Why can I not extend 'any' in Typescript?

Question

I want to make a placeholder interface for classes that use an external library. So I made:

export interface IDatabaseModel extends any
{
}

I'd rather do this and add methods later (e.g. findAll()) than mark all my classes as 'any' and have to manually search and replace 'any' in hundreds of specific places with 'IDatabaseModel' later.

But the compiler complains it cannot find 'any'.

What am I doing wrong?

Answer 1

With an alias for any

I want to make a placeholder interface

Use an alias:

type IDatabaseModel = any;

Or with an index signature [Edit from August, 2017]

Since TypeScript 2.2, the property access syntax obj.propName is allowed with an index signature:

interface AnyProperties {
    [prop: string]: any
}

type MyType = AnyProperties & {
    findAll(): string[]
}

let a: MyType
a.findAll() // OK, with autocomplete
a.abc = 12 // OK, but no autocomplete

Answer 2

2020 TLDR:

instead of extends any try extends Record<string, any>.

Long version:

In certain cases extends any still works in TypeScript. However, in TypeScript 3.9, it is interpreted in a more conservative way.

@Paleo creates AnyProperties interface in his answer. That can now be rewritten with Record as extends Record<string, any>.

Answer 3

The any type is a way to opt-out of type checking and so it doesn't make sense to use it in an environment where type checking is done.

You could instead add your methods to the interface as you use them:

export interface IDatabaseModel
{
    findAll(): any;
}

Edit: See Paleo's answer for using a type alias in the meantime.

Answer 4

"any" is a type and, unlike other basic types, doesn't appear to have an interface. Given it is a wildcard of sorts, I can't imagine what an interface to "any" could consist of.