从集合中仅收集特定类型值的键 [英] Collect only keys for values of a certain type from a collection

问题描述

例如假设我们有

type AnObject = {
  a: string
  b: number
  c: string
  d: number
}

type ExtractKeysOfType<O extends {[K: string]: any}, T> = ///...

type StringKeys = ExtractKeysOfType<AnObject, string> // 'a' | 'c'

它们是实现 ExtractKeysOfType 的方法吗?

Is they're a way to achieve ExtractKeysOfType ?

推荐答案

您可以在此处结合使用映射类型和条件类型:

You can use mapped types combined with conditional types here:

type ExtractKeysOfType<T, Target> = {
  [K in keyof T]: T[K] extends Target ? K : never
}[keyof T];

这基本上是通过遍历 T 类型中的每个键来工作的.T[K] 是否扩展了我们的 Target 类型?如果是这样，那就太好了，并且属性值就是那个 Key.如果不是，则该键的类型为 never.

This essentially works by going over each key in type T. Does T[K] extend our Target type? if so, great, and the property value is simply that Key. If not, that key has type never.

对于您的情况，这种中介类型如下所示:

This intermediary type, for your case, looks like:

{
    a: "a";
    b: never;
    c: "c";
    d: never;
}

然后，该中间类型被 T 的键再次索引.这将产生您想要的联合，因为 never 类型在这里被编译器忽略.

Then, that intermediary type is indexed again by keys of T. This will produce your desired union, since the never types are ignored by the compiler here.

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