Typescript：以目标字符串开头时排除属性键

Question

我想当属性键以目标字符串开头时，应从原始类型中过滤新的对象类型。

产地类型：

type Origin = {
    a: string,
    b: string,
    _c: string,
    _d: string,
}

我想要的结果类型：

// type Result = SomethingWork<Origin, '_'>;
type Result = {
    a: string,
    b: string
};

Origin 类型具有动态属性键。所以如果直接使用类型，这是不正确的'_c' | '_d'

Answer 1

现在可以使用模板文字类型实现这一点

type FilterNotStartingWith<Set, Needle extends string> = Set extends `${Needle}${infer _X}` ? never : Set

例子：

type Origin = {
  a: string,
  b: string,
  _c: string,
  _d: string,
}

type FilteredKeys = FilterNotStartingWith<keyof Origin, '_'>
type NewOrigin = Pick<Origin, FilteredKeys>
/*
type NewOrigin = {
  a: string;
  b: string;
}
*/

Answer 2

我想对现有答案进行稍微简化。

type Origin = {
  a: string,
  b: string,
  _c: string,
  _d: string,
}

type OriginWithoutUnderscoreProps = Omit<Origin, `_${string}`>
/*
type OriginWithoutUnderscoreProps = {
  a: string;
  b: string;
}
*/

包含的逆过程也有效。请注意，keyof Origin这里需要 ，因为对其第二个参数Pick有约束，而没有。keyofOmit

type OriginWithUnderscoreProps = Pick<Origin, keyof Origin & `_${string}`>
/*
type OriginWithUnderscoreProps = {
  _c: string;
  _d: string;
}
*/

Answer 3

不能用前缀来做

但

import { ITSDiff, ITSPickMember } from 'ts-type';

export type Origin = {
    a: string,
    b: string,
    _c: string,
    _d: string,
}

export type Result = Pick<Origin, ITSDiff<keyof Origin, '_c' | '_d'>>

let a: Result;

a.a.padEnd(1)

a._c // err

export type OriginOnlyHasPrefix = {
    _c: string,
    _d: string,
}

export type Result3 = Pick<Origin, ITSDiff<keyof Origin, keyof OriginOnlyHasPrefix>>

let a3: Result3;

a3.a.padEnd(1)

// @ts-ignore err when without @ts-ignore
a3._c;