有没有一种方法可以在不创建特定类型的情况下使用具有复杂验证的断言函数？

Question

I'm receiving data from api and I'd like to add some validations on front-end and get correct types after that validations, but really specific ones. For example, suppose I'm getting data with type { keyA?: { property?: string | null | undefined }, keyB?: string | null } | null { keyA?: { property?: string | null | undefined }, keyB?: string | null } | null { keyA?: { property?: string | null | undefined }, keyB?: string | null } | null and I'd like to be sure property is a non empty string, while keyB could be null (just to illustrate I don't want to validate every field, it's not a common NonNull~ case). Is there a way to do it in another function scope (that receives this data) without manually creating a type?

I tried creating a function:

function assertData(data: DType): asserts data {
  if(!data) throw new Error('no data')
  if(!data.keyA) throw new Error('keyA undefined');
  if(!data.keyA.property) throw new Error('property undefined');

After calling the function, I can use data (not null value), but the keyA can be undefined . I expected to able to use data.keyA.property without problems and already typed as string.

Answer 1

You could modify the function to have the following signature:

function assertData(data: DType): asserts data is DType & { 
  keyA: { property: string } 
} {
  if(!data) throw new Error('no data')
  if(!data.keyA) throw new Error('keyA undefined');
  if(!data.keyA.property) throw new Error('property undefined')
}

We assert that data is DType but also that keyA and keyA.property are not undefined.

declare const a: DType
assertData(a)
a.keyA.property.charAt(0)

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