Resolved: Shorthand for conditionally creating an array out of an optional value?

In this post, we will see how to resolve Shorthand for conditionally creating an array out of an optional value?

Question:

I have a function that optionally accepts an array. I have other variables that may be undefined, but if they are defined I would like to create an array from it:

function test(array?: string[]) {
    console.log(array);
}

let value: string | undefined;
value = 'abc';

const temp = value ? [value] : undefined; // Shorthand?
test(temp)

Is there a shorthand way of creating the temp variable with an an array of strings or pass undefined? I tried to do the following but didn’t achieve it:

// If value is 1, true, or 'some string' it will take that value not the array
const temp = value && [value];

// Passes back an empty array if value is undefined
const temp = [value].filter(Boolean),

Best Answer:

As it was pointed out in the comments, just create a helper function for this:

const arrOrUndef = <T,>(val: T | undefined): T[] | undefined => val !== undefined ? [val] : undefined;

Then just use:

let value: string | undefined = 'abc';
test(arrOrUndef(value))

TS Playground Link
Also, thanks to pilchard in the comments for noting that it is better to use a generic, rather than just the string type in your question or the any type. Also, JDB for noting you must use !== undefined, rather than simply checking for a falsy value (as the original question does).

If you have better answer, please add a comment about this, thank you!

Source: Stackoverflow.com

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