3 replies

zod foreign key validation

This is probably a question that has been answered to death but I can't find anything that really helps so apologies if that's the case

I'm trying to validate a create mutation for a album table and I want to have a field that specifies the episode id foreign key that the album is discussed in

Getting the error 'Type 'string' has no properties in common with type 'MainEpisodeCreateNestedOneWithoutAlbumsInput'.'

Prisma schema looks something like this (cut out all the unnecessary stuff):

model MainEpisode {
  id String @id @default(cuid())
  name String
  albums Album[]
}

model Album {
  id String @id @default(cuid())
  name String
  mainEpisodeId String
  mainEpisode MainEpisode @relation(fields: [mainEpisodeId], references: [id])
}

model MainEpisode {
  id String @id @default(cuid())
  name String
  albums Album[]
}

model Album {
  id String @id @default(cuid())
  name String
  mainEpisodeId String
  mainEpisode MainEpisode @relation(fields: [mainEpisodeId], references: [id])
}

Is there a way, other than just calling the id a string, that I can specify the shape of the foreign key?

TTC

Theo's Typesafe Cult•4y ago•

3 replies

hyhy

zod foreign key validation

model MainEpisode {
  id String @id @default(cuid())
  name String
  albums Album[]
}

model Album {
  id String @id @default(cuid())
  name String
  mainEpisodeId String
  mainEpisode MainEpisode @relation(fields: [mainEpisodeId], references: [id])
}

model MainEpisode {
  id String @id @default(cuid())
  name String
  albums Album[]
}

model Album {
  id String @id @default(cuid())
  name String
  mainEpisodeId String
  mainEpisode MainEpisode @relation(fields: [mainEpisodeId], references: [id])
}

Is there a way, other than just calling the id a string, that I can specify the shape of the foreign key?

zod foreign key validation

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