我不确定这是可能的,但我非常接近。
如果我有这个物体/形状:
export const initialState: State = {
foods: {
filter: '',
someForm: {
name: '',
age: 2,
},
someFormWithoutAnAge: {
name: '',
},
}
};
declare function pickAForm<R extends keyof State, S extends keyof State[R]>(key1: R, key2: S): void;
该功能效果很好,我可以使用类型安全调用pickAForm("foods", "someForm")
,如果我pickAForm("foods", "somePropertyThatDoesntExist")
但是,我想增加额外的安全性,因为你只能挑选具有特定形状的物品。示例:someForm
应该有效但someFormWithoutAnAge
应该失败,因为您选择必须具有年龄属性。像这样:
declare function pickAFormWithAge<R extends keyof State, S extends keyof State[R], T extends State[R][S] & {age: number}>(key1: R, key2: S): void;
但我完全不确定如何去做。总结一下:
pickAFormWithAge('foods', 'someForm') // Passes
pickAFormWithAge('foods', 'someFormWithoutAge') // Should fail, does not look like {age: number}
pickAFormWithAge('foods', 'blah') // Should fail, not a key
答案 0 :(得分:2)
我能做到这一点的唯一方法是:
一个。约束数据结构以匹配字符串文字,而不是相反 湾将数据结构作为函数参数传递。
const state = {
foods: {
filter: '',
someForm: {
name: 'Some form',
age: 2
},
someFormWithoutAnAge: {
name: 'other form',
priority: 10
}
}
};
interface HasAge { age: number }
// IMPLEMENTATION
function getForm<O extends {[P in P1]: {[P in P2]: HasAge}}, P1 extends string, P2 extends string>(o: O, p1: P1, p2: P2) {
return (o[p1] as any)[p2] as any;
}
// USAGE
const form1 = getForm(state, 'foods', 'someForm'); // GOOD
const form2 = getForm(state, 'foods', 'someFormWithoutAnAge'); // ERROR
const form3 = getForm(state, 'foods', 'blah'); // ERROR
更简单,更灵活的解决方案是使用常规代码。 pickAForm
采用函数而不是字符串文字。
const form1 = pickAForm((state) => state.foods.someForm);
// ...or...
const form1 = pickAForm(() => initialState.foods.someForm);