Is it Possible to Expand Union Type Aliases in TypeScript?

One challenge I'm facing is limiting string fields to specific values during compile time, while also allowing these values to be extendable. Let's explore a simplified example:

type Foobar = 'FOO' | 'BAR';

interface SomeInterface<T extends Foobar> {
  amember: T;

  [key: string]: string; // must remain in place
}

// testing it out

const yes = {
    amember: 'FOO'
} as SomeInterface<'FOO'>; // compiles successfully

//    const no = {
//        amember: 'BAZ'
//    } as SomeInterface<'BAZ'>; // Expected error: Type '"BAZ"' does not satisfy the constraint 'Foobar'

// all good so far
// Now here comes the issue

abstract class SomeClass<T extends Foobar> {
  private anotherMember: SomeInterface<T>;
}

type Foobarbaz = Foobar | 'BAZ';

class FinalClass extends SomeClass<Foobarbaz> { // Oops, this doesn't work anymore
}

The generated error message states:

Type 'Foobarbaz' does not satisfy the constraint 'Foobar'. Type '"BAZ"' is not assignable to type 'Foobar'.

So my question is: How can I restrict a 'type' in TypeScript to only certain strings, yet allow for extension with additional strings? Or is there a better solution that I might be missing?

Currently using Typescript 2.3.4 but willing to upgrade to 2.4 if it offers any solutions.

Answer №1

It seems like there might be a misunderstanding in the use of the term "extendable" compared to the keyword extends. When referring to an object as "extendable," it suggests the ability to broaden its scope to accommodate more values. On the other hand, when something extends a type, it means restricting the range to accept fewer values.

SomeInterface<T extends Foobar>
can essentially fall into one of these four categories:

  • SomeInterface<'FOO'|'BAR'>: amember is limited to either 'FOO' or 'BAR'
  • SomeInterface<'FOO'>: amember must only be 'FOO'
  • SomeInterface<'BAR'>: amember should solely hold the value 'BAR'
  • SomeInterface<never>: amember cannot be assigned any value

It's possible that this may not align with your intended usage, but ultimately, only you can determine that.


Alternatively, if you wish for SomeInterface<T> where T can always be either FOO or BAR, along with other possible string values, TypeScript lacks direct support for setting a lower boundary on T. For instance, using

SomeInterface<T extends super Foobar extends string>
, which does not adhere to TypeScript syntax.

However, if your primary concern lies in defining the type of amember rather than T, and you desire amember to accept either FOO or BAR, along with additional string values, you could specify it this way:

interface SomeInterface<T extends string = never> {
  amember: Foobar | T;
  [key: string]: string; 
}

Within this setup, T represents any added literals permitted. Should no extras be allowed, utilize never or exclude the type parameter (with default set to never).

See this example in practice:

const yes = {
    amember: 'FOO'
} as SomeInterface; // valid, 'FOO' matches Foobar

const no = {
    amember: 'BAZ'
} as SomeInterface; // invalid, 'BAZ' doesn't match Foobar

abstract class SomeClass<T extends string> {
  private anotherMember: SomeInterface<T>;
}

class FinalClass extends SomeClass<'BAZ'> { 
} // acceptable, 'BAZ' has been included

// To confirm:
type JustChecking = FinalClass['anotherMember']['amember']
// Result: 'BAZ' | 'FOO' | 'BAR'

Have I addressed your query? Trusting this information proves useful.

Answer №2

In order to accomplish your goal, you have the option to utilize intersection using the & symbol.

type Foobar = 'FOO' | 'BAR';
type FoobarBaz = Foobar | 'BAZ'; // or: 'BAZ' | Foobar

https://i.sstatic.net/7wqIJ.png

Answer №3

How can I restrict a 'type' in typescript to only certain strings, yet allow it to be expanded with additional strings?

I'm not sure what your specific issue is, but one approach could be introducing another generic parameter and limiting the type by specifying that it extends that parameter. Typescript 2.3 now supports default types for generic parameters, so with 'Foobar' as the default, you can use 'SomeInterface' with a single argument as usual. When you need it to extend something else, simply provide that explicitly:

type Foobar = 'FOO' | 'BAR';

interface SomeInterface<T extends X, X extends string=Foobar> {
  amember: T;

  [key: string]: string; // must remain
}

// testing

const yes = {
    amember: 'FOO'
} as SomeInterface<'FOO'>; // compiles correctly


abstract class SomeClass<T extends X, X extends string=Foobar> {
  private anotherMember: SomeInterface<T, X>;
}

type Foobarbaz = Foobar | 'BAZ';

class FinalClass extends SomeClass<Foobarbaz, Foobarbaz> { 
}

UPDATE

After further consideration, one solution might involve representing union types like 'Foobar' as 'keyof' an artificial interface type used solely for key representation (value types are irrelevant). By extending this interface, you effectively expand the set of keys:

interface FoobarKeys { FOO: { }; BAR: { } };

type Foobar = keyof FoobarKeys;

interface SomeInterface<X extends FoobarKeys = FoobarKeys> {
  amember: keyof X;

  [key: string]: string; // must remain
}



abstract class SomeClass<X extends FoobarKeys = FoobarKeys> {
    protected anotherMember: SomeInterface<X> = {
        amember: 'FOO'
    };

    protected amethod(): void { 
        this.bmethod(this.anotherMember); // no error
    }

    protected bmethod(aparam: SomeInterface<X>): void { 

    }
}

// extension example

interface FoobarbazKeys extends FoobarKeys { BAZ: {} };
type Foobarbaz = keyof FoobarbazKeys;

class FinalClass extends SomeClass<FoobarbazKeys> {
    private f() {
        this.bmethod({amember: 'BAZ'})
    } 
}

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