打字稿通用联合类型

Question

我正在尝试创建一个简单的 switch 函数，它采用第一个参数，该参数必须是字符串的并集和一个对象，该对象具有基于第一个参数并集的键并且可以 return 任何值。

export const mySwitch = <T extends string>(value: T, possibilities: {[key in T]: any}): any => {
    return possibilities[value];
};

典型用法是

let option: "val1" | "val2" | "val3" = "val1";
// should returns s1
// Impossible should be type-checked as an error since it's not part of the option union type
mySwitch(option, {val1: "s1", val2: "s2", val3: "s3", impossible: "impossible"});

我的问题出现是因为泛型类型 T 必须是 string 才能用作对象键。我不知道你怎么知道 T 是 string.

的并集

我尝试了 T extends string 但没有成功。

Answer 1

T extends string 版本似乎运行良好。它不允许 impossible，但您不想禁止它吗，因为如果参数永远不能具有该值，那么该选项将毫无用处？：

export const mySwitch = <T extends string>(value: T, possibilities: {[key in T]: any}): any => {
    return possibilities[value];
};


declare let option: "val1" | "val2" | "val3";
mySwitch(option, {val1: "s1", val2: "s2", val3: "s3", impossible: "impossible"});

play

如果你想允许额外的键，你可以单独声明 case 对象（绕过多余的属性检查并允许你重用 case 对象）


declare let option: "val1" | "val2" | "val3";
const casses = {val1: "s1", val2: "s2", val3: "s3", impossible: "impossible"}
mySwitch(option, casses);

play

或者你可以稍微改变你的类型，这样泛型类型参数就是 case 对象，值将被键入为 keyof T:

export const mySwitch = <T>(value: keyof T, possibilities: T): any => {
    return possibilities[value];
};


declare let option: "val1" | "val2" | "val3";
mySwitch(option, {val1: "s1", val2: "s2", val3: "s3", impossible: "impossible"});

play

另外一个更好的选择是保留 case 对象的类型而不是使用 any:

export const mySwitch = <T, K extends keyof T>(value: K, possibilities: T): T[K] => {
    return possibilities[value];
};


declare let option: "val1" | "val2" | "val3";
mySwitch(option, {val1: 1, val2: "s2", val3: "s3", impossible: false});  // returns string | number

play

编辑：

要同时保留正确的 return 类型和错误（如果联合中不存在的可能性），您可以使用此方法：

const mySwitch = <T extends Record<K, any>, K extends string>(value: K, possibilities: T & Record<Exclude<keyof T, K>, never>): any => {
    return possibilities[value];
};

let option: "val1" | "val2" | "val3" = (["val1", "val2", "val3"] as const)[Math.round(Math.random() * 2)]
mySwitch(option, {val1: "s1", val2: "s2", val3: "s3" });
mySwitch(option, {val1: "s1", val2: "s2", val3: "s3", impossible: "" }); //err on impossible

play

请注意，由于 typescript 会控制流分析，因此您需要确保 option 不只是类型作为您分配的实际常量，而不是您指定的类型注释

打字稿通用联合类型

Typescript generic union type

typescript

typescript-generics

typescript-typings