13 replies

[drizzle-zod] How to make all fields of a schema derived from a table required ?

I just tried the plugin and I'm happy to move from manually creating zod schemas.
However, I came into a use case where a schema must have all nullable fields in the table required.

Assume we have these in users table:

  const users = mysqlTable("users", {
  ...
  speciality: varchar('speciality', { length: 255 }),
  level: tinyint('level'),
  ...
});

  const users = mysqlTable("users", {
  ...
  speciality: varchar('speciality', { length: 255 }),
  level: tinyint('level'),
  ...
});

The generated schema will have this type

type Schema = {
speciality: string | null | undefined,
level: number | null | undefined
}

type Schema = {
speciality: string | null | undefined,
level: number | null | undefined
}

Is there a way to make it like this ?

type Schema = {
speciality: string ,
level: number 
}

type Schema = {
speciality: string ,
level: number 
}

Drizzle Team•3y ago•

13 replies

Ahmed

[drizzle-zod] How to make all fields of a schema derived from a table required ?

  const users = mysqlTable("users", {
  ...
  speciality: varchar('speciality', { length: 255 }),
  level: tinyint('level'),
  ...
});

  const users = mysqlTable("users", {
  ...
  speciality: varchar('speciality', { length: 255 }),
  level: tinyint('level'),
  ...
});

The generated schema will have this type

type Schema = {
speciality: string | null | undefined,
level: number | null | undefined
}

type Schema = {
speciality: string | null | undefined,
level: number | null | undefined
}

Is there a way to make it like this ?

type Schema = {
speciality: string ,
level: number 
}

type Schema = {
speciality: string ,
level: number 
}

[drizzle-zod] How to make all fields of a schema derived from a table required ?

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