Is it possible to create two interface variations; one that includes all optional fields and another that includes all required fields, without repeating yourself in the process?

I developed a specific interface named IPreferences. Here's how it's constructed:

export interface IPreferences { 
    genres: Genres[],
    singers: Singer[],
    volume: number
}

As I provide users with the ability to modify their preferences by updating one or more fields, I introduced another interface called IPreferenceUpdateRequest which is structured like this:

export interface IPreferencesUpdateRequest { 
    genres?: Genres[],
    singers?: Singer[],
    volume?: number
}

However, I've realized that having two very similar interfaces is not an efficient approach.

Is there a way for me to handle this situation while adhering to the principle of DRY ?

Answer №1

A solution to your question is to utilize the Partial<IPreferences> method. This method is part of the TypeScript standard library and is a type of mapped type that transforms all properties of T into optional versions:

export type IPreferencesUpdateRequest = Partial<IPreferences>

We hope this solution is helpful to you. Good luck with your project!

Answer №2

A more efficient way to achieve this in TypeScript is by using the Partial<T> type. Here's an example:

const partialUpdate: Partial<IPreferences> = { /* your properties here */ };

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