这是最适合我制作下拉菜单的方法：

handleClick = () => {
    document.getElementById("myDrop").classList.toggle("showing");
}

render() {

    return (
        <div className="courses">
            <div class="dropdownBody">
                <button onClick={this.handleClick} onBlur={this.handleClick} class="dropbtn">Dropdown</button>
                <div id="myDrop" class="dropdown-content">
                    <a href="#home">Home</a>
                    <a href="#about">About</a>
                    <a href="#contact">Contact</a>
                </div>
            </div>
        </div>
    )
}

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。

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>
  );
}

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。

以上答案中没有一个对我有效，所以我最终做了如下：

从“React”导入React，｛Component｝；

/**
 * Component that alerts if you click outside of it
 */
export default class OutsideAlerter extends Component {
  constructor(props) {
    super(props);

    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 (!event.path || !event.path.filter(item => item.className=='classOfAComponent').length) {
      alert('You clicked outside of me!');
    }
  }

  render() {
    return <div>{this.props.children}</div>;
  }
}

OutsideAlerter.propTypes = {
  children: PropTypes.element.isRequired,
};

这里的其他答案都不适合我。我试图在模糊上隐藏一个弹出窗口，但由于内容是绝对定位的，所以即使单击内部内容，onBlur也会启动。

以下是一个对我来说行之有效的方法：

// Inside the component:
onBlur(event) {
    // currentTarget refers to this component.
    // relatedTarget refers to the element where the user clicked (or focused) which
    // triggered this event.
    // So in effect, this condition checks if the user clicked outside the component.
    if (!event.currentTarget.contains(event.relatedTarget)) {
        // do your thing.
    }
},

希望这有帮助。