获取泛型参数的类型

我编写了一个小函数来更好地处理类型。

function evaluate(variable: any, type: string): any {
    switch (type)
    {
        case 'string': return String(variable);
        case 'number': return isNumber(variable) ? Number(variable) : -1;
        case 'boolean': {
            if (typeof variable === 'boolean')
                return variable;

            if (typeof variable === 'string')
                return (<string>variable).toLowerCase() === 'true';

            if (typeof variable === 'number')
                return variable !== 0;

            return false;
        }
        default: return null;
    }
}

function isNumber(n: any): boolean {
    return !isNaN(parseFloat(n)) && isFinite(n);
}

我尝试对泛型进行相同的操作，但不知道如何从泛型参数获取类型。这是可能的？

你无法消除type字符串，但您可以通过添加重载使您的函数在类型方面更加智能和可用：

function evaluate(variable: any, type: 'string'): string;
function evaluate(variable: any, type: 'number'): number;
function evaluate(variable: any, type: 'boolean'): boolean;
function evaluate(variable: any, type: string): unknown {
    ...
    default: throw Error('unknown type');
}

const myBool = evaluate('TRUE', 'boolean'); // myBool: boolean
const myNumber = evaluate('91823', 'number'); // myBool: boolean
evaluate('91823', 'qwejrk' as any); // RUNTIME ERROR (violated types)

const mysteryType = 'number' as 'boolean' | 'number';
const myMystery = evaluate('91823', mysteryType); // COMPILER ERROR, no overload matches.

游乐场链接 https://www.typescriptlang.org/play?#code/GYVwdgxgLglg9mABAUwG4EMA2J1WQCgwCcZ0AjTZALkXTAE8AaRKegB2sQHIBnKEsAHMuAShp8BggNwAoUJFgIUGbLgLFSFTnSYt2nLmBABbMsiKiaR0+dnzo8JGiw48hdCXKUaO5qw40XGRwcJR0lojBoch0duAOSs6qbhpe2gx++uL8MEJiiOAA1mBwAO5IAN4yiDWIPKUwUBAAFoj4-sgi1bVVtX2IEOg8yNwSucI0RMhQIERIAMo5Qu6eWiKy-bWDw9zWZhaT07NIMDwAcib7K5qUIogA-IgXNkTXaXc0ALQAjBubA0MRkEQmEwFwaL1-n0YMA2h04LDUlpEABeNGjJbCLpQqFTGZzRBIygAOigcAAMmVzABhQH4O5olHcfggZBcP44xB446IACEvKJyA5-QAvt1NgATZDAdAgTBQQ74pBGTCYYVisVyeKKE7nS7mfBgHwMfJRUGISE1bkE3mnM7oM74NgeYYAMUwcFwhpEdwAZL7EKdXblGgQwOsZJqIAg+IhjPQAEIg1HKFxqfBcAAqACUAKoAUS4zGB0XC60QAHoK3HEyCaGaYmAZNGwLH4899imkq4CFwAJzfAAcACYAMxF3b6izlqs1pOhesgxsybvp-tDscTrgAR1KyAAVkRClxaDxaCapJXq3mzpmAJIAWXziHz2ezAHls21UPBMGoJXoHA8F0zYxlANZ8OY9CZvoKaGFOJ5DNwDbhIgAA+k4vOyoGtuB8YPvQkFEPQXYqD2GYDiO47MPGRHQfoM7VtS74PgACne5Kvi+b6fswJSIHAqDmB66AAcYuAtMgPDEjIQA

请注意，不再存在空情况，因为不可能知道未知情况是否存在string类型实际上可能包含一个有效值，例如'number'在编译时。

这对于大多数人来说已经足够了。

然而...

上面请注意，mysteryType 联合不起作用。如果出于某种原因你真的真的希望它起作用，你可以使用条件类型：

function evaluate<T extends string>(variable: any, type: T):
    T extends 'string' ? string :
    T extends 'number' ? number :
    T extends 'boolean' ? boolean :
    never;
function evaluate(variable: any, type: string): unknown {
    ...
    default: throw Error('unknown type');
}

const mysteryType = 'number' as 'boolean' | 'number';
const myMystery = evaluate('91823', mysteryType); // myMystery: number | boolean

游乐场链接 https://www.typescriptlang.org/play?#code/GYVwdgxgLglg9mABAUwG4EMA2J1WQHgBUUAPPMAEwGdEqoAnGMAcwD4AKDR9AI02QBcidGACeAGkRRRAB0GJCASgEAoROoWly1RAHI6jFrsQB+WgybNEqjZuRlklGrrAgAtj2T1jZ1x6-WahrE9trOPHBw-CI+iBFRyCKBtmBoXgDcKqCQsAgoGNi4yJzo3HzyIhJSsvIGlsqI4ADWYHAA7kgA3kHqVG0wUBAAFojs0nKKPYjdtrYQ6FTIenVGQvTIUCD0SADKFiwlZfyKmbMa84t6fp7eaxtbSDBUAHLuN4cwvMemiK-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

此外，如果您在谷歌上搜索了这个问题并且想知道如何获得T from MyClass<T>，这也是可能的：

class MyClass<T> {}

type GetMyClassT<C extends MyClass<any>> = C extends MyClass<infer T> ? T : unknown;

const myInstance = new MyClass<"hello">();
let x: GetMyClassT<typeof myInstance>; // x: "hello"