打字稿:从元组类型中删除条目 [英] Typescript: Remove entries from tuple type

问题描述

不确定这是否可行，但我希望能够定义一种类型，将元组转换为:[number, string, undefined, number] 到 [number, string, number](即过滤掉undefined).

not sure if this is possible, but I would like to be able to define a type that converts tuples like: [number, string, undefined, number] to [number, string, number] (ie filter out undefined).

我想过这样的事情:

type FilterUndefined<T extends any[]> = {
    [i in keyof T]: T[i] extends undefined ? /* nothing? */ : T[i];
}

可悲的是，我很确定没有办法实现这一点.

Sadly I am am pretty sure that there is no way to achieve this.

推荐答案

TS 4.1

元组上的过滤器操作现在正式 可能:

type FilterUndefined<T extends unknown[]> = T extends [] ? [] :
    T extends [infer H, ...infer R] ?
    H extends undefined ? FilterUndefined<R> : [H, ...FilterUndefined<R>] : T

让我们做一些测试来检查它是否按预期工作:

Let's do some tests to check, that it is working as intended:

type T1 = FilterUndefined<[number, string, undefined, number]> 
// [number, string, number]
type T2 = FilterUndefined<[1, undefined, 2]> // [1, 2]
type T3 = FilterUndefined<[undefined, 2]> // [2]
type T4 = FilterUndefined<[2, undefined]> // [2]
type T5 = FilterUndefined<[undefined, undefined, 2]> // [2]
type T6 = FilterUndefined<[undefined]> // []
type T7 = FilterUndefined<[]> // []

更多信息

• 递归条件类型 #40002 (TS 4.1)
• 可变元组类型 #39094 (TS 4.0)

More infos

• Recursive conditional types #40002 (TS 4.1)
• Variadic tuple types #39094 (TS 4.0)

游乐场

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