Error in TypeScript: The object may be null when using the window.open method

Is there a way to implement this code in Typescript?

window.open(externalUrl, '_blank').focus();

Encountering the following TypeScript error:

Object is possibly 'null'.ts(2531)

I attempted the following solution without success:

if (typeof window !== 'undefined' && window && externalUrl !== '' && window.open(externalUrl, '_blank')) {
      window.open(externalUrl, '_blank').focus();
    }
  

Answer №1

If you are using an integrated development environment (IDE), you may notice a red squiggly line highlighting the problematic part of your expression:

  window.open(externalUrl, '_blank').focus(); // issue
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
//Object is possibly 'null'

The error message indicates that

window.open(externalUrl, '_blank')
could potentially be null, not necessarily the entire window object. Therefore, in addressing this particular error, you do not need to check if window itself is null, assuming there is one available in your JavaScript runtime environment.

This error aligns with the information provided in the MDN documentation for Window.open(), stating that a return value of null signifies an inability to open the window. To tackle this issue, you should examine the return value and only invoke focus() if it is not null. Attempting to do so differently like below will still trigger an error:

if (window.open(externalUrl, '_blank')) {
    window.open(externalUrl, '_blank').focus(); // still error
}

The compiler does not assume that a successful first call to window.open() guarantees success for subsequent calls. While it might seem improbable that the second call would fail when the first one doesn't, there is no certainty. Additionally, opening two windows simultaneously may not align with your intentions.


So, how can we handle this situation correctly? The solution that works well with most TypeScript versions involves storing the result in a variable and then verifying it:

const w = window.open(externalUrl, '_blank');
if (w) {
    w.focus(); // now okay
}

For TypeScript 3.7 and later, you have access to the optional chaining operator, ?.. This allows you to express the above logic more succinctly as follows:

window.open(externalUrl, '_blank')?.focus(); // also works

In due time, the optional chaining operator is expected to become a part of JavaScript itself. Until then, the code is transpiled into JavaScript similar to:

// Transpiled JavaScript 
(_a = window.open(externalUrl, '_blank')) === null || _a === void 0 ? void 0 : _a.focus();

This approach stores the result of window.open() in a variable, akin to what we did with

w</code earlier, ensuring that <code>focus()
is only invoked if the result is neither null nor undefined. Either method should suit your needs adequately.


I hope this explanation proves helpful; best of luck!

Link to Code Playground

Answer №2

After reading the informative response from @jcalz, I discovered another neat one-liner that utilizes the exclamation mark !:

window.open(externalUrl, '_blank')!.focus();

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