Typescript泛型验证指定了keyof的所有键

您可以采取的一种方法是将constraint也添加到T，这样就可以有效地禁止除Q或M之外的任何键的属性(使用the impossible never type:

function createService<
    T extends Record<Exclude<keyof T, Q | M>, never>,
    Q extends keyof T,
    M extends Exclude<keyof T, Q>
>(obj: T, queries: readonly Q[], mutations: readonly M[]) { }


interface Test {
    a: () => void;
    b: () => void;
}
const obj: Test = { a: () => { }, b: () => { } };
    
createService(obj, ["a"], ["b"]); // OK
createService(obj, ["a", "b"], []); // OK

createService(obj, ["a", "b"], ["b"]); // Error
createService(obj, ["a"], []); // Error
createService(obj, [], ["b"]); // Error

现在，这起到了预期的作用.

请注意，这并没有考虑到所有可能的情况.TypeScrip本身并不代表exhaustive arrays的概念.类型为(T1 | T2 | T3)[]的数组不一定都有类型为T1、T2和T3的元素.它根本不能保证有任何元素.所以有could个人这样做:

const aArr: "a"[] = [];
const bArr: "b"[] = [];
createService(obj, aArr, bArr);

想必这不是您关心的用例.如果是，那么唯一可能的方法是列举每一种可能的可接受的tuple输入:

type AllPossiblePartialTuples<T, U = T> =
    [T] extends [never] ? readonly [] : T extends any ?
    readonly [T, ...AllPossiblePartialTuples<Exclude<U, T>>] | 
    AllPossiblePartialTuples<Exclude<U, T>>
    : never;

type AllPossibleTuples<T, U = T> =
    [T] extends [never] ? readonly [] : T extends any ?
    readonly [T, ...AllPossibleTuples<Exclude<U, T>>] : never;

declare function createService<
    T extends object,
    const Q extends AllPossiblePartialTuples<keyof T>
>(
    obj: T,
    queries: Q,
    mutations: AllPossibleTuples<
        Exclude<keyof T, Q extends readonly (infer V)[] ? V : never>
    >
): void;

const obj: Test = { a: () => { }, b: () => { } };

createService(obj, ["a"], ["b"]); // OK
createService(obj, ["a", "b"], []); // OK

createService(obj, ["a", "b"], ["b"]); // Error
createService(obj, ["a"], []); // Error
createService(obj, [], ["b"]); // Error

但只有在T只有少量密钥的情况下，这才是可信的.即使是数量不多的键，这些键的所有可能排列的并集也非常大，并且会导致性能问题或关于复杂性限制的编译器错误，或者两者兼而有之:

// okay-ish
createService(
    { a: 0, b: 1, c: 2, d: 3, e: 4, f: 5 },
    ['a', 'c', 'e'],
    ['b', 'd', 'f']
);

// 🔥💻🔥 COMPILER EATS UP CPU RESOURCES
createService(
    { a: 0, b: 1, c: 2, d: 3, e: 4, f: 5, g: 6, h:7},
    ['a', 'c', 'e', 'g'],
    ['b', 'd', 'f', 'h']
);

// ☠🔥💻🔥☠ EVENTUALLY ISSUES AN ERROR ABOUT COMPLEXITY
createService(
    { a: 0, b: 1, c: 2, d: 3, e: 4, f: 5, g: 6, h:7, i: 8, j: 9, k: 10},
    ['a', 'c', 'e', 'g', 'i', 'k'],
    ['b', 'd', 'f', 'h', 'j']
);

因此，我可能建议您使用第一种方法，除非您有非常特殊的用例.

Playground link to code个