函数重载似乎是一个很好的解决方案.
首先,我们创建一个名为Overwrite
的类型来更改某些属性的类型.
type Overwrite<T, U> = Pick<T, Exclude<keyof T, keyof U>> & U extends infer O ? {
[K in keyof O]: O[K]
} : never
现在我们可以添加所有可能的函数重载:
function serializeDates<T extends HasBothValues>(obj: T): Overwrite<T, {updatedAt: string, createdAt: string}>
function serializeDates<T extends HasUpdatedAt>(obj: T): Overwrite<T, {updatedAt: string}>
function serializeDates<T extends HasCreatedAt>(obj: T): Overwrite<T, {createdAt: string}>
function serializeDates<T extends AllInputVariants>(obj: T) {
const retObj: any = { ...obj };
if (obj.createdAt) {
retObj.createdAt = obj.createdAt.toISOString();
}
if (obj.updatedAt) {
retObj.updatedAt = obj.updatedAt.toISOString();
}
return retObj;
};
Playground
让我们看看它是否有效:
const t1 = serializeDates({
createdAt: new Date(),
})
// const t1: {
// createdAt: string;
// }
const t2 = serializeDates({
updatedAt: new Date()
})
// const t2: {
// updatedAt: string;
// }
const t3 = serializeDates({
createdAt: new Date(),
updatedAt: new Date()
})
// const t3: {
// updatedAt: string;
// createdAt: string;
// }
const t4 = serializeDates({
createdAt: new Date(),
updatedAt: new Date(),
a: 123,
b: "123"
})
// const t4: {
// a: number;
// b: string;
// updatedAt: string;
// createdAt: string;
// }
再想想,我们可以做得更好.让我们修改Overwrite
,只覆盖T
上实际存在的T
属性.
type Overwrite<T, U> = (Pick<T, Exclude<keyof T, keyof U>> & Pick<U, Extract<keyof U, keyof T>>) extends infer O ? {
[K in keyof O]: O[K]
} : never
现在我们不再需要函数重载:
function serializeDates<
T extends { updatedAt?: Date, createdAt?: Date }
>(obj: T): Overwrite<T, {updatedAt: string, createdAt: string}> {
const retObj: any = { ...obj };
if (obj.createdAt) {
retObj.createdAt = obj.createdAt.toISOString();
}
if (obj.updatedAt) {
retObj.updatedAt = obj.updatedAt.toISOString();
}
return retObj as any;
};
Playground