JSON utility type about zod HOT 6 CLOSED

Now that the requirement is 4.1+, any chance z.json() gets revisited?

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@ggoodman I decided to add this snippet to the README so it's easier to find 👍 Thanks again!

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Yikes! When I copied the snippet above into the README I renamed Json to jsonSchema (to avoid confusion with the built-in JSON class) but I forgot to change it everywhere 😬 Here's the correct snippet:

type Literal = boolean | null | number | string;
type Json = Literal | { [key: string]: Json } | Json[];
const literalSchema = z.union([z.string(), z.number(), z.boolean(), z.null()]);
const jsonSchema: z.ZodSchema<Json> = z.lazy(() =>
  z.union([literalSchema, z.array(jsonSchema), z.record(jsonSchema)]),
);

jsonSchema.parse({
  // data
})

Thanks for pointing this out! I fixed this in the README as well.

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Love this! I'd add it to the standard API (z.json()) but the recursive type alias required TypeScript 3.7. Currently everything else in the lib only requires 3.2+ so I don't want to bump the version without a really compelling reason. But thanks for sharing!

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Hey @ggoodman or @vriad, thanks for this helpful snippet.

I'm slowly getting more comfortable with Zod. But every now and then I feel like I am just missing the point of something.

Now is such a time.

When using this snippet with say jsonSchema.parse(data)

I get this in the log.

    types/util.ts:190:68 - error TS2693: 'Literal' only refers to a type, but is being used as a value here.

    190 export const jsonSchema: z.ZodSchema<Json> = z.lazy(() => z.union([Literal, z.array(Json), z.record(Json)]));
                                                                           ~~~~~~~
    types/util.ts:190:85 - error TS2693: 'Json' only refers to a type, but is being used as a value here.

    190 export const jsonSchema: z.ZodSchema<Json> = z.lazy(() => z.union([Literal, z.array(Json), z.record(Json)]));
                                                                                            ~~~~
    types/util.ts:190:101 - error TS2693: 'Json' only refers to a type, but is being used as a value here.

    190 export const jsonSchema: z.ZodSchema<Json> = z.lazy(() => z.union([Literal, z.array(Json), z.record(Json)]));

After trying to work this out I realised I am just playing whack a mole.

Any direction would be appreciated :)

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Sharing here in case it will help somebody else, for our application, we had to change the string to be a symbol to handle some of our more complex types:

type Json = Literal | { [key: symbol]: Json } | Json[];

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