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
如果您需要typescript版本:
import React, { useRef, useEffect } from "react";
interface Props {
ref: React.MutableRefObject<any>;
}
export const useOutsideAlerter = ({ ref }: Props) => {
useEffect(() => {
const handleClickOutside = (event: MouseEvent) => {
if (ref.current && !ref.current.contains(event.target as Node)) {
//do what ever you want
}
};
// Bind the event listener
document.addEventListener("mousedown", handleClickOutside);
return () => {
// Unbind the event listener on clean up
document.removeEventListener("mousedown", handleClickOutside);
};
}, [ref]);
};
export default useOutsideAlerter;
如果您想扩展它以关闭模态或隐藏某些内容,也可以执行以下操作:
import React, { useRef, useEffect } from "react";
interface Props {
ref: React.MutableRefObject<any>;
setter: React.Dispatch<React.SetStateAction<boolean>>;
}
export const useOutsideAlerter = ({ ref, setter }: Props) => {
useEffect(() => {
const handleClickOutside = (event: MouseEvent) => {
if (ref.current && !ref.current.contains(event.target as Node)) {
setter(false);
}
};
// Bind the event listener
document.addEventListener("mousedown", handleClickOutside);
return () => {
// Unbind the event listener on clean up
document.removeEventListener("mousedown", handleClickOutside);
};
}, [ref, setter]);
};
export default useOutsideAlerter;
对于那些需要绝对定位的人,我选择的一个简单选项是添加一个包装器组件,该组件的样式是以透明背景覆盖整个页面。然后可以在这个元素上添加一个onClick来关闭内部组件。
<div style={{
position: 'fixed',
top: '0', right: '0', bottom: '0', left: '0',
zIndex: '1000',
}} onClick={() => handleOutsideClick()} >
<Content style={{position: 'absolute'}}/>
</div>
现在,如果您在内容上添加一个单击处理程序,那么事件也将传播到上面的div,从而触发handlerOutsideClick。如果这不是您想要的行为,只需停止处理程序上的事件进程。
<Content style={{position: 'absolute'}} onClick={e => {
e.stopPropagation();
desiredFunctionCall();
}}/>
`
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。
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>
);
}