Typescript - Getting the type of properties in my interface to avoid using any

Question

I have an interface like:

interface OktaUser {
    someId: number;
    email: string;
    email_verified: boolean;
}

I then have some method to get these properties out as requested:

async getOktaUserProperty(key: keyof OktaUser): Promise<any> {
        // Do some stuff

        return oktaUser[key];
}

Is there any way I can avoid using any in the return, I could manually do Promise<number | string | boolean> but makes it a little cumbersome to maintain.

Any ideas?

Thanks.

Shaun Luttin · Accepted Answer · 2020-02-18 03:46:14Z

1

Try this playground link. It uses several features of TypeScript including generics and index types.

interface OktaUser {
    someId: number;
    email: string;
    email_verified: boolean;
}

function getOktaUserProperty<TKey extends keyof OktaUser>(key: TKey): OktaUser[TKey] {
    const x: OktaUser = {
        someId: 10,
        email: 'foo',
        email_verified: true,
    };

    return x[key]
}

const y = getOktaUserProperty('email'); // string
const z = getOktaUserProperty('someId'); // number

answered Feb 18, 2020 at 3:46

Shaun Luttin

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2 Comments

userMod2 Over a year ago

Awesome that works - but I have no idea what it means lol So are we declaring a generic function, with some TKey being a property of OktaUser, - however at that stage why is the extends keyword needed?

Shaun Luttin Over a year ago

@userMod2 It takes some time to get the hang of it. There are detail in the index types link. Reading that section and understanding what it means will take you a long way to becoming a TypeScript expert.

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