Set every attribute inside a Typescript interface as non-mandatory

Question

Set every attribute inside a Typescript interface as non-mandatory

I have defined an interface within my software:

interface Asset {
  id: string;
  internal_id: string;
  usage: number;
}

This interface is a component of another interface named Post:

interface Post {
  asset: Asset;
}

In addition, there is an interface designed for a post draft, where the asset object can be incompletely constructed:

interface PostDraft {
  asset: Asset;
}

My goal is to allow a PostDraft object to contain a partially complete asset object while still enforcing type checks on the available properties (rather than using any).

Essentially, I am seeking a method to generate the following structure:

interface AssetDraft {
  id?: string;
  internal_id?: string;
  usage?: number;
}

without fully redefining the original Asset interface. Is there a technique to achieve this? If not, what would be the most efficient approach to organizing my types in this scenario?

typescript

Answer 1

Answer №1

In TypeScript versions prior to 2.1, achieving this isn't possible without creating an additional interface featuring optional properties. However, the use of mapped types in TypeScript 2.1 and above makes this task achievable.

To accomplish this, leverage the inherent Partial<T> type provided by TypeScript.

interface PostDraft {
    asset: Partial<Asset>;
}

Consequently, all attributes within the asset object become optional, enabling actions like:

const postDraft: PostDraft = {
    asset: {
        id: "some-id"
    }
};

Understanding Partial<T>

The Partial<T> concept is a mapped type that renders each property within the specified type as optional (via the ? symbol). Its definition can be found here.

type Partial<T> = {
    [P in keyof T]?: T[P];
};

For further insights on mapped types, refer to resources available here and within the TypeScript handbook here.

Enhancing with Deep Partiality

If a deeply partial implementation working recursively on objects is desired, TS version 4.1 and beyond offer the following type structure:

type DeepPartial<T> = {
    [P in keyof T]?: T[P] extends object ? DeepPartial<T[P]> : T[P];
};

Answer 2

In TypeScript versions prior to 2.1, achieving this isn't possible without creating an additional interface featuring optional properties. However, the use of mapped types in TypeScript 2.1 and above makes this task achievable.

To accomplish this, leverage the inherent Partial<T> type provided by TypeScript.

interface PostDraft {
    asset: Partial<Asset>;
}

Consequently, all attributes within the asset object become optional, enabling actions like:

const postDraft: PostDraft = {
    asset: {
        id: "some-id"
    }
};

Understanding Partial<T>

The Partial<T> concept is a mapped type that renders each property within the specified type as optional (via the ? symbol). Its definition can be found here.

type Partial<T> = {
    [P in keyof T]?: T[P];
};

For further insights on mapped types, refer to resources available here and within the TypeScript handbook here.

Enhancing with Deep Partiality

If a deeply partial implementation working recursively on objects is desired, TS version 4.1 and beyond offer the following type structure:

type DeepPartial<T> = {
    [P in keyof T]?: T[P] extends object ? DeepPartial<T[P]> : T[P];
};

Answer 3

Answer №2

In order to create a specific AssetDraft interface, I can utilize both the extends and Partial keywords:

interface Asset {
  id: string;
  internal_id: string;
  usage: number;
}

interface AssetDraft extends Partial<Asset> {}

Answer 4

In order to create a specific AssetDraft interface, I can utilize both the extends and Partial keywords:

interface Asset {
  id: string;
  internal_id: string;
  usage: number;
}

interface AssetDraft extends Partial<Asset> {}

Answer 5

Answer №3

In the interface, properties can be either optional or mandatory. It is not possible to mix and match the same property as both optional and mandatory within the same interface.

One solution is to create an interface with optional properties for AssetDraft, and then a class with mandatory properties for Asset:

interface AssetDraft {
    id?: string;
    internal_id?: string;
    usage?: number;
}

class Asset {
    static DEFAULT_ID = "id";
    static DEFAULT_INTERNAL_ID = "internalid";
    static DEFAULT_USAGE = 0;

    id: string;
    internal_id: string;
    usage: number;

    constructor(draft: AssetDraft) {
        this.id = draft.id || Asset.DEFAULT_ID;
        this.internal_id = draft.internal_id || Asset.DEFAULT_INTERNAL_ID;
        this.usage = draft.usage || Asset.DEFAULT_USAGE;
    }
}

The default values in this example are set as static members, but you could obtain them differently or handle missing values by throwing an error.

Using this method makes it easier to work with JSON data received from servers or similar sources. The interfaces represent the data structure, while the classes act as models constructed using the JSON data as initial values.

Answer 6

In the interface, properties can be either optional or mandatory. It is not possible to mix and match the same property as both optional and mandatory within the same interface.

One solution is to create an interface with optional properties for AssetDraft, and then a class with mandatory properties for Asset:

interface AssetDraft {
    id?: string;
    internal_id?: string;
    usage?: number;
}

class Asset {
    static DEFAULT_ID = "id";
    static DEFAULT_INTERNAL_ID = "internalid";
    static DEFAULT_USAGE = 0;

    id: string;
    internal_id: string;
    usage: number;

    constructor(draft: AssetDraft) {
        this.id = draft.id || Asset.DEFAULT_ID;
        this.internal_id = draft.internal_id || Asset.DEFAULT_INTERNAL_ID;
        this.usage = draft.usage || Asset.DEFAULT_USAGE;
    }
}

The default values in this example are set as static members, but you could obtain them differently or handle missing values by throwing an error.

Using this method makes it easier to work with JSON data received from servers or similar sources. The interfaces represent the data structure, while the classes act as models constructed using the JSON data as initial values.

Answer 7

Answer №4

Aside from David Sherret's response, I would like to provide my own insights on how the implementation can be done directly without using the Partial<T> type, in order to enhance clarity on the topic.

interface IAsset {
  id: string;
  internal_id: string;
  usage: number;
}

interface IPost {
  asset: IAsset;
}

interface IPostDraft {
  asset: { [K in keyof IAsset]?: IAsset[K] };
}

const postDraft: IPostDraft = {
  asset: {
    usage: 123
  }
};

Answer 8

Aside from David Sherret's response, I would like to provide my own insights on how the implementation can be done directly without using the Partial<T> type, in order to enhance clarity on the topic.

interface IAsset {
  id: string;
  internal_id: string;
  usage: number;
}

interface IPost {
  asset: IAsset;
}

interface IPostDraft {
  asset: { [K in keyof IAsset]?: IAsset[K] };
}

const postDraft: IPostDraft = {
  asset: {
    usage: 123
  }
};

Answer 9

Answer №5

Why not consider utilizing

interface AssetDraft = { id?: string, internal_id?: string; usage?: number; }

and then expanding it for the Asset object as follows:

interface Asset extends Required<AssetDraft> {}

?

While unconventional, is there a discernible distinction in this alternate approach?

Answer 10

Why not consider utilizing

interface AssetDraft = { id?: string, internal_id?: string; usage?: number; }

and then expanding it for the Asset object as follows:

interface Asset extends Required<AssetDraft> {}

?

While unconventional, is there a discernible distinction in this alternate approach?

Answer 11

Answer №6

Have you ever considered forcefully instantiating an empty object?

const newDraft = <PostDraft>{}
newDraft.id = 123
newDraft.internal_id = 456
newDraft.usage = 789

If this is a necessity for you, one potential solution could be to create a d.ts interface based on a predefined template that encompasses both optional and typed properties.

As previously mentioned by Nitzan, in Typescript interfaces, properties can either be optional or mandatory.

Answer 12

Have you ever considered forcefully instantiating an empty object?

const newDraft = <PostDraft>{}
newDraft.id = 123
newDraft.internal_id = 456
newDraft.usage = 789

If this is a necessity for you, one potential solution could be to create a d.ts interface based on a predefined template that encompasses both optional and typed properties.

As previously mentioned by Nitzan, in Typescript interfaces, properties can either be optional or mandatory.

Set every attribute inside a Typescript interface as non-mandatory

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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