我在 Typescript 将默认值传递给类似于 lodash 的 pick
的函数时遇到问题。
该函数接受一个已知(非通用)接口(interface)的对象和一组从该对象中选取和返回的键。
函数的常规(无默认参数)声明工作正常,但是,我似乎无法将数组设置为选择属性的参数的默认值。
interface Person {
name: string;
age: number;
address: string;
phone: string;
}
const defaultProps = ['name', 'age'] as const;
function pick<T extends keyof Person>(obj: Person, props: ReadonlyArray<T> = defaultProps): Pick<Person, T> {
return props.reduce((res, prop) => {
res[prop] = obj[prop];
return res;
}, {} as Pick<Person,T>);
}
const testPerson: Person = {
name: 'mitsos',
age: 33,
address: 'GRC',
phone: '000'
};
如果您删除默认值 = defaultProps
,它会成功编译,并且从示例调用返回的类型也是正确的,例如:const testPick = pick(testPerson, ['name'] );
但是,设置默认值会产生以下错误:
Type 'readonly ["name", "age"]' is not assignable to type 'readonly T[]'.
Type '"name" | "age"' is not assignable to type 'T'.
'"name" | "age"' is assignable to the constraint of type 'T', but 'T' could be instantiated with a different subtype of constraint 'keyof Person'.
Type '"name"' is not assignable to type 'T'.
'"name"' is assignable to the constraint of type 'T', but 'T' could be instantiated with a different subtype of constraint 'keyof Person'.
如何将默认值成功传递给 props
参数?
Typescript Playground 链接 here
更新
玩了一会儿之后,我尝试使用条件类型并设法使函数签名正常工作,但现在无法正确识别 reduce 的问题:
interface Person {
name: string;
age: number;
address: string;
phone: string;
}
const defaultProps = ['name', 'age'] as const;
type DefaultProps = typeof defaultProps;
type PropsOrDefault<T extends keyof Person> = DefaultProps | ReadonlyArray<T>;
type PickedPropOrDefault<T extends PropsOrDefault<keyof Person>> = T extends DefaultProps ? Pick<Person, DefaultProps[number]> : Pick<Person, T[number]>;
function pick<T extends keyof Person>(obj: Person, props: PropsOrDefault<T> = defaultProps): PickedPropOrDefault<PropsOrDefault<T>> {
return props.reduce<PickedPropOrDefault<PropsOrDefault<T>>>((res, prop) => {
res[prop] = obj[prop];
return res;
}, {} as PickedPropOrDefault<PropsOrDefault<T>>);
}
const testPerson: Person = {
name: 'mitsos',
age: 33,
address: 'GRC',
phone: '000'
};
const result = pick(testPerson) // Pick<Person, "name" | "age">
const result2 = pick(testPerson, ['phone']) // Pick<Person, "phone">
const result3 = pick(testPerson, ['abc']) // expected error
最佳答案
您可以重载 pick
功能:
interface Person {
name: string;
age: number;
address: string;
phone: string;
}
const defaultProps = ['name', 'age'] as const;
type DefaultProps = typeof defaultProps;
function pick(obj: Person): Pick<Person, DefaultProps[number]>
function pick<Prop extends keyof Person, Props extends ReadonlyArray<Prop>>(obj: Person, props: Props): Pick<Person, Props[number]>
function pick<T extends keyof Person>(obj: Person, props = defaultProps) {
return props.reduce((res, prop) => ({
...res,
[prop]: obj[prop]
}), {} as Pick<Person, T>);
}
const testPerson = {
name: 'mitsos',
age: 33,
address: 'GRC',
phone: '000'
};
const result = pick(testPerson) // Pick<Person, "name" | "age">
const result2 = pick(testPerson, ['phone']) // Pick<Person, "phone">
const result3 = pick(testPerson, ['abc']) // expected error
你可以找到更高级的pick
在我的 article 中打字和其他答案:
First , second , third
更新
props
有问题此代码中的参数:
function pick<T extends keyof Person>(obj: Person, props: PropsOrDefault<T> = defaultProps): PickedPropOrDefault<PropsOrDefault<T>> {
return props.reduce((res, prop) => {
return {
...res,
[prop]: obj[prop]
}
}, {});
}
PropsOrDefault
可能等于此类型:type UnsafeReduceUnion = DefaultProps | ReadonlyArray<'phone' | 'address'>
您可能已经注意到,联合中的这些数组是完全不同的。他们没有任何共同点。
如果你想调用reduce
:
declare var unsafe:UnsafeReduceUnion;
unsafe.reduce()
你会得到一个错误,因为reduce
是not callable
关于typescript - 如何在 Typescript 中将默认值设置为具有类型 keyof(来自显式类型)的参数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/69680545/