In TypeScript how do I declare a generic with 2 arguments, where the first is a key of the second that returns a specific type?

Question

I'm looking to have a generic class like this one:

export abstract class Foo<ID extends keyof T, T> {
    bar(T t) {
        const s: string = t[ID];
    }
}

Obviously the code above can't infer the type of t[ID] and we're getting an implicit any. How can I enforce using generics T[ID] will be a string?

This isn't valid TypeScript, I assume you mean `T t` to be `t: T`, but where do you expect the value for `t[ID]` to come from? Is `id` an argument or class property? — Aaron Beall, Apr 13 '18 at 18:00

jcalz · Accepted Answer · 2018-04-13T15:25:45.820

Your code doesn't compile so I changed it to something that will:

export abstract class Foo<ID extends string, T extends Record<ID, string>> {

  // add a property of type ID.
  constructor(public id: ID) { } 

  // change the Java-esque (T t) to (t: T)
  bar(t: T) {
    //use the id property to index t; can't use type ID at runtime
    const s: string = t[this.id]; 
  }

}

The way to enforce the relationship you want is: instead of constraining ID to keyof T, you constrain T to Record<ID, string> where Record<K, V> is a type from the standard TypeScript library describing any object with keys from K and values from V. Hope that helps. Good luck.

In TypeScript how do I declare a generic with 2 arguments, where the first is a key of the second that returns a specific type?

1 Answers1