Question

我正在尝试创建一个简单的switch函数，该函数需要一个第一个参数，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"});

发生我的问题是因为通用类型T必须是string才能用作对象键。我不知道如何告诉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。它禁止使用impossible，但是您不想禁止它吗，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

如果要允许额外的键，则可以分别声明case对象（绕过多余的属性检查并允许您重复使用case对象）


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

play

或者您可以稍微更改类型，以使泛型类型参数成为case对象，并且该值将被键入为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

更好的选择是保留case对象中的类型，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：

要保留正确的返回类型和错误（如果工会中不存在任何可能性），可以使用以下方法：

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，因为typescript确实控制流分析，所以您需要确保option不仅是作为您分配的实际常量的类型，还不仅仅是您指定的类型注释

打字稿通用联合类型

1 个答案: