Typescript check if property in object in typesafe way

Question

The code

const obj = {};
if ('a' in obj) console.log(42);

Is not typescript (no error). I see why that could be. Additionally, in TS 2.8.1 "in" serves as type guard.

But nevertheless, is there an way to check if property exists, but error out if the property is not defined in the interface of obj?

interface Obj{
   a: any;
}

I'm not talking about checking for undefined...

Halfsole

Possible duplicate of How to determine whether an object has a given property in JavaScript – Ensanguine

Not a duplicate. – Ailin

Answer 1

You don't get an error because you use a string to check if the property exists.

You will get the error this way:

interface Obj{
   a: any;
}

const obj: Obj = { a: "test" };

if (obj.b)          // this is not allowed
if ("b" in obj)     // no error because you use string

If you want type checking to work for string properties you could add index signatures using this example

Chromophore

But "if (obj.b)" also disallows undefined. There is a difference between property not existing and it being undefined. I JUST want to check if the property exists in a typesafe way. – Halfsole

The type system already defines if the property exists or not, so there is no need to check it yourself. Type guards are used when an object can be of multiple types, such as: const obj: Thing | OtherThing. – Chromophore

No if i have an interface { a?: any; }, I do not know if the property exists. And if I get an object from a server I don't control (e.g.), I never know what property exists. I just want to discriminate in a typesafe way between actual undefined or property not existing at all! – Halfsole

If you don't know what your JSON data looks like, you can use console.log(o.hasOwnProperty("a")) to see if a property exists at all. If it exists but it's undefined, then it will still return true. – Chromophore

still is not a typesafe check but your answer is stell the best because only one. and i can always write a function to do a typesafe "in" check myself (with typescript keyof syntax) – Halfsole

I should add an example of index signatures to get type safety with obj[“test”] – Chromophore

Very useful to verify interfaces! basarat.gitbook.io/typescript/type-system/typeguard#in – List

Answer 2

The following handle function checks hypothetical server response typesafe-way:

/**
 * A type guard. Checks if given object x has the key.
 */
const has = <K extends string>(
  key: K,
  x: object,
): x is { [key in K]: unknown } => (
  key in x
);

function handle(response: unknown) {
  if (
    typeof response !== 'object'
    || response == null
    || !has('items', response)
    || !has('meta', response)
  ) {
    // TODO: Paste a proper error handling here.
    throw new Error('Invalid response!');
  }

  console.log(response.items);
  console.log(response.meta);
}

Playground Link. Function has should probably be kept in a separate utilities module.

Orient

This is not type-safe though. If you define response: {'a': number}, has('items', response) does not yield an expected compile-time error. – Bosomy

Erm... I've published the solution for cases when you're unable to control the type of arguments of the function handle. For example, this could be API endpoint handler -- everyone can accidentally send invalid data there. That's why handle accepts the argument of type unknown (which means "everything, I don't know what"). Type-safety here means that you're unable (in compile time) to use response without checking its type. E.g. here in line 24 response.somethingElse was used without type-check and TypeScript complains about that. – Orient

Oh, so you meant "type-safe" in a completely different way than I thought. Cool one, thanks for the response! – Bosomy

Answer 3

You can implement your own wrapper function around hasOwnProperty that does type narrowing.

function hasOwnProperty<T, K extends PropertyKey>(
    obj: T,
    prop: K
): obj is T & Record<K, unknown> {
    return Object.prototype.hasOwnProperty.call(obj, prop);
}

I found this solution here: TypeScript type narrowing not working when looping

Usage:

const obj = {
    a: "what",
    b: "ever"
} as { a: string }

obj.b // Type error: Property 'b' does not exist on type '{ a: string; }'

if (hasOwnProperty(obj, "b")) {
    // obj is no longer { a: string } but is now
    // of type { a: string } & Record<"b", unknown>
    console.log(obj.b)
}

The limitation with this approach is that you only get a Record back with the single key added that you specified. This might be fine for some needs, but if you need a more general solution then I suggest a library like Zod which can validate a complex object and give you the full type: https://github.com/colinhacks/zod

Hydrodynamics

I'm baffled that the built-in property checks don't already narrow types like this. I'm adding this to my default "include in all my TypeScript projects" code. – Fahland

Can you explain what obj is T & Record<K, unknown> means please? – Irradiance

I keep getting Type '<Whatever>' is not assignable to type 'never' errors when I do if (!has(o, p) { o[p] = x } when using your signature instead of const has = (obj: { [key: string]: any; }, prop: string) => { which is what I had before. I have to keep doing (o[p] as Whatever) = x – Irradiance

Docs for type predicates, intersection types, and the record type – Mopboard