Typescript match first uppercased letter on type level

Question

I want to transform a string literal from camelCase to snake_case, like:

type CamelCaseStr = "helloWorldAgain"
type _ = ToSnakeCase<CamelCaseStr> // "hello_world_again"

Here is my Solution:

type str = "helloWorldAgain";
type UpperCollection = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
type Chars<T extends string> = T extends `${infer Char}${infer Rest}` ? Char | Chars<Rest> : never;
type UpperLetters = Chars<UpperCollection>;

type ToSnakeCase<T extends string> = 
T extends `${infer LowerStr}${UpperLetters}${infer AfterFirstBigLetter}` ? 
T extends `${LowerStr}${infer WithFirstBigLetter}` ? 
`${LowerStr}_${ToSnakeCase<Uncapitalize<WithFirstBigLetter>>}` : 
LowerStr : 
T;
type _ = ToSnakeCase<str>; 
// type _ = "helloWorld_again" | "helloWorld_world_again" | "hello_again" | "hello_world_again"

I attempted to extract a LowerStr until firstly encountered an uppercased letter, therefore extract first capitalized str, then Uncapitalize it and use ToSnakeCase again. However, the result is wrong, so I wrote some test code.

type TestToSnakeCase<T extends string> = 
T extends `${infer LowerStr}${UpperLetters}${infer AfterFirstBigLetter}` ? 
T extends `${LowerStr}${infer WithFirstBigLetter}` ? 
[LowerStr, WithFirstBigLetter] : 
never
: T;
type __ = TestToSnakeCase<str>; 
// type __ = ["helloWorld" | "hello", "Again" | "WorldAgain"];

The problem seems to be reduced down to shut discrimination. I searched and found this link useful, with which I tried again.

type ShutDiscrimination<T extends string> = 
[T] extends [`${infer LowerStr}${UpperLetters}${infer AfterFirstBigLetter}`] ? 
[T] extends [`${LowerStr}${infer WithFirstBigLetter}`] ? 
`${LowerStr}_${ShutDiscrimination<Uncapitalize<WithFirstBigLetter>>}` : 
LowerStr : 
T;
type ___ = ShutDiscrimination<str>;
// type ___ = "helloWorld_again" | "helloWorld_worldAgain" | "hello_again" | "hello_worldAgain"

Is there anything critical I missed to solve this problem? I hope I expressed my idea well, formatting is quite cumbersome:)

playground link

Answer 1

The main issue with your code is that you are using UpperLetters directly in the conditional clause where you infer. Because UpperLetters can only be a single character, you should instead infer the character then check if it extends UpperLetters. This will flatten the unions you see.

type UpperLetter = "a" | "b";
type Parse<S extends string> = S extends
  `${infer $Start}${infer $UpperLetter}${infer $Rest}`
  ? $UpperLetter extends UpperLetter ? true : false
  : false;

Anyways, I attempted the problem myself without using a static UpperLetter union by checking if Capitalize<S> is equal to S. Here it is:

type CamelToSnakeCase<S extends string, $Acc extends string = ""> =
  //
  S extends `${infer $Ch}${infer $Rest}`
    ? S extends Capitalize<S>
      ? CamelToSnakeCase<$Rest, `${$Acc}_${Lowercase<$Ch>}`>
    : CamelToSnakeCase<$Rest, `${$Acc}${$Ch}`>
    : $Acc;

type foo = CamelToSnakeCase<"helloAwesomeWorld">; // "hello_awesome_world"
type foo2 = CamelToSnakeCase<"">; // ""
type foo3 = CamelToSnakeCase<"bruhSOCool">; // "bruh_s_o_cool"

type SnakeToCamelCase<S extends string, $Acc extends string = ""> =
  //
  S extends `${infer $Ch}${infer $Rest}`
    ? $Ch extends "_"
      ? $Rest extends `${infer $Ch}${infer $Rest}`
        ? SnakeToCamelCase<$Rest, `${$Acc}${Capitalize<$Ch>}`>
      : ""
    : SnakeToCamelCase<$Rest, `${$Acc}${$Ch}`>
    : $Acc;

type asdf = SnakeToCamelCase<"hello_awesome_world">; // "helloAwesomeWorld"
type asdf2 = SnakeToCamelCase<"bruh_s_o_cool">; // "bruhSOCool"
type asdf3 = SnakeToCamelCase<"">; // ""
type asdf4 = SnakeToCamelCase<"so_cool_bro">; // "soCoolBro"

TypeScript Playground Link

Edit: It was not the union that fixed your code. It was the fact that I inferred the next letter when I did ${infer $Ch}${infer $Rest} then checked if it was uppercase via Capitalize. You were doing ${infer LowerStr}${UpperLetters}${infer AfterFirstBigLetter} which computes the value of the string for each branch in UpperLetters which results in having issues with multiple types for LowerStr and AfterFirstBigLetter. Here's an example:

type UpperLetter = "A" | "B" | "C" | "D" // ..

// This is going to go through each computed type in the extends clause and come up with multiple 
// answers in the union for after the various characters.

type bad = "ABCD" extends `${string}${UpperLetter}${infer $After}` ? $After : never; // "" | "D" | "BCD" | "CD"

// Now, let's fix the code by writing an infer character to force the extends clause to only have one possible combination.
// This can be done by replacing `UpperLetter` with `infer $Ch` because any `string` / `infer $Ch` and removing the leading
// `${string}` part.

type fixed = "ABCD" extends `${infer $Ch}${infer $After}` ? $Ch extends UpperLetter ? $After : never : never; // "BCD"

TypeScript Playground Link

Answer 2

Here is my solution, inspired by @sno2

I've focused there is only just one way comparing 'target character' with recursions for no Alphabet Collection.

export type CamelCaseToSnakeCase<
  T extends string,
  Joiner extends '' | '_' = ''
> = T extends `${infer Character}${infer Rest}`
  ? Character extends Uppercase<Character>
    ? `${Joiner}${Lowercase<Character>}${CamelCaseToSnakeCase<Rest, '_'>}`
    : `${Character}${CamelCaseToSnakeCase<Rest, '_'>}`
  : '';