Using zod to serialize data and remove data not present in the scheme

[derolf]

Let's assume I have a schema like:

const Person = z.object({name: z.string();});

Is there a way to use the Person-scheme to turn an object into a JSON while removing any additional values that are not present in the schema?

So, Person.serialize({name: 'Foo', secret: 'Bar'}) should serialize to a JSON and remove secret

[scotttrinh]

By default, schemas will strip unknown keys, so this should work:

const json = JSON.stringify(Person.parse({ name: "Foo", secret: "Bar" }));

[JacobWeisenburger]

@scotttrinh suggestion works, or you can also use .transform

const Person = z.object( { name: z.string() } )
const personToString = Person.transform( JSON.stringify )

personToString.parse( { name: 'Foo' } )                // => {"name":"Foo"}
personToString.parse( { name: 'Foo', secret: 'Bar' } ) // => {"name":"Foo"}
personToString.parse( { name: 42 } )                   // throws error

More Info:

• Unrecognized keys
• .transform