Correct way to use a third party Class/type in a Zod schema object?

[callumlocke]

I'm trying to do this, but TypeScript complains:

  z.object({
    slug: z.string().min(1).max(255),
    title: z.string().max(255),
    timeStart: Timestamp,
  })

Error:

Type 'typeof Timestamp' is missing the following properties from type 'ZodType<any, ZodTypeDef>': _type, _def, parse, is, and 5 more.

I basically want to make a schema that asserts that timeStart is an instance of the class Timestamp (which itself is a class I imported from a third party library).

How can I do this? Sorry if it's obvious.

Thank you for writing this library by the way, it's by far the best solution for TypeScript runtime checking I've found (and I've tried many).

[colinhacks]

Hey Callum,

It's definitely not obvious. It may make sense to add a method specifically for this use case, I'll add it to the list.

In the meantime, it's already achievable but it's not super pretty:

const TimestampSchema: z.ZodType<Timestamp> = z.any().refine(x => x instanceof Timestamp);

z.any() lets anything through, but with the .refine makes sure that it's an instance.

You can "hard code" the type you want to validate by casting to z.ZodType<YourTypeHere>. Inference works as you'd expect:

type Timestamp = z.infer<typeof TimestampSchema>;
const time: Timestamp = new Timestamp(); // works fine

[callumlocke]

Your suggestion using z.ZodType and .refine() works for me, thanks.

Yes, it's not super pretty at first glance. It does make sense now I see it, but it feels like a clever trick. I agree that something like z.instanceOf(Foo) would be a good addition for my specific use case. But I can see myself needing to use that z.any().refine(fn) trick again for other special cases. Maybe it would be good to provide a wrapper with a more intuitive name, like z.custom()? I guess it would also need to take a generic to set the type, something like:

const x: = z.custom<MyType>((value: any) => {
  // examine value and return true/false
}))

z.infer<x> // MyType

One more thing while I'm here: I think the name z.ZodType<T> might be better as z.Schema<T>, as you're using the word 'schema' throughout the readme. (Unless I've misunderstood what this utility is?)

Just to reiterate, I am in love with this library! Thank you for writing it. It's made writing TypeScript a joy.

[colinhacks]

Lots of great ideas here! Just added all three of these things to my todo list 🤙

I follow up here when they're implemented.