映射类型:删除可选修饰符 [英] Mapped Types: removing optional modifier

问题描述

鉴于此代码:

interface Foo{
  one?: string;
  two?: string;
}

type Foo2 = {
  [P in keyof Foo]: number;
}

我希望 Foo2 的类型是 { one: number;二:数量；} 然而，它似乎保留了可选的修饰符 { one?: number;二?:数；}

I would expect the type of Foo2 to be { one: number; two: number; } However, instead it seems to keep the optional modifier { one?: number; two?: number; }

使用映射类型时是否可以删除可选修饰符?

Is it possible to remove the optional modifier when using mapped types?

推荐答案

在 Typescript 2.8 中，您可以显式消除修饰符:

In Typescript 2.8 you can explicitly eliminate the modifier:

type Foo2 = {
  [P in keyof Foo]-?: number;
}

或者使用新版本中内置的 Required 类型.

Or use the Required type that is built into newer versions.

如果您使用的是旧版本，则可以使用以下解决方法:

If you are using an older version you can use this workaround:

type Helper<T, TNames extends string> = { [P in TNames]: (T & { [name: string]: never })[P] };
type Foo3 = Helper<Foo, keyof Foo>;

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