listener, { capture: true });
    };
  }, []);

  return (
    <div  ref={ref}>
      ....
    </div>
  );
};

我对所有其他答案最担心的是必须从根/父级向下过滤单击事件。我发现最简单的方法是简单地设置一个具有位置的同级元素:fixed,下拉列表后面的z索引1,并处理同一组件内固定元素上的单击事件。将所有内容集中到给定组件。

示例代码

#HTML
<div className="parent">
  <div className={`dropdown ${this.state.open ? open : ''}`}>
    ...content
  </div>
  <div className="outer-handler" onClick={() => this.setState({open: false})}>
  </div>
</div>

#SASS
.dropdown {
  display: none;
  position: absolute;
  top: 0px;
  left: 0px;
  z-index: 100;
  &.open {
    display: block;
  }
}
.outer-handler {
    position: fixed;
    top: 0;
    left: 0;
    right: 0;
    bottom: 0;
    opacity: 0;
    z-index: 99;
    display: none;
    &.open {
      display: block;
    }
}
import { RefObject, useEffect } from 'react';

const useClickOutside = <T extends HTMLElement>(ref: RefObject<T>, fn: () => void) => {
    useEffect(() => {
        const element = ref?.current;
        function handleClickOutside(event: Event) {
            if (element && !element.contains(event.target as Node | null)) {
                fn();
            }
        }
        document.addEventListener('mousedown', handleClickOutside);
        return () => {
            document.removeEventListener('mousedown', handleClickOutside);
        };
    }, [ref]);
};

export default useClickOutside;

聚会晚了一点,但我在使用React时遇到了一些问题。选择下拉菜单,因为单击的选项将不再包含在我希望在onClick被激发时单击的父项中。

我通过以下方式解决了这个问题:

componentDidMount() {
    document.addEventListener('mousedown', this.onClick );
}

componentWillUnmount() {
    document.removeEventListener('mousedown', this.onClick );
}

onClick = (event) => {
    if(!event.path.includes(this.detectOutsideClicksDiv)) {
        // Do stuff here
    }
}