CODEWITHSUNDEEP
typescriptconditional-types
Simon_Weaver
Simon_Weaver

Reputation: 146000

Compile time check if typescript interface has one or more properties

I need to find a way to check if a typescript interface has one or more properties (property names are unknown) at compile time.

So for example given the following definitions:

export type Cat = {};
export type Dog = { barking: boolean };

I need a conditional type HasAnyProperties<T> which will give me:

type catHasProperties = HasAnyProperties<Cat>;   // false   (because Cat is {})
type dogHasProperties = HasAnyProperties<Dog>;   // true    (because Dog has one or more properties)

To be clear I do not want:

This may seem like an odd request, but what I'm actually doing is first filtering a type after which I need to know if anything was left over. I then use that value to optionally add a new property on a mapped type. But that's the easy part!

The following attempts don't work:

// This always returns true
type HasAnyProperties<T> = T extends { [key: string]: any } ? true : false;

It would be a lot easier if I knew the names of the properties, but I don't.

My guess if this is possible is that it will look something like RequireAtLeastOne.

Upvotes: 0

Views: 840

Answers (1)

Simon_Weaver
Simon_Weaver

Reputation: 146000

Turns out it's as simple as this:

 export type HasAnyProperties<T> = {} extends T ? false : true;

At least for my case. If anyone with a similar need finds issues with this please comment or add a new answer.

Upvotes: 2

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