Fix incorrect type inference: ensure ZodString instead of generic ZodType<string> on convexToZod function types #692
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…nvexToZod function output
ianmacartney
requested changes
Jul 18, 2025
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| : V extends VLiteral<any, any> | ||
| ? z.ZodLiteral<V["value"]> |
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Could this be VLiteral<infer T, any> instead of using V["value"]?
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| : V extends VObject<any, any, any, any> | ||
| ? z.ZodObject<{ | ||
| [K in keyof V["fields"]]: ZodValidatorFromConvex< |
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Same here for extracting the type instead of using V["fields"] - though maybe your version avoids some inference challenges I'm unaware of? wdyt?
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| : V extends VString<any, any> | ||
| ? z.ZodString |
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Do we lose branded types here? e.g. the brandedStringhelper results in aVString<string & {_brand: "foo" },type - can we pass that through to a more specificZodString` type?
| @@ -1393,7 +1444,9 @@ export function convexToZod<V extends GenericValidator>( | |||
| throw new Error(`Unknown convex validator type: ${convexValidator.kind}`); | |||
| } | |||
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| return isOptional ? z.optional(zodValidator) : zodValidator; | |||
| return isOptional | |||
| ? (z.optional(zodValidator) as ZodValidatorFromConvex<V>) | |||
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are optional types going through? I fear above that we're losing that.
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What This PR Does
This PR fixes an issue where the inferred type of a Zod schema is too generic (
ZodType<string, ZodTypeDef, string>), which prevents chaining string-specific methods such as.min(),.email(), etc.Problem
When conditionally constructing a schema, the inferred type ends up being the generic
ZodType<string, ZodTypeDef, string>instead of the more specificZodString. As a result, methods like.min()are not recognized by TypeScript:Closes #671