Theo's Typesafe Cult•2y ago

Is it possible to dynamically generate zod schema?

Imagine a form where a user can add/remove fields of different types, how can we dynamically generate zod schema for that form?

7 Replies

Neto•2y ago

generate the basic zod structure such as z.string() then merge them probably its just impossible the get the correct type later

shikishikichangchangOP•2y ago

what do you mean

Neto•2y ago

https://zod.dev/?id=merge

GitHub

TypeScript-first schema validation with static type inference

Neto•2y ago

const input = {elem1: 'string', elem2: 'string'}

const getZodElem = (t: string) => {
  if (t === 'string') {
    return z.string()
  }
  ...
}

// map input into zod schema

const input = {elem1: 'string', elem2: 'string'}

const getZodElem = (t: string) => {
  if (t === 'string') {
    return z.string()
  }
  ...
}

// map input into zod schema

shikishikichangchangOP•2y ago

ok let me try that thank you so much!

Pijus•16mo ago

@shikishikichangchang what was final solution that you did? I've started gathering ideas today, wondering if there is any clean implementation

Tomate Malicioso•15mo ago

You can generate the zod validator but you can't infer the type from it , as it'd be done on runtime and not during the TypeScript compilation step. If I understand the implications correctly, you can't pick the type from the generated validator and, for example, have it as an argument or return type from another function. But you can definitely validate around using your dynamically generated zod validator.

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Is it possible to dynamically generate zod schema?