How to restrict object types based on a value in the object?

user_13210593 0
Viewed 74

For now I'm using this type alias:

export type Country = {
  name: 'USA' | 'Russia' | 'Bulgaria' | 'Romania' | 'Austria' | 'Great Britain' | 'Italy',
  code: 'us' | 'ru' | 'bg' | 'ro' | 'au' | 'gb' | 'it'
};

But it allows to create incorrect objects:

{ name: 'USA', code: 'ru'}

What is the best way to make a type for this purpose?

typescript javascript
2 Answers

You could use a discriminated union approach:

type Country = {
    name: 'USA',
    code: 'us',
} | {
    name: 'Russia',
    code: 'ru',
} | {
    name: 'Bulgaria',
    code: 'bg',
} | {
    name: 'Romania',
    code: 'ro',
} | {
    name: 'Austria',
    code: 'au',
} | {
    name: 'Great Britain',
    code: 'gb',   
} | {
    name: 'Italy',
    code: 'it',   
};

// works
const myCountry: Country = {
    name: 'USA',
    code: 'us',
};

// does not
const myCountry2: Country = {
    name: 'USA',
    code: 'ru',
};

More reading here

You could also use a paramater map, but then you would have to specify the paramater on the interface:

type CountryMap = {
    USA: 'us',
    Russia: 'ru',
    Bulgaria: 'bg',
    Romania: 'ro',
    Austria: 'au',
    'Great Britain': 'gb',
    Italy: 'it',
};

interface ICountry<T extends keyof CountryMap> {
    name: T;
    code: CountryMap[T];
}

// works
const myICountry: ICountry<'USA'> = {
    name: 'USA',
    code: 'us'
};

// does not
const myICountry2: ICountry<'USA'> = {
    name: 'USA',
    code: 'ru',
};

Working typescript playground for both examples here

user_119492780

That isn't good to constuct such types manually. Provide single type definging courtry codes (you already got it) and accompany it with exported funciton of type CountryCode -> CountryName. You may want to require some other parametrs that define naming strategy (localization, short/full names, etc.).

user_74938400
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