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I'm quite new to TypeScript and I need to declare a type like:

type U = { [key: string]: number }; // any string for the keys
type T = { [key: string]: U };      // any string for the keys too

Example of use:

const test: T = {
    'aaa': {
        'xxx': 777,
        'yyy': 888,
        'zzz': 999  // error expected
    },
    'bbb': {
        'xxx': 444,
        'yyy': 555,
        'ooo': 666  // error expected
    }
};

Is there a way, a better and probably more complex signature, to ensure that all U properties in T will have the same keys?

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6
  • are you saying you want only xxx, yyy and zzz or those 3 plus whatever else? Commented Apr 29, 2020 at 17:17
  • I don't understand the use case. Is ooo a typo and you want to see an error there? Is it intended but you want to see an error that zzz is missing? Are the particular keys xxx, yyy and zzz important or do you want the type to enforce that whatever the keys are of the nested properties they are the same across each property? Etc etc etc. I'm hoping you can write out some more examples of what things should and should not be allowed. Depending on how complex it is, your requirements might not be expressible as a specific type and you will instead need a generic constraint. Commented Apr 29, 2020 at 17:21
  • Sorry like you said I would like "to enforce that whatever the keys, they are the same across each property". Commented Apr 29, 2020 at 17:24
  • I think that's only going to work as a generic constraint like this. Does that work for you? Commented Apr 29, 2020 at 17:28
  • @jcalz I don't really like the need to use a function, but that's brilliant!!! That was the idea :) Commented Apr 29, 2020 at 19:22

2 Answers 2

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1

you can define an interface requiring some keys with an index signature:

interface U { 
  [key: string]: number;
  xxx: number;
  yyy: number;
  zzz: number; 
};

if you just want two objects to be required to have the SAME keys, you could do something with generics

interface BaseType<T> {
  xxx: T;
  yyy: T;
  zzz: T; 
}

type U = BaseType<number>;
type T = BaseType<U>;
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4 Comments

Sorry for not being clear enough, my English is not really good, xxx, yyy & zzz are just examples, they can be any string, I just need to ensure that they will be the same in each.
so you want type U and T to be constrained to have the same keys?
No I need every U property in T to have the same property keys.
ah ok got it i think
1

Define U with concrete properties you need and you will get desired behavior:

type U = {
    "xxx": number,
    "yyy": number;
    "zzz": number;
}
type T = { [key: string]: U }

That way you get an error on ooo:

const test: T = {
    'aaa': {
        'xxx': 777,
        'yyy': 888,
        'zzz': 999
    },
    'bbb': {
        'xxx': 444,
        'yyy': 555,
        'ooo': 666  // error expected here
    }
};

There is Playground link with full example.

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