英文:
Make all object properties of a type optional
问题
根据给定的类型 T，如何使所有属性的类型为 object 变为可选的？
因此，对于以下类型：
interface MyObject {
name: string;
age: number;
favorites: {
color: string
}
}
type MyNewObject = ObjectPropertiesOptional<MyObject>;
它将返回：
interface MyNewObject {
name: string;
age: number;
favorites?: {
color: string
}
}
英文:
Given a type T how can I make all the properties that are object optional?
So for a type:
interface MyObject {
name: string;
age: number;
favorites: {
color: string
}
}
type MyNewObject = ObjectPropertiesOptional<MyObject>;
it returns
interface MyNewObject {
name: string;
age: number;
favorites?: {
color: string
}
}
答案1
得分: 4
以下是可能的实现：
type ObjectPropertiesOptional<T> = (
Partial<T> & { [K in keyof T as T[K] extends object ? never : K]: T[K] }
) extends infer O ? { [K in keyof O]: O[K] } : never;
整个实现都被包裹在(...) extends infer O ? { [K in keyof O]: O[K] } : never中，这是一个让编译器将相对丑陋的交叉类型（例如{foo: string} & {bar: number}）转换为单一对象类型（例如{foo: string; bar: number}）的技巧。这与https://stackoverflow.com/q/57683303/2887218中描述的Expand<T>相同的技术。
实际实现是将Partial<T>（使用PartialT的重映射版本进行交叉，该版本仅包含非object属性。（通过将键重映射为never，我们将其抑制）。
因此，对于{foo: string, bar: object}，它将变成{foo?: string, bar?: object} & {foo: string}，这等效于{foo: string, bar?: object}（并通过extends infer O技巧转换为它）。
让我们来测试一下：
interface MyObject {
name: string;
age: number;
favorites: {
color: string
}
}
type MyNewObject = ObjectPropertiesOptional<MyObject>;
/*
type MyNewObject = {
name: string;
age: number;
favorites?: {
color: string;
} | undefined;
}
*/
看起来不错。`name`和`age`属性仍然是必需的，但`favorites`属性现在是可选的。
英文:
Here's one possible implementation:
type ObjectPropertiesOptional<T> = (
Partial<T> & { [K in keyof T as T[K] extends object ? never : K]: T[K] }
) extends infer O ? { [K in keyof O]: O[K] } : never;
The whole thing is wrapped in (...) extends infer O ? { [K in keyof O]: O[K] } : never which is a trick to make the compiler convert a relatively ugly interesection type like {foo: string} & {bar: number} and turn it into a single object type like {foo: string; bar: number}. This is the same technique as Expand<T> as described in https://stackoverflow.com/q/57683303/2887218.
The actual implementation is to intersect Partial<T> (using the Partial<T> utility type to make every property optional) with a remapped version of T that only includes non-object properties. (By remapping a key to never we suppress it).
So for {foo: string, bar: object}, it would become {foo?: string, bar?: object} & {foo: string}, which is equivalent to {foo: string, bar?: object} (and converted to it by the extends infer O trick).
Let's test it out:
interface MyObject {
name: string;
age: number;
favorites: {
color: string
}
}
type MyNewObject = ObjectPropertiesOptional<MyObject>;
/*
type MyNewObject = {
name: string;
age: number;
favorites?: {
color: string;
} | undefined;
}
*/
Looks good. The name and age properties are still required, but the favorites property is now optional.
答案2
得分: 0
据我所知，在TypeScript中目前还不可能实现这个。
这是Github讨论的链接。
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