Is there a way in TypeScript to say "Include" (opposite of Exclude)?

user_1096967 0
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I have a String Union that looks like: type AlphabetLike = 'a' | 'b' | 'c' | 'zeta' | 'beta' | 'gamma' | 'mu'; I want to be able to construct the type type Alphabet = 'a' | 'b' | 'c'. I could do this by using Exclude and removing zeta, beta, gamma, and mu but that's a lot of things to remove and if more values are added to AlphabetLike, it changes Alphabet. I want to know if my value from "AlphabetLike" is ever removed from that type.

Is there a way to do an Include instead of an Exclude?

angular typescript reactjs strong-typing
3 Answers

There is an Extract<T, U> utility type which acts like the opposite of the Exclude<T, U> utility type:

type AlphabetLike = 'a' | 'b' | 'c' | 'zeta' | 'beta' | 'gamma' | 'mu';

type Alphabet = Extract<AlphabetLike, 'a' | 'b' | 'c' | 'zzz'>;
// type Alphabet = "a" | "b" | "c"

Note that nothing forces the second argument (called U here) to either Extract<T, U> or Exclude<T, U> to contain only elements from the first argument (called T here). Hence that 'zzz' above doesn't have an effect on anything.

Playground link to code

user_28872180

Instead of custructing Alphabet as an exclusion of types from AlphabetLike, construct AlphabetLike as a union of different alphabet types.

type Alphabet = 'a' | 'b' | 'c';
type GreekAlphabet = 'zeta' | 'beta' | 'gamma' | 'mu';
type AlphabetLike = Alphabet | GreekAlphabet;
user_24973350

A double "exclude" can act as an include.

First, create a list of all of the items you want to exclude if you were using exclude.

type GreekAlphabet = Exclude<AlphabetLike, 'a' | 'b' | 'c'>

This results in a set that is: 'zeta' | 'beta' | 'gamma' | 'mu' We can use that set with Exclude to give us back the values we want:

type Alphabet = Exclude<AlphabetLike, GreekAlphabet>

This results in: 'a' | 'b' | 'c' which is effectively Include<AlphabetLike, 'a'|'b'|'c'> and will protect you if an option in use is removed from "AlphabetLike".

user_10969670
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