提问人：Valentin Vignal 提问时间：8/28/2023 更新时间：8/28/2023 访问量：20
对象保持可为空性的强制转换字段
Cast fields of object keeping nullability
问：
我正在尝试创建一个类型，该类型将对象的所有属性强制转换为 。这是我现在所拥有的：DateToNumberDatenumber
type LiteralDateToNumber<T> = T extends Date
? number
: T extends Date | null
? number | null
: T extends Date | undefined
? number | undefined
: T extends Date | null | undefined
? number | null | undefined
: T;
/**
* @description Converts all `Date` properties to `number`.
*/
type DateToNumber<T> = {
[K in keyof T]: LiteralDateToNumber<T[K]>;
};
我想保留 ，这里所有带有 或 的检查。nullability| null| undefined
这大多有效，但是使用其他类型时会出现一个问题，而不是可以为空的类型：Date
type MyType = {
aDate: Date;
aString: string;
bDate: Date | null;
bString: string | null;
cDate: Date | undefined;
cString: string | undefined;
dDate: Date | null | undefined;
dString: string | null | undefined;
eDate?: Date;
eString?: string;
fDate?: Date | null;
fString?: string | null;
gDate?: Date | undefined;
gString?: string | undefined;
hDate?: Date | null | undefined;
string?: string | null | undefined;
};
type MyCastedType = DateToNumber<MyType>;
在这种情况下，等效于：MyCastedType
type MyCastedType = {
aDate: number;
aString: string;
bDate: number | null;
bString: string | number | null;
cDate: number | undefined;
cString: string | number | undefined;
dDate: number | null | undefined;
dString: string | number | null | undefined;
eDate?: number | undefined;
eString?: string | number | undefined;
fDate?: number | null | undefined;
fString?: string | number | null | undefined;
gDate?: number | undefined;
gString?: string | number | undefined;
hDate?: number | null | undefined;
hString?: string | number | null | undefined;
};
它有几个问题，我希望它等同于：
type MyCastedType = {
aDate: number;
aString: string;
bDate: number | null; // And not `bString: string | number | null;`
bString: string | null;
cDate: number | undefined;
cString: string | undefined; // And not `cString: string | number | undefined;`
dDate: number | null | undefined;
dString: string | null | undefined; // And not `dString: string | number | null | undefined;`
eDate?: number; // And not `eDate?: number | undefined;`
eString?: string; // And not `eString?: string | number | undefined;`
fDate?: number | null; // And not `fDate?: number | null;`
fString?: string | null; // And not `fString?: string | number | null | undefined;`
gDate?: number | undefined;
gString?: string | undefined; // And not `gString?: string | number | undefined;`
hDate?: number | null | undefined;
string?: string | null | undefined; // And not `hString?: string | number | null | undefined;`
};
如何实现此类型？DateToNumber
答：
1赞
shantr
8/28/2023
#1
应将 LiteralDateToNumber 简化为：
type LiteralDateToNumber<T> = T extends Date ? number : T;
当您为泛型提供类型联合时，它实际上会分发它
type A = LiteralDateToNumber<Date | null>;
// is equivalent to
type B = LiteralDateToNumber<Date> | LiteralDateToNumber<null>;
// => number | null
你的问题是，用作条件的联合也将被分配使用LiteralDateToNumber
type A<T> = T extends Date| null ? number : T;
// is equivalent to saying if T extends either Date or null (not exactly Date | null)
// => equivalent to
type B<T> = T extends Date ? number : T extends null ? number : T;
type Test = Test<null>;
// => null does extends Date | null
// => number;
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