CODEWITHSUNDEEP
typescriptgenerics
Murphy4
Murphy4

Reputation: 1531

Accessing type of a property from an extended generic

Consider the following setup:

genericModel.ts

export default class GenericModel<TModelId> {
    id!: TModelId;

    constructor(defaultId: any){
        this.id = defaultId;
    }
}

genericStore.ts

class GenericStore<TModel extends GenericModel<any>> {

    @observable loadedModels = observable.map<typeof(TModel.id), TModel>();

    constructor(service: TService){
        this.service = service;
    }

}

export default GenericStore;

I know that you cannot call typeof on a generic and why -- this is only to illustrate intent. The observable decorator is from mobx, but it is just a unique structure that requires the type of the model id to use as an identifier, and the model type itself. Now I know I can make it work like this:

class GenericStore<TModelId, TModel extends GenericModel<TModelId, any>, TService> {
    @observable loadedModels = observable.map<TModelId, TModel>();

    constructor(service: TService){
        this.service = service;
    }

}

export default GenericStore;

But when I am extending the generic store, the type of the model is already declared, so I am thinking there should be a way to infer it without having to declare the TModelId explicitly. I'm new to generics so this may just not be possible but if that's the case I just want confirmation.

Upvotes: 0

Views: 27

Answers (1)

elderapo
elderapo

Reputation: 374

I think TModel["id"] is what you're looking for. Complete example:

class GenericModel<ID> {
    constructor(public id: ID) { }
}

class GenericStore<TModel extends GenericModel<any>> {
    public aaa!: TModel["id"];
}


const store1 = new GenericStore<GenericModel<number>>();

store1.aaa; // number

Upvotes: 1

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