Before wrote this post, I saw this post, but i'm not able to link all of the code to mine.
This is my toggle component:
<ToggleContent
toggle={show => (
<div>
<button type="button" onClick={show} className={styles.acronym}>
{acronym}
</button>
</div>
)
}
content={show => (
<LogoutCard onClick={show} acronym={acronym} name={name} />
)}
/>
and this is the inside of ToggleContent
function ToggleContent({ toggle, content }) {
const [isShown, setIsShown] = useState(false);
const hide = () => setIsShown(false);
const show = () => setIsShown(!isShown);
return (
<Fragment>
{toggle(show)}
{isShown && content(hide)}
</Fragment>
);
}
and this is the wrapper of LogoutCard inside the props content
import React, { useRef, useEffect } from "react";
/**
* Hook that alerts clicks outside of the passed ref
*/
function useOutsideAlerter(ref) {
/**
* Alert if clicked on outside of element
*/
function handleClickOutside(event) {
if (ref.current && !ref.current.contains(event.target)) {
alert("You clicked outside of me!");
}
}
useEffect(() => {
// Bind the event listener
document.addEventListener("mousedown", handleClickOutside);
return () => {
// Unbind the event listener on clean up
document.removeEventListener("mousedown", handleClickOutside);
};
});
}
/**
* Component that alerts if you click outside of it
*/
export default function OutsideAlerter(props) {
const wrapperRef = useRef(null);
useOutsideAlerter(wrapperRef);
return <div ref={wrapperRef}>{props.children}</div>;
}
Problem
The problem is that I'm able to print the alert, but i'm not able to close the popup because I'm not able to pass the show
value, that's in the only allowed to close and open the little popup.
Question
How can I close the popup ?