题目
中文
实现 MapTypes<T, R>
, 它将对象 T
中的类型转换为有如下结构的类型 R
定义的类型
type StringToNumber = {
mapFrom: string; // 类型为 string 的键(key)
mapTo: number; // 会被转换为 number
};
示例:
type StringToNumber = { mapFrom: string; mapTo: number };
MapTypes<{ iWillBeANumberOneDay: string }, StringToNumber>; // 结果 { iWillBeANumberOneDay: number; }
注意, 用户可能会传入联合(union)
类型
type StringToNumber = { mapFrom: string; mapTo: number };
type StringToDate = { mapFrom: string; mapTo: Date };
MapTypes<{ iWillBeNumberOrDate: string }, StringToDate | StringToNumber>; // gives { iWillBeNumberOrDate: number | Date; }
如果类型在字典中不存在, 那就让它保持原样
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];
};