如何使模式中的所有键都不是必需的

Question

I'm looking for a simple way to transform a yup schema to the same schema without any required field.

const requiredSchema = yup.object().shape({
  name: yup.string().required(),
  description: yup.string().required(),
})
// to
const notRequiredSchema = yup.object().shape({
  name: yup.string(),
  description: yup.string(),
})

I am looking for a function to apply on either requiredSchema or notRequiredSchema . The idea looks kind of like the Partial key word for types (TS) but for yup objects basically.

Answer 1

Yup supports this in the 1.0.0 beta versions . If you're not ok with using the beta version you could snag the code they're using to implement "partial" for object schemas.

This is the gist of it:

function partial(objectSchema) {
  const partial: any = {};
  for (const [key, schema] of Object.entries(objectSchema.fields)) {
    partial[key] =
    "optional" in schema && schema.optional instanceof Function
    ? schema.optional()
    : schema;
  }
  
  objectSchema.fields = partial

  return  objectSchema
}

The typescript seems like it might be kind of a mess to hook up if you implement it yourself.