TanStackT
TanStack•5mo ago•
3 replies
clean-aquamarine

React-aria-components & createLink

heya,

I'm working with RAC and tanstack router, and have moved away from RAC's RouterProvider pattern to createLink, but a few things have come up.

ListBoxItem can potentially be a link, but not always, if passed a href prop it renders an 'a' tag, otherwise its a div.

However, if I wrap my styled version of ListBoxItem in createLink it's always a 'a', and the to prop is always required.

export function ListBoxItem(props: ListBoxItemProps) { // can render 'a' or 'div'
    const textValue =
        props.textValue ||
        (typeof props.children === "string" ? props.children : undefined);
    return (
        <AriaListBoxItem {...props} textValue={textValue} className={itemStyles}>
            {composeRenderProps(props.children, (children) => (
                <>
                    {children}
                    <div className="absolute left-4 right-4 bottom-0 h-px bg-white/20 forced-colors:bg-[HighlightText] hidden [.group[data-selected]:has(+[data-selected])_&]:block" />
                </>
            ))}
        </AriaListBoxItem>
    );
}
export const ListBoxItemLink = createLink(ListBoxItem); // will always render 'a'


And with Link,
function StyledLink(props: Props) {
    return (
        <AriaLink
            {...props}
            className={composeRenderProps(props.className, (className, renderProps) =>
                styles({ ...renderProps, className, variant: props.variant }),
            )}
        />
    );
}

export const Link = createLink(StyledLink);

I originally only exported the 'Link' component from my styled implementation, but in one place I needed to link externally via a href but typescript marks to a required prop which doesn't support external links.


I know the answer is 'you created a link, obviously it's gonna be a link' and just have 2 components, one for internal links, one for external links / standard behaviours is fair, but maybe someone has a better idea?

Thanks!
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