TypeScript: how to make a property conditional from an existing type?

Question

So let's say I have

type Link = {
  text: string;
  link: string;
}

interface BigLink extends Link {
  some: number;
  something: string;
  else: string;
}

but I have a variable that shares all these properties except that the link property can be optional. I don't want to create a whole new type just to change one field to link?: string. Is there a way to do something like

type OptionalBigLink = BigLink & { link?: string }

so that I can overwrite the link field and make it optional. ^ The above code still throws an error when I don't pass in the link property.

Dean · Accepted Answer · 2022-07-14 18:32:57Z

3

Yes you can.

type OptionalBigLink = Partial<Pick<BigLink, "link">> & Omit<BigLink, "link">;

Some tests in the playground

answered Jul 14, 2022 at 18:32

Dean

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Tobias S. · Accepted Answer · 2022-07-14 18:33:45Z

3

You can use Omit to remove the property from BigLink and then do the intersection.

type OptionalBigLink = Omit<BigLink, "link"> & { link?: string }

answered Jul 14, 2022 at 18:33

Tobias S.

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1 Comment

Hackerman Over a year ago

Thank you this works perfectly as well - might be easier to use for my case since link is a simple scalar type.

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