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

为了扩展Ben Bud给出的公认答案,如果您使用的是样式化组件,那么这样传递引用会给您一个错误,例如“this.wrapperRef.contains is not a function”。

在注释中,建议的修复方法是用div包装样式化的组件,并将ref传递到那里。尽管如此,在他们的文档中,他们已经解释了这一点的原因,以及在样式化组件中正确使用ref:

将ref属性传递给样式化组件将为您提供StyledComponent包装器的实例,但不会传递给底层DOM节点。这是由于裁判的工作方式。不可能直接在包装器上调用DOM方法,如focus。要获取对实际包装的DOM节点的引用,请将回调传递给innerRef属性。

像这样:

<StyledDiv innerRef={el => { this.el = el }} />

然后您可以在“handleClickOutside”函数中直接访问它:

handleClickOutside = e => {
    if (this.el && !this.el.contains(e.target)) {
        console.log('clicked outside')
    }
}

这也适用于“onBlur”方法:

componentDidMount(){
    this.el.focus()
}
blurHandler = () => {
    console.log('clicked outside')
}
render(){
    return(
        <StyledDiv
            onBlur={this.blurHandler}
            tabIndex="0"
            innerRef={el => { this.el = el }}
        />
    )
}

其他回答

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/

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>
  );
};

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>
  );
}

使用OnClickOutside Hook-反应16.8+

创建通用useOnOutsideClick函数

export const useOnOutsideClick = handleOutsideClick => {
  const innerBorderRef = useRef();

  const onClick = event => {
    if (
      innerBorderRef.current &&
      !innerBorderRef.current.contains(event.target)
    ) {
      handleOutsideClick();
    }
  };

  useMountEffect(() => {
    document.addEventListener("click", onClick, true);
    return () => {
      document.removeEventListener("click", onClick, true);
    };
  });

  return { innerBorderRef };
};

const useMountEffect = fun => useEffect(fun, []);

然后在任何功能组件中使用钩子。

const OutsideClickDemo = ({ currentMode, changeContactAppMode }) => {

  const [open, setOpen] = useState(false);
  const { innerBorderRef } = useOnOutsideClick(() => setOpen(false));

  return (
    <div>
      <button onClick={() => setOpen(true)}>open</button>
      {open && (
        <div ref={innerBorderRef}>
           <SomeChild/>
        </div>
      )}
    </div>
  );

};

链接到演示

部分灵感来自于@pau1itzgerald的回答。

这已经有很多答案了,但它们没有解决e.stopPropagation()和阻止单击要关闭的元素之外的react链接的问题。

由于React有自己的人工事件处理程序,您无法将文档用作事件侦听器的基础。在这之前,您需要e.stopPropagation(),因为React使用文档本身。如果改用document.querySelector('body')。您可以防止点击React链接。下面是我如何实现单击外部并关闭的示例。这使用ES6和React 16.3。

import React, { Component } from 'react';

class App extends Component {
  constructor(props) {
    super(props);

    this.state = {
      isOpen: false,
    };

    this.insideContainer = React.createRef();
  }

  componentWillMount() {
    document.querySelector('body').addEventListener("click", this.handleClick, false);
  }

  componentWillUnmount() {
    document.querySelector('body').removeEventListener("click", this.handleClick, false);
  }

  handleClick(e) {
    /* Check that we've clicked outside of the container and that it is open */
    if (!this.insideContainer.current.contains(e.target) && this.state.isOpen === true) {
      e.preventDefault();
      e.stopPropagation();
      this.setState({
        isOpen: false,
      })
    }
  };

  togggleOpenHandler(e) {
    e.preventDefault();

    this.setState({
      isOpen: !this.state.isOpen,
    })
  }

  render(){
    return(
      <div>
        <span ref={this.insideContainer}>
          <a href="#open-container" onClick={(e) => this.togggleOpenHandler(e)}>Open me</a>
        </span>
        <a href="/" onClick({/* clickHandler */})>
          Will not trigger a click when inside is open.
        </a>
      </div>
    );
  }
}

export default App;