Is it possible to alter the type without compromising on the optional state of the type?

I am interested in modifying the data types of specific fields to a customized type.

For instance, I would like to transform the b field of TSomeType into a custom type, resulting in TSomeTypeConverted type.

// initial type
type TSomeType = {
    a: boolean;
    b: string;
}
// desired outcome
type TSomeTypeConverted = {
    a: boolean;
    b: TCustomType;
}

A similar transformation can be achieved using the TConvertToCustomType method below.

type TConvertToCustomType<T, K extends keyof T> = Omit<T, K> & {[P in K]: TCustomType};

type TSomeTypeConverted = TConvertToCustomType<TSomeType, 'b'>
/* type TSomeTypeConverted = {
    a: boolean;
    b: TCustomType;
} */

However, this process results in all optional values becoming required.

// prior to modification
type TSomeType = {
    a: boolean;
    b: string;
    c?: string; // optional
}
type TSomeTypeConverted = TConvertToCustomType<TSomeType, 'b' | 'c'>
/* type TSomeTypeConverted = {
    a: boolean;
    b: TCustomType;
    c: TCustomType; // no longer optional
} */

Is there a way to change only the data type while preserving the optional status?

Answer №1

Utilizing Mapped Types in conjunction with Conditional Types is a more efficient approach than using Omit and object:

type ConvertToCustomType<T, K extends keyof T> = { [P in keyof T]: P extends K ? CustomType : T[P] };

type SomeTypeConverted = ConvertToCustomType<SomeType, 'b' | 'c'>
/* type SomeTypeConverted = {
    a: boolean;
    b: CustomType;
    c?: CustomType; // maintains optional property
} */

Test it out on Playground

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