Skip to content
Discussion options

You must be logged in to vote

Hello @LufyCZ

  1. You can do z.string().regex(/^0x.+$/) and the Open API depiction would be:
type: string
pattern: ^0x.+$
  1. You can use Custom brands approach for taking a full control on depiction.

I hope it helps. Let me know what you think.

Replies: 1 comment 1 reply

Comment options

You must be logged in to vote
1 reply
@LufyCZ
Comment options

Answer selected by RobinTail
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Category
Q&A
Labels
None yet
2 participants