json datatype only validates in a formData object, not a regular object

[jdgamble555]

Description
If I use dataType: 'json' on the client, and pass the data to the server, the data only gets validated if it is a request object, not a regular object.

For example:

const form = await superValidate(request, valibot(mySchema));

This schema has no problems validating. The request object contains a formData object. You can see __superform_json only gets checked using a formData object on this line.

But what if I need to modify the data first?

Example:

const _formData = await request.formData();

// https://github.com/fabian-hiller/decode-formdata
const formData = decode(_formData, {
    arrays: ['tags'],
    files: ['photo']
});

const form = await superValidate(formData, valibot(mySchema));

You can see now the formData is no longer a request object, but a regular parsed object. Now, __superform_json won't get parsed at all.

Since superValidate accepts BOTH a request object and a regular object, so should the json validator. Otherwise, there is no way to modify the form data before passing it to the superValidate function. This means there are NO work arounds.

J

[ciscoheat]

If you transform the form data to an object, that's what will be validated, not the request. It could come from a DB or anywhere else.

It's not meant to combine FormData and the __superform_json data (which is encoded with devalue). The validation schema is for one of them, not both.

But if you want to bend the rules, you need to call superValidate twice:

// This will validate __superform_json
const form1 = await superValidate(_formData, valibot(superFormSchema));

// This will validate the transformed FormData
const form = await superValidate(formData, valibot(formDataSchema));

So YES, there are workarounds. 😀

[jdgamble555]

I think using devalue directly is what I was looking for. Thanks!