using an const array as a source for a string literal union

Given this:

const weatherIcons = [
  'sun',
  'rain',
  'snow',
] as const

const weatherIcons = [
  'sun',
  'rain',
  'snow',
] as const

How can I create a type like

type Weather = {
  icon: typeof weatherIcons[number]
}

type Weather = {
  icon: typeof weatherIcons[number]
}

Full example: https://www.typescriptlang.org/play/?#code/JYWwDg9gTgLgBAbzgLzgXzgMyhEcDkyEAJvgLABQoksicMAnmAKbpY574CGUA1oy3IVKAYwgA7AM7wA7sy4wAFsygBJMVLgBeOAG1KcApICu4-ABoDBKF2BnLFQ-kniIMi5QC6cLpLgbpSiCKAPhkAGURZRAubRQAOggAIwArZhEYAAoEKxhmcAAuBPFjECSVTIBKB0NgDSLkeOYSkEy5BWU1AOrKNErKAVYALUjo2J1Gu0wVAB5BiEwUUfyuAD4BpmHlmIB1eSUVdQk4kaiV3Xw6iXxPSgB6O8M4AD0AfmDRCWkfbfH6TeyuXyYCK+Ba5SgHkccCu4lB0igdgA5hY4A84DMALSYwyvPG9foUQZwACCvzi80WXF+8SmKg2LFJvz2HUOGjiZLOMQusJu90ehjeQA

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4 Replies

ssalbdivad•7mo ago

const weatherIcon = type("===", ...weatherIcons)

const weatherIcon = type("===", ...weatherIcons)

You can also inline it with a tuple expression like this:

const t = type({
    icon: ["===", ...weatherIcons]
})

const t = type({
    icon: ["===", ...weatherIcons]
})

though generally I'd recommend defining types like that as standalone first then referencing them in your definitions so they're more reusable.

AlexWayneOP•7mo ago

Nice. So the '===' just infers the type from what follows it?

ssalbdivad•7mo ago

Yeah, it accepts spread args and validates at runtime that the input === one of them So it can be used for other references like symbols and objects as well if you want reference equality

AlexWayneOP•7mo ago

nice. Thank you.

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using an const array as a source for a string literal union

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