Typescript 通用联合类型

Question

我正在尝试创建一个简单的开关 function ，它采用第一个参数，该参数必须是字符串和 object 的联合，它具有基于第一个参数联合的键并且可以返回任何值。

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"});

出现我的问题是因为泛型类型T必须是string才能用作 object 键。 我不知道你怎么能告诉T是string的并集。

我试过T extends string没有成功。

Answer 1

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

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"}); 

玩

如果您想允许额外的键，您可以单独声明案例 object （绕过多余的属性检查并允许您重用案例对象）

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

玩

或者你可以稍微改变你的类型，这样泛型类型参数就是 object，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"}); 

玩

另外一个更好的选择是保留案例 object 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

玩

编辑：

如果联合中不存在可能性，要保留正确的返回类型和错误，您可以使用以下命令：

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

玩

请注意，因为 typescript 进行控制流分析，您需要确保option不仅仅是类型作为您分配的实际常量，而不是您指定的类型注释