Typescript: define a primitive within another

user_8635165 0
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The issue is that when I receive a response back from an API, it is the most inconsistent response I think I've dealt with so far. Anyways, what I am trying to do is give more structure with Typescript. For example:

interface Response {
  foo: string
}

The problem that I have is that EVERYTHING is a string: foo could be "true" || "123" || "undefined" || "null"

but there are some values that I know will be there. So I know that this field will always be of type string but within that string is going to be a number

what I would like is to create a class or an interface that looks like this:

interface Response {
  foo: string<number>
}

because now I just have notes next to everything like:

interface Response {
  service: string // string("true")
}
template-literals typescript
1 Answers

Consider this example:

type Primitives = boolean | number | null | undefined

type MakeString<T extends Primitives> = `${T}`

// "undefined" | "null" | `${number}` | "false" | "true"
type Result = MakeString<Primitives>

Playground

Please see distributive-conditional-types if you are confused by T extends any

TL;TR Wrapping T into template literal string distributes the union.

Thanks @Clarity

user_84952540
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