创建Trim类型的正确方法

Question

我读到了类型记录4.5beta的博客文章，作者实现了一个TrimLeft实用程序类型来显示Tail-Recursion Elimination on Conditional Types，所以我决定实现一个Trim实用程序类型(为了学习目的)，我的问题是，你会做什么，你会怎么做？

type Trim<T extends string> =
    T extends 
        `${infer RestOne} ${infer RestTwo}` 
        ? Trim<`${RestOne}${RestTwo}`>
        : T extends ` ${infer RestThree} ${infer RestFour}`
           ? Trim<`${RestThree}${RestFour}`>
           : T extends ` ${infer RestFive} ${infer RestSix} `
             ? Trim<`${RestFive}${RestSix}`>
             : T extends `${infer RestSeven} ${infer RestEight} `
               ? Trim<`${RestSeven}${RestEight}`>
               : T;


type TestOne = Trim<" foo"> ; // "foo"
type TestTwo = Trim< " foo "> ; // "foo"
type TestThree = Trim<" foo bar "> ; // "foobar"
type TestThree = Trim<"   foo bar "> ; // "foobar"
type TestFour = Trim<"foo bar ">; // "foobar"

basically, it trims out all space in `Trim<T>`

Answer 1

考虑一下这个例子：

type Separator = ' ';

type Trim<T extends string, Acc extends string = ''> =
  (T extends `${infer Char}${infer Rest}`
    ? (Char extends Separator
      ? Trim<Rest, Acc>
      : Trim<Rest, `${Acc}${Char}`>)
    : (T extends ''
      ? Acc
      : never)
  )




type TestOne = Trim<" foo">; // "foo"
type TestTwo = Trim<" foo ">; // "foo"

type TestThree = Trim<" foo bar ">; // "foobar"
type TestThree2 = Trim<"   foo bar ">; // "foobar"
type TestFour = Trim<"foo bar ">; // "foobar"

游乐场

Trim -遍历整个字符串。如果它找到Separator (空空间)，它只是省略它，并将Acc与不带分隔符的部分字符串连接起来。