[SOLVED] How to generate value from type?


Given the following TypeScript type:

type options = 
    | {label: "a", fields: Array<Object>}
    | {label: "b", fields: Array<Object>}
    | {label: "c", fields: Array<Object>}

Is there a way to use the options type to generate a value (e.g. ["a", "b", "c"]) that can be used at runtime?

If so, how can ["a", "b", "c"] be generated from the options type?


As others have suggested – you can’t generate a value from a type, as the concepts of types don’t exist at runtime.

What you could do instead, is generate your types from values. The typeof operation is helpful here

const labels = ["a", "b", "c"] as const; 

type PossibleValues = typeof labels[number]; // type PossibleValues = "a" | "b" | "c"
type Options = {label: PossibleValues, fields: Array<unknown>}; 

const a: Options = {
    label: "a", 
    fields: [], 

const notRight: Options = {
    label: "fobar", //Type '"fobar"' is not assignable to type '"a" | "b" | "c"'.(2322)
    fields: [], 

Playground Link

What’s happening here:

const labels = ["a", "b", "c"] as const; 

We need to use the as const assertion here to tell typescript that labels isn’t an array of any ol’ strings, it’s an array of these specific strings.

type PossibleValues = typeof labels[number];

We are doing two things here:

  • First we are saying ‘what is the type of labels‘, which is Array<"a" | "b" | "c">.
  • Next we say, ‘what is the type of of the things accessible by number‘. (ie. what are all the possible types of the arrays contents. Remember an array also has prototype functions like .forEach etc).

Now we can create our Options type.

Answered By – dwjohnston

Answer Checked By – Terry (BugsFixing Volunteer)

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