type X<T>=T extends `${string}${'*('}${infer A}${')+'}${string}${'*('}${infer A}${')+'}${string}`?A:never
type Y=X<'g*(a12)+gggggg*(h23)+'> // 'a12' | 'h23'
type z=X<'g*(a12)+gggggg*(h23)+gggggg*(5hgf)+'> // 'a12' | 'h23' but without '5hgf'
My objective is to extract all literals from a string based on certain pattern. In the code above, I wanna extract literals with prefix '*(' and postfix ')+'. Hence, type Y = 'a12' | 'h23'.
But the problem is a string can have any number of matched-pattern-literal and I do not know how to let typescript to extract all the matched literals.
For the code example, Typescript only extracts 2 literals because I write two times of ${string}${'*('}${infer A}${')+'}. If I write type z=X<'g*(a12)+gggggg*(h23)+gggggg*(5hgf)+'>, I still get type Z = 'a12' | 'h23''. Ideally, I should get type Z = 'a12' | 'h23'|''5hgf''
How to make typescript to 'iterate' through the string to get all the desired literals?
Thank you!
There's no iterative approach to this, but you can use an equivalent recursive approach. And if you write it as a tail-recursive form, then it will work for quite long strings, if you have them. Here's one way to do it:
type X<T extends string, A extends string = never> =
T extends `${string}${'*('}${infer U}${')+'}${infer R}` ?
X<R, A | U> : A;
We're still writing X<T> to parse string via template literal types, but now there's an accumulator type parameter A to keep track of the union of matches we've already collected. If you are given a string that matches your pattern, the type U will be the first matching portion, and the type R will be the rest of the string. Then we recurse with X<R, A | U>, which adds U to the existing A accumulator. Otherwise there are no matches, and we return A.
Let's test it out:
type Y = X<'g*(a12)+gggggg*(h23)+'>
// ^? type Y = "a12" | "h23"
type Z = X<'g*(a12)+gggggg*(h23)+gggggg*(5hgf)+'>
// ^? type Z = "a12" | "h23" | "5hgf"
Looks good.