Typescript, how to pass "Object is possibly null" error?

When you declare const overlayEl = useRef(null); Makes the type it comes out as is null because that's the best possible inference it can offer with that much information, give typescript more information and it will work as intended.

Try....

 const overlayEl = useRef<HTMLDivElement>(null);

Alternatively some syntax sugar for if you don't care for when its undefined is to do something like this.

const overlayEl = useRef(document.createElement("div"))

using the above syntax all common DOM methods just return defaults such as "0" i.e overlayEl.offsetWidth, getBoundingClientRect etc.

Usage:

if(overlayEl.current) {
    // will be type HTMLDivElement NOT HTMLDivElement | null
    const whattype = overlayEl.current; 
}

The way this works is typescripts static analysis is smart enough to figure out that if check "guards" against null, and therefore it will remove that as a possible type from the union of null | HTMLDivElement within those brackets.

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const overlayEl = useRef() as MutableRefObject<HTMLDivElement>;

It will cast overlayEl to an initiated MutableRefObject that is the returning value of useRef:

function useRef<T = undefined>(): MutableRefObject<T | undefined>;

Yet in this case, the compiler will always think that overlayEl has a value.

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Add a type to the ref as mentioned by @Shanon Jackson:

const linkRef = useRef<HTMLLinkElement>(null);

And then, make sure you check for null value before using current:

if (linkRef.current !== null) {
  linkRef.current.focus();
}

This will satisfy Typescript. Whereas either by itself wouldn't.

Using any or casting in order to "trick" the compiler defeats the purpose of using Typescript, don't do that.

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If you want to "pass/skip" then this will do it const overlayEl: any = useRef(null);

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I think this is more succinct than the other answers here:

const ModalOverlay = (props: any[]) => {
  const overlayEl = useRef<HTMLDivElement>(null);
  useEffect(() => {
    overlayEl.current!.focus();
  });
  return <div {...props} ref={overlayEl} />;
}

You specify the type of the reference, and you state you know it's not null.

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You can also use optional chaining which is introduced in ES2020 instead of "if" statement for cleaner code

const myRef = useRef<HTMLLinkElement>(null);

myRef.current?.focus();

You can check its browser support at caniuse.

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If you really know that in executing time you dont have a error here then just put :

 (overlayEl as any).current 

If not, better use:

    if (typeof overlayEl !== 'undefined' &&
      typeof overlayEl.current !== 'undefined' &&
      overlayEl.current === null) {          
      return;
    }

    // Or

    try {
      // you code here ...
      // This is fine way to check by order -> parent.parent.FinalInstance
      // Also try & catch will handle all bad situation about current error
      overlay &&  overlayEl.current && overlayEl.current.focus();

    } catch(e){
       console.log("Real null >> ", e);     
    }

  // Suggest if i am wrong in syntax somewhere ,this is fast answer ;)

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Sometimes we useRef not only to hold an element but a value like data. somehow when I check if(something.current) return something.current not working even if i add && something.current!=null so i found that:

something.current! which says to typescript i know there is a value and bypass this issue.

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if you want to give no chance to typescript use this :

const iframe = useRef<HTMLIFrameElement | null>(null);

     if (
   typeof iframe !== "undefined" &&
   typeof iframe.current !== "undefined" &&
   iframe.current !== null
 ) {
   iframe?.current?.contentWindow='';
   );
 }

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Depends on what you prefer but a nice syntax and structure might be to create an interface:

interface divEL {
  current: HTMLDivElement | null;
}

This will keep your declaration clear and short and you can reuse the divEl interface for similar useRef hooks.

const overlayEl: divEL = useRef(null);

Then add focus() like this:

overlayEl.current!.focus();

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