CODEWITHSUNDEEP
typescripttypescript-generics
AbhishekGowda28
AbhishekGowda28

Reputation: 1054

Typescript - type to check if the string contains only specific characters

How to have type check for strings that we know will contains only certain values. Example: const binary = '1010000101000'; We know that binary values represented in decimal would only be 1's & 0's. To have a better type check, what would be a good type definition for these kinds of values.

type Binary = '0' | '1'; wouldn't work, because these would be representing only single characters of the string. But the idea is how to have an interface/type for the whole string that we know would contain only certain types of characters in a string.

The question is not about choosing interface for binary values, it's how to declare/define types for predefined string values.

Upvotes: 4

Views: 2915

Answers (1)

Matthieu Riegler
Matthieu Riegler

Reputation: 55514

You can use a recursive typing :

type BinDigit = "0" | "1"

type OnlyBinDigit<S> =
    S extends ""
        ? unknown
        : S extends `${BinDigit}${infer Tail}`
            ? OnlyBinDigit<Tail>
            : never




function onlyBinDigit<S extends string>(s: S & OnlyBinDigit<S>) {
}



onlyBinDigit("01010101010011"); // OK 
onlyBinDigit("010101012"); // NOK

To explain a bit the typing here, OnlyBinDigit is recursive type.

  • Base case : empty string returns the type unknown.
  • Recurcivity : using template literals we can split the string and make a recursive call on the tail until we reach the empty string.
  • In non-recursive part the template literal, if the first character doesn't match the type (here '0' | '1'), it will return never thus failing the typecheck.

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Upvotes: 8

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