Question

假设我有以下内容：

export const ContentType = {
    Json: "application/json",
    Urlencoded: "application/x-www-form-urlencoded",
    Multipart: "multipart/form-data",
};

export interface RequestOptions {
    contentType: string,
}

const defaultOptions: Partial<RequestOptions> = {
    contentType: ContentType.Json,
};

如何限制contentType，以便只使用ContentType中声明的密钥？

Answer 1

这是我从TypeScript 2.1开始的首选方式：

export const contentTypes = {
    "application/json": true,
    "application/x-www-form-urlencoded": true,
    "multipart/form-data": true
};

type ContentType = keyof typeof contentTypes;

export interface RequestOptions {
    contentType: ContentType;
}

const defaultOptions: Partial<RequestOptions> = {
    contentType: "application/json",
};

Try it in TypeScript Playground

在对象文字中定义一组有效字符串。
使用keyof创建包含该对象的键的类型。这是一个字符串联合类型。
键入您的字符串属性为该联合类型，编译器将只允许这些字符串。

Answer 2

您可以使用string literal types。不要将contentType键入string，而是将其输入为可能值的文字类型。

export interface RequestOptions {
    contentType: "application/json" | "multipart/form-data" | "multipart/form-data:"
}

为避免重复常量，您可以单独声明它们，并将contentType键入typeof ConstantString：

export const ContentTypeJson = "application/json";
export const ContentTypeUrlencoded = "application/x-www-form-urlencoded";
export const ContentTypeMultipart = "multipart/form-data";


export interface RequestOptions {
    contentType: typeof ContentTypeJson| typeof ContentTypeUrlencoded | typeof ContentTypeMultipart,
}

const defaultOptions: Partial<RequestOptions> = {
    contentType: ContentTypeJson,
};

如何将值限制为对象的键？

2 个答案: