Creating a type guard for a custom type in TypeScript

Objective: Developing a type guard for a unique custom type.

Introducing my custom data type:

type AppProviders =  'box' | 'dropbox' | 'google';

This represents my initial endeavor to create a type guard, though it appears repetitive as I specify allowed values twice:

type AppProviders =  'box' | 'dropbox' | 'google';
const appProviders: AppProviders[] = [ 'box', 'dropbox', 'google' ];

function isAppProviders(provider): provider is AppProviders {
    return appProviders.includes(provider)
}

Is there an improved method for implementing a type guard with custom literal types?

Appreciate your insights

typescripttypeguards

Answer №1

The error you're encountering is due to:

https://www.typescriptlang.org/docs/handbook/advanced-types.html#type-aliases

type Easing = "ease-in" | "ease-out" | "ease-in-out";

By using string literal types, you can specify the precise value that a string should have. These types work well with union types, type guards, and type aliases, allowing you to achieve enum-like behavior with strings.

I had to research this further... came across an example at 

export const APP_PROVIDERS = ['a', 'b'] as const;
export type AppProviders = typeof APP_PROVIDERS[number];

export function isAppProviders(unknownString: string): unknownString is AppProviders {
  return (APP_PROVIDERS as unknown as string[]).includes(unknownString);
}

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