Represent `Map`s as records

[MOZGIII]

Maps are not handled properly. They can't be parsed directly from JSON, so it is not clear how to represent them. Currently, they are represented as [K, V][], but it would be great to have the ability to use { [K]: V } instead. It matters not that K must always be a string with this representation, since zod has a neat z.preprocess feature to fix it before parsing.

[StefanTerdell]

Maps are not handled properly

it is not clear how to represent them

Pick one :)

It doesn't matter that it has a preprocess if it can't be guaranteed to be used all the time.

[MOZGIII]

Pick one :)

Ok, fair.

It doesn't matter that it has a preprocess if it can't be guaranteed to be used all the time.

I'd be fine with an opt-in, since either way we build schema explicitly - maybe we could just have a configuration option?

[MOZGIII]

I ended up using this hack:

const coercedMap = <K extends ZodTypeAny, V extends ZodTypeAny>(
  keySchema: K,
  valueSchema: V,
) =>
  z.preprocess(
  (arg, ctx) =>
      arg === undefined
        ? new Map()
        : new Map(
            Object.entries(arg as {}).map(([key, val]) => [
              keySchema.parse(key, {
                path: [...ctx.path, key, "<key>"],
              }),
              val,
            ]),
          ),
    // For JSON Schemas pretend we are an object rather than a map.
    tweaks.jsonSchemaHack
      ? (z.record(keySchema, valueSchema) as never)
      : z.map(keySchema, valueSchema),
  );

tweaks.jsonSchemaHack is a global that is toggled on when we export schemas, but is off for everything else.

What I'm looking for is to get rid of the tweaks.jsonSchemaHack - if it was possible to just tell zod-to-json-schema I am using z.map but I want it to look like a z.record that would be great.

[StefanTerdell]

Sure, wth.

[StefanTerdell]

Done in master

[MOZGIII]

Sweet! Thanks!