Object index key type in Typescript

I defined my generic type as

interface IDictionary<TValue> {
    [key: string|number]: TValue;
}

But TSLint's complaining. How am I supposed to define an object index type that can have either as key? I tried these as well but no luck.

interface IDictionary<TKey, TValue> {
    [key: TKey]: TValue;
}

interface IDictionary<TKey extends string|number, TValue> {
    [key: TKey]: TValue;
}

type IndexKey = string | number;

interface IDictionary<TValue> {
    [key: IndexKey]: TValue;
}

interface IDictionary<TKey extends IndexKey, TValue> {
    [key: TKey]: TValue;
}

None of the above work.

So how then?

You can achieve that just by using a IDictionary<TValue> { [key: string]: TValue } since numeric values will be automatically converted to string.

Here is an example of usage:

interface IDictionary<TValue> {
    [id: string]: TValue;
}

class Test {
    private dictionary: IDictionary<string>;

    constructor() {
       this.dictionary = {}
       this.dictionary[9] = "numeric-index";
       this.dictionary["10"] = "string-index"

       console.log(this.dictionary["9"], this.dictionary[10]);
    }
}
// result => "numeric-index string-index"

As you can see string and numeric indices are interchangeable.

In javascript the keys of object can only be strings (and in es6 symbols as well).
If you pass a number it gets converted into a string:

let o = {};
o[3] = "three";
console.log(Object.keys(o)); // ["3"]

As you can see, you always get { [key: string]: TValue; }.

Typescript lets you define a map like so with numbers as keys:

type Dict = { [key: number]: string };

And the compiler will check that when assigning values you always pass a number as a key, but in runtime the keys in the object will be strings.

So you can either have { [key: number]: string } or { [key: string]: string } but not a union of string | number because of the following:

let d = {} as IDictionary<string>;
d[3] = "1st three";
d["3"] = "2nd three";

You might expect d to have two different entries here, but in fact there's just one.

What you can do, is use a Map:

let m = new Map<number|string, string>();
m.set(3, "1st three");
m.set("3", "2nd three");

Here you will have two different entries.