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How do I define a typescript type with a repeating structure?

I have a type that looks something like this:

type Location=`${number},${number};${number},${number};...`

Is there a Utility type like Repeat<T> that can do this for me? like this:

type Location=Repeat<`${number},${number};`>
over 4 years ago · Santiago Trujillo
4 Respostas
Responde à pergunta

0

You can directly choose type string if you want as:

type myLocation = string;
const myLocation: myLocation = "123123123";

console.log(myLocation);

else if you want to narrow it down then better to declare it const as:

const myLocation = "123123123";
console.log(myLocation);

then your type will be the value of the myLocation

enter image description here

over 4 years ago · Santiago Trujillo Relatório

0

I don't think there is a way to define an infinite repeating pattern for a type that you use on variable declaration.

However, a type guard on a function can check that a string matches an infinite pattern, like so (playground):

type MatchesPattern<Pattern extends string, Current extends string> = Current extends `` ? string : (Current extends `${Pattern}${infer Rest}` ? MatchesPattern<Pattern, Rest> : never);

type LocationPattern = `${number},${number};`;

declare function onlyAcceptsLocation<L extends string & IsLocation, IsLocation = MatchesPattern<LocationPattern, L>>(location: L): void;

onlyAcceptsLocation("12,34;56,78;"); // 👍 matches 

onlyAcceptsLocation("12,34;56,78"); // ⚠️

onlyAcceptsLocation("12'34;56,78;"); // ⚠️

onlyAcceptsLocation("1,2,3;45,67;"); // ⚠️
over 4 years ago · Santiago Trujillo Relatório

0

You can use an as const assertion to get a string literal type from a template literal, like this:

TS Playground

const str1 = 'str1';
const str2 = `${str1},${str1};${str1},${str1}` as const; // type is "str1,str1;str1,str1"
over 4 years ago · Santiago Trujillo Relatório

0

IT WORKS ONLY IN TS >=4.5

It is possible to create standalone type.

Please see this example:

type Coordinates = `${number},${number};`

type MAXIMUM_ALLOWED_BOUNDARY = 50

type Last<T extends string[]> = T extends [...infer _, infer Last] ? Last : never;

type ConcatPrevious<T extends any[]> = Last<T> extends string ? `${Last<T>}${Coordinates}` : never

type Mapped<
    N extends number,
    Result extends Array<unknown> = [Coordinates],
    > =
    (Result['length'] extends N
        ? Result
        : Mapped<N, [...Result, ConcatPrevious<Result>]>
    )


// type MyLocation = 
// | `${number},${number};` 
// | `${number},${number};${number},${number};` 
// | `${number},${number};${number},${number};${number},${number};` 
// | `${number},${number};${number},${number};${number},${number};${number},${number};` 
// | `${number},${number};${number},${number};${number},${number};${number},${number};${number},${number};` 
// | ... 44 more ... 
// | `${number},${number};${number},${number};${number},${number};${number},${number};${number},${number};${number},${number}; ....

type MyLocation = Mapped<MAXIMUM_ALLOWED_BOUNDARY>[number]

const myLocation1: MyLocation = '45,56;67,68;' // ok
const myLocation2: MyLocation = '45,56;67,68;1,2;3,4;5,6;7,8;9,10;' // ok
const myLocation3: MyLocation = '45,56;67,68;1,2;3,4;5,6;7,8;9,10,' // expected error

Playground

Mapped types is an utility type which represents a while loop. It iterates until length of Result will reach N. In other words Mapped<10> - will iterate 10 times. See this example in pure js:

const mapped = (N: number, Result: any[] = []): string => {
    if (N === Result.length) {
        return Result.join('')
    }
    
    const x = Math.random();
    const y = Math.random()
    return mapped(N, [...Result, `${x},${y};`])
}

It is hard to represent unions in js, thats why I have used join(''). I hope it is clear how it works.

If you want to increase MAXIMUM_ALLOWED_BOUNDARY to 500 it will heat your CPU so be careful.

As you might have noticed, it is impossible in type script to represent recursive pattern for type but it is possible to create big enough union.

Please keep in mind that there are some drawbacks of ${number} type. You are allowed to use numbers with leading zero like here:

const x: `${number}` = '01'.

Useful links:

  1. Here you can find an example how you can create a range of numbers using this pattern.

  2. Here you can find a PR where this feature was introduced

  3. Here and here you can find my articles which are related to this pattern

over 4 years ago · Santiago Trujillo Relatório
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