Can you declare a object literal type that allows unknown properties in typescript?

Question

Essentially I want to ensure that an object argument contains all of the required properties, but can contain any other properties it wants. For example:

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Answer 1

This can be most easily accomplished by defining your object before the function call:

function foo(bar: { baz: number }) : number {
    return bar.baz;
}

const obj = { baz: 1, other: 2 };

foo(obj);

Answer 2

Yes, you can. Try this:

interface IBaz {
    baz: number;
    [key: string]: any;
}

function foo(bar: IBaz) : number {
    return bar.baz;
}

foo({ baz: 1, other: 2 });

Answer 3

Well, i hate answering my own questions, but the other answers inspired a little thought... This works:

function foo<T extends { baz: number }>(bar: T): void {
    console.log(bar.baz);
}

foo({baz: 1, other: 2});

Answer 4

If the known fields are coming from a generic type the way to allow wildcards is with T & {[key: string]: unknown}, any fields that are known must fit with the type's constraints and other fields are allowed (and considered type unknown)

Here is a sample:

type WithWildcards<T> = T & { [key: string]: unknown };

function test(foo: WithWildcards<{baz: number}>) {}

test({ baz: 1 }); // works
test({ baz: 1, other: 4 }); // works
test({ baz: '', other: 4 }); // fails since baz isn't a number

Then if you have a generic type T you can allow wildcard fields with WithWildCards<T>