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

https://stackoverflow.com/a/42234988/9536897解决方案在手机上不起作用。

您可以尝试:

  // returns true if the element or one of its parents has the class classname
  hasSomeParentTheClass(element, classname) {
    if(element.target)
    element=element.target;
    
    if (element.className&& element.className.split(" ").indexOf(classname) >= 0) return true;
    return (
      element.parentNode &&
      this.hasSomeParentTheClass(element.parentNode, classname)
    );
  }
  componentDidMount() {
    const fthis = this;

    $(window).click(function (element) {
      if (!fthis.hasSomeParentTheClass(element, "myClass"))
        fthis.setState({ pharmacyFocus: null });
    });
  }

在视图中,将className赋给特定元素。

其他回答

我使用了这个模块(我与作者没有关联)

npm install react-onclickout --save

const ClickOutHandler=require('act-on-clickout');ExampleComponent类扩展React.Component{单击退出(e){if(hasClass(e.target,'忽略我'))返回;alert('用户在组件外部单击!');}render(){返回(<ClickOutHandler onClickOut={this.onClickOut}><div>点击我的外部</分区></ClickOutHandler>);}}

它做得很好。

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>;
  }
}

或者:

const onClickOutsideListener = () => {
    alert("click outside")
    document.removeEventListener("click", onClickOutsideListener)
  }

...

return (
  <div
    onMouseLeave={() => {
          document.addEventListener("click", onClickOutsideListener)
        }}
  >
   ...
  </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>
  );
}