英文:
How to make all keys of a type required while keeping undefined as possible value?
问题
我想将类型的键/值映射如下：
required_key: string | undefined到required_key: string | undefined（相同）required_key: string到required_key: string（相同）optional_key?: string | undefined到optional_key: string | undefined（去除键中的 ?）optional_key?: string到optional_key: string | undefined（去除键中的 ? 并添加 undefined 作为可能的值）
interface Person {
firstName: string | undefined;
middleName?: string;
lastName?: string | undefined;
}
// 我需要的是：
// interface Person {
// firstName: string | undefined;
// middleName: string | undefined;
// lastName: string | undefined;
// }
type Normalize<T> = { [K in keyof T]-?: T[K] };
// 使用 Normalize<Person> 后得到的结果：
// interface Person {
// firstName: string | undefined;
// middleName: string;
// lastName: string;
// }
const test: Normalize<Person> = {
firstName: undefined,
middleName: undefined, // ts 错误，类型为 string，不允许 undefined
lastName: undefined, // ts 错误，类型为 string，不允许 undefined
};
在 TypeScript 中是否可能实现这个要求？似乎 -? 移除了键和值中的 undefined，但我不知道如何在键是可选的情况下添加 undefined 值选项。
我阅读了这个问题/回答：https://stackoverflow.com/questions/72402413/is-it-possible-to-make-a-property-required-yet-preserve-undefined，我了解到 exactOptionalPropertyTypes 编译器标志，但它仍然没有解决如何将 key?: string 转换为 key: string | undefined。
英文:
I want to map key/values of a type as follows:
required_key: string | undefinedtorequired_key: string | undefined(same)required_key: stringtorequired_key: string(same)optional_key?: string | undefinedtooptional_key: string | undefined(removed ? from key)optional_key?: stringtooptional_key: string | undefined(removed ? from key and added undefined as possible value)
interface Person {
firstName: string | undefined;
middleName?: string;
lastName?: string | undefined;
}
// What I need:
// interface Person {
// firstName: string | undefined;
// middleName: string | undefined;
// lastName: string | undefined;
// }
type Normalize<T> = { [K in keyof T]-?: T[K] };
// What I get with Normalize<Person>:
// interface Person {
// firstName: string | undefined;
// middleName: string;
// lastName: string;
// }
const test: Normalize<Person> = {
firstName: undefined,
middleName: undefined, // ts err, type = string, undefined not allowed
lastName: undefined, // ts err, type = string, undefined not allowed
};
Is it possible in typescript? It seems that -? is removing undefined from both key and value and I don't know how to add undefined value option if key is optional.
I read this Q/A https://stackoverflow.com/questions/72402413/is-it-possible-to-make-a-property-required-yet-preserve-undefined and I know about exactOptionalPropertyTypes compiler flag but it still does not solve how to convert key?: string to key: string | undefined.
答案1
得分: 1
以下是翻译好的部分：
type AllOptionalKeys<T> = { [K in keyof T]-?: undefined extends T[K] ? K : never; }[keyof T];
type AllNonOptionalKeys<T> = { [K in keyof T]-?: undefined extends T[K] ? never : K; }[keyof T];
type Normalize<T> = { [K in AllOptionalKeys<T>]: T[K] | undefined } & { [K in AllNonOptionalKeys<T>]: T[K]; };
英文:
Not sure if this is the best solution, but you can achieve it with the following:
type AllOptionalKeys<T> = { [K in keyof T]-?: undefined extends T[K] ? K : never; }[keyof T];
type AllNonOptionalKeys<T> = { [K in keyof T]-?: undefined extends T[K] ? never : K; }[keyof T];
type Normalize<T> = { [K in AllOptionalKeys<T>]: T[K] | undefined } & { [K in AllNonOptionalKeys<T>]: T[K]; };
The two intermediate types allow to strip the ? from the keys, but keep track of which are optional.
Then it constructs an union type, where all optional keys have a value of T[K] | undefined, and all the non optional keys have T[K].
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