Check if variable is a specific interface type in a typescript union

Question

Is it possible to create a typeguard, or something else that accomplishes the same purpose, to check if a variable is a specific interface type in a typescript union?

Answer 1

In TypeScript 2 you can use Discriminated Unions like this:

interface Foo {
    kind: "foo";
    a:string;
}
interface Bar {
    kind: "bar";
    b:string;
}
type FooBar = Foo | Bar;
let thing: FooBar;

and then test object using if (thing.kind === "foo").

If you only have 2 fields like in the example I would probably go with combined interface as @ryan-cavanaugh mentioned and make both properties optional:

interface FooBar {
    a?: string;
    b?: string
}

Note that in original example testing the object using if (thing.a !== undefined) produces error Property 'a' does not exist on type 'Foo | Bar'.

And testing it using if (thing.hasOwnProperty('a')) doesn't narrow type to Foo inside the if statement.

@ryan-cavanaugh is there a better way in TypesScript 2.0 or 2.1?

Answer 2

Since Typescript 1.6 you can use user-defined type guards:

let isFoo = (object: Foo| Bar): object is Foo => {
    return "a" in object;
}

See https://www.typescriptlang.org/docs/handbook/advanced-types.html#user-defined-type-guards and https://github.com/microsoft/TypeScript/issues/10485

Answer 3

typeof doesn't do this. It always return "string", "number", "boolean", "object", "function", or "undefined".

You can test for object properties with a test like if(thing.a !== undefined) { or if(thing.hasOwnProperty('a')) {.

Note that you could make an object that had both a string a and a string b, so be aware of that possibility.