Employ both mandatory and partial input methods

I'm having trouble grasping the concept of Advanced Types in Typescript.

I'm looking to create a type that includes both mandatory and optional properties, with the properties easily identifiable in a list.

Currently, I have a type for required properties and another type for optional properties.

type BaseProperties = 
    | 'price'
    | 'cost'
    | 'location';

type Features =
    | 'radio'
    | 'wings'
    | 'tires'
    | 'rockets'
    | 'slushie_machine';

The desired type I want to achieve is:

type WithFeatures = {
    price: string;
    cost: string;
    location: string;
    radio?: string | number;
    wings?: string | number;
    tires?: string | number;
    rockets?: string | number;
    slushie_machine?: string | number;
};

Additionally, I want an array of:

public ThingsWithFeatures: WithFeatures[] = [];

I attempted to use:

type WithFeatures = Required<BaseProperties> & Partial<Features>;

...but unfortunately, it did not yield the desired outcome.

What steps should I take to properly utilize required and partial to achieve the WithProperties type as specified?

Answer №1

Issue

The key here is not the actual BaseProperties and Features themselves (as they are just string literal types and Required<"foo"> is simply "foo"), but rather the properties within an object that bear their names.

Resolution

To address this, transform the property names into object types that utilize them. The initial type will consist of an object where the keys are BaseProperties and the values are of type string. This can be achieved with the following code snippet:

Record<BaseProperties, string>

Since all properties are mandatory by default, the Required helper is not necessary. However, the Partial helper is required for the second type.

Partial<Record<Features, string>>

Finally, the ultimate structure is an intersection (&) of these two types.

type WithFeatures = Record<BaseProperties, string> & Partial<Record<Features, string>>;

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