Map a union to a string literal union of the type names

I have a type like this (in this case, the user would be a custom interface):

export type ArgumentTypes = string | number | boolean | user;

now, I would like to have a type of

export type ArgumentTypeNames = "string" | "number" | "boolean" | "user";

I’ve tried to search around myself for a while, but I have not been able to find a way to get the name. I tried to do ${ArgumentTypes} but that didn’t work.

I’m pretty new to advanced types ( Mapped, conditional, and so on ) so I would love an explanation of the answer, or of potential solutions.

Answer

AFAIK, as of today’s version of TypeScript, you can’t do that without an explicit mapping of string to "string", number to "number" etc. existing somewhere in your program. The mapping might look like this:

interface TypeMapping {
  string: string;
  number: number;
  boolean: boolean;
  user: user;
}

However, not all is lost. You could make this mapping the source of truth, and derive other types from it:

type ArgumentTypeNames = keyof TypeMapping;
type ArgumentTypes = TypeMapping[ArgumentTypeNames];

Try it.