TypeScript extend all value types of an interface / type

Question

Given an interface of known keys and values, is it possible to tell TypeScript that I want to have a type that has all the same keys of my original interface with value types extended via union of the original type and another type. Pseudo (sadly non-functional) TypeScript code that describes the desired behaviour:

type FancyCSSObject = { [key: keyof CSSObject]: CSSObject[key] | OtherType };

The above is not valid TypeScript, so my question is, how to express this in valid TypeScript code?

Dániel Emőd Kovács · Accepted Answer · 2020-01-17 11:09:05Z

2

Before I finished posting this, I realised the solution to this problem, however seeing that it's sort of a corner-case that's not documented in its entirety, I will share my solution, in case someone else needs this:

type FancyCSSObject = {
  [key in keyof CSSObject]: CSSObject[key] | OtherType;
};

I was close to the solution. I hope this helps someone.

answered Jan 17, 2020 at 11:09

Dániel Emőd Kovács

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