How to check if optional object property exists?

December 7, 2022

I’m having trouble getting TypeScript to recognise that an optional property of an object is defined.

type OuterKeys = 'a' | 'b';

type Inner = {
    value: '';
}

type Outer = Partial<Record<OuterKeys, Inner>>;

const thisGivesError = (key: OuterKeys, outer: Outer) => {
    if (outer[key]) {
        console.log(outer[key].value);
    }
}

Even though I explicitly check if outer[key] is defined, I’m getting error when trying to access outer[key].value: Object is possibly 'undefined'.

How can I work around it? The only way I found is assigning outer[key] to a variable and having another if, like below. Is this the only way to go?

const thisWorks = (key: OuterKeys, outer: Outer) => {
    const o = outer[key];
    if (o) {
      console.log(o.value);
    }    
}

The playground with the code can be found here

Edit: I don’t want to use ! operator, as it feels like a workaround – why would I tell TypeScript that this value is set using non-null assertion, if I already checked for its existence with if? Same for .? – I already have the if, so this value for sure exists and is not optional.