-
Hi, const TooltipTrigger = React.forwardRef<HTMLButtonElement>(
(props, forwardedRef) => (
<button disabled {...props} ref={forwardedRef}>
someButtonText
</button>
)
);
interface TooltipProps {
info: string;
}
const Tooltip = ({ info }: TooltipProps) => {
return (
<TooltipPrimitive.Provider>
<TooltipPrimitive.Root>
<TooltipPrimitive.Trigger asChild>
<TooltipTrigger />
</TooltipPrimitive.Trigger>
<TooltipPrimitive.Content sideOffset={4}>
<p>{info}</p>
</TooltipPrimitive.Content>
</TooltipPrimitive.Root>
</TooltipPrimitive.Provider>
);
}; It works just fine.
Any help would be appreciated. |
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Answered by
joaom00
Aug 22, 2023
Replies: 2 comments 2 replies
-
See Extending a primitive for how to type |
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1 reply
Answer selected by
hhkim0729
-
I'm not sure if it's correct, but I fixed my code like this and the error has gone. const TooltipTrigger = React.forwardRef<
React.ElementRef<typeof TooltipPrimitive.Trigger>,
React.ComponentPropsWithoutRef<typeof TooltipPrimitive.Trigger> & {
children: React.ReactNode;
asChild?: boolean;
}
>((props, forwardedRef) => (
<TooltipPrimitive.Trigger {...props} ref={forwardedRef} />
));
interface TooltipProps {
info: string;
}
const Tooltip = ({ info }: TooltipProps) => {
return (
<TooltipPrimitive.Provider>
<TooltipPrimitive.Root>
<TooltipTrigger asChild>
<button disabled>
some button text
</button>
</TooltipTrigger>
<TooltipPrimitive.Content>
<p>{info}</p>
</TooltipPrimitive.Content>
</TooltipPrimitive.Root>
</TooltipPrimitive.Provider>
);
}; |
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1 reply
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See Extending a primitive for how to type