Do not allow extra properties with zod parse

Question

I'm using zod for validation. It seems like if I define a schema and then parse() some input with some extra properties that aren't even in the schema, zod parses the input as valid but just removes those keys.

import { z } from 'zod'

const schema = z.object({
  foo: z.string(),
  bar: z.number()      
})

// this validates fine, printing { foo: 'hello', bar: 1 }
console.log(schema.parse({ foo: 'hello', bar: 1, baz: true }))

However, extra input properties is not something I'd like to ignore, instead I'd like to throw a useful error when that happens, reporting the keys of the extra properties.

Is there a way to do that with zod?

score 26 · Accepted Answer · answered Dec 19 '21 at 21:23

26

You can use the strict option:

const schema = z.object({
  foo: z.string(),
  bar: z.number()      
}).strict();

answered Dec 19 '21 at 21:23

nemesv

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Ah great, thanks! Even throws an error with a code `unrecognized_keys` and an array of those keys. – davnicwil Dec 19 '21 at 21:32
It works, but how to make it as default? I don't want to pass .strict to every z.object. – Renan Coelho Jun 28 '23 at 14:18
Using `z.strictObject` instead of `z.object` solved my issue. – Renan Coelho Jun 28 '23 at 14:44

Do not allow extra properties with zod parse

1 Answers1