通过Typescript检查对象界面

Question

Is there a best way to check an object against a typescript infterface?

I am working between two schemas, where one extends the other. I need to convert the extension back to base, and am doing so by deleting props from the extension to shape it to the original base class. An example:

interface Base {
  prop1: string
  prop2?: string
  prop3?: string
}

interface Extended extends Base {
  prop4: string
  prop5: string
}

So in order to get Extended back to Base , I need to delete props prop4 and prop5

This is of course easy enough to do, but, I need to ensure that the final result conforms to the Base class, since deleting props does not tell the compiler anything , and the extension will pass the Base type check.

I have looked at using a type guard, and other alternatives , however, I have not found a way to do this which accommodates for the optional properties found in Base , or a seemingly efficient way to do it.

Here is a basic plan for the type guard as well as handling optional props:

function isBase(arg: any): arg is Base {
  return arg
    && arg.prop1
    && (arg.prop2 || true)
    && (arg.prop3 || true)
}

Besides being a lot of work for a lot of props, it's brittle. Soon as Base changes, this technique no longer works as intended. I feel like there is (should be) some Typescript way via utility types that could check the props against the Base interface, so it will always work as intended.

Update

I found this . Looks very close but could not get it working with interfaces.

Answer 1

If you need to keep Base as a standalone property, using composition instead of inheritance might be a better approach, for instance:

interface Base {
  prop1: string
  prop2?: string
  prop3?: string
}

interface Extended {
  data: Base
  prop4: string
  prop5: string
}

Do whatever is needed with Extended but only send data to database.