題目
中文
實現 MapTypes<T, R>, 它將對象 T 中的類型轉換爲有如下結構的類型 R 定義的類型
type StringToNumber = {
mapFrom: string; // 類型爲 string 的鍵(key)
mapTo: number; // 會被轉換爲 number
};
示例:
type StringToNumber = { mapFrom: string; mapTo: number };
MapTypes<{ iWillBeANumberOneDay: string }, StringToNumber>; // 結果 { iWillBeANumberOneDay: number; }
注意, 用戶可能會傳入聯合(union)類型
type StringToNumber = { mapFrom: string; mapTo: number };
type StringToDate = { mapFrom: string; mapTo: Date };
MapTypes<{ iWillBeNumberOrDate: string }, StringToDate | StringToNumber>; // gives { iWillBeNumberOrDate: number | Date; }
如果類型在字典中不存在, 那就讓它保持原樣
type StringToNumber = { mapFrom: string; mapTo: number };
MapTypes<
{ iWillBeANumberOneDay: string; iWillStayTheSame: Function },
StringToNumber
>; // // gives { iWillBeANumberOneDay: number, iWillStayTheSame: Function }
English
Implement MapTypes<T, R> which will transform types in object T to different types defined by type R which has the following structure
type StringToNumber = {
mapFrom: string; // value of key which value is string
mapTo: number; // will be transformed for number
};
Examples:
type StringToNumber = { mapFrom: string; mapTo: number };
MapTypes<{ iWillBeANumberOneDay: string }, StringToNumber>; // gives { iWillBeANumberOneDay: number; }
Be aware that user can provide a union of types:
type StringToNumber = { mapFrom: string; mapTo: number };
type StringToDate = { mapFrom: string; mapTo: Date };
MapTypes<{ iWillBeNumberOrDate: string }, StringToDate | StringToNumber>; // gives { iWillBeNumberOrDate: number | Date; }
If the type doesn't exist in our map, leave it as it was:
type StringToNumber = { mapFrom: string; mapTo: number };
MapTypes<
{ iWillBeANumberOneDay: string; iWillStayTheSame: Function },
StringToNumber
>; // // gives { iWillBeANumberOneDay: number, iWillStayTheSame: Function }
答案
type MapTypes<T, R extends { mapFrom: unknown; mapTo: unknown }> = {
[P in keyof T]: T[P] extends R['mapFrom']
? R extends R
? Equal<T[P], R['mapFrom']> extends true
? R['mapTo']
: never
: never
: T[P];
};