I'm trying to type an object so that its' keys are all of a specific type, but in such a way that I still get intellisense when I access the the main object.

interface ObjectWithKeysOf<T>
{
    [key: string]: T;
}

const TEST: ObjectWithKeysOf<number> =
{
    prop1: 1,
    prop2: 2
};

Given the above, I hoped that the following would work, but it doesn't. Intellisense doesn't suggest prop1 as a property, and the code fails to compile.

const aNumber = TEST.prop1;

Is this possible?

CodePudding user response：

You can use the satisfies operator — introduced in TypeScript v4.9 — to constrain a type while preserving its specifics (including IntelliSense autocompletion):

Try it out in the TS Playground:

const test = {
    prop1: 1,
    prop2: 2,
} satisfies Record<string, number>;

test.prop1;
   //^? (property) prop1: number

test.prop2;
   //^? (property) prop2: number

test.prop3; /*
     ~~~~~
Property 'prop3' does not exist on type '{ prop1: number; prop2: number; }'. Did you mean 'prop1'?(2551) */

The same example, but combined with a const assertion:

TS Playground

const test = {
    prop1: 1,
    prop2: 2,
} as const satisfies Record<string, number>;

test.prop1;
   //^? (property) prop1: 1

test.prop2;
   //^? (property) prop2: 2

test.prop3; /*
     ~~~~~
Property 'prop3' does not exist on type '{ readonly prop1: 1; readonly prop2: 2; }'. Did you mean 'prop1'?(2551) */

See also the type utility Record<Keys, Type>.