getTitle

Returns the title of the schema.

If multiple titles are defined, the last one of the highest level is returned. If no title is defined, undefined is returned.

const title = v.getTitle<TSchema>(schema);

Generics

• TSchema extends BaseSchema<unknown, unknown, BaseIssue<unknown>> | BaseSchemaAsync<unknown, unknown, BaseIssue<unknown>> | SchemaWithPipe<readonly [BaseSchema<unknown, unknown, BaseIssue<unknown>>, ...(PipeItem<any, unknown, BaseIssue<unknown>> | TitleAction<unknown, string>)[]]> | SchemaWithPipeAsync<readonly [BaseSchema<unknown, unknown, BaseIssue<unknown>> | BaseSchemaAsync<unknown, unknown, BaseIssue<unknown>>, ...(PipeItem<any, unknown, BaseIssue<unknown>> | PipeItemAsync<any, unknown, BaseIssue<unknown>> | TitleAction<unknown, string>)[]]>

Parameters

• schema TSchema

Returns

• title string | undefined

Examples

The following examples show how getTitle can be used.

Get title of schema

Get the title of a username schema.

const UsernameSchema = v.pipe(
  v.string(),
  v.regex(/^[a-z0-9_-]{4,16}$/iu),
  v.title('Username'),
  v.description(
    'A username must be between 4 and 16 characters long and can only contain letters, numbers, underscores and hyphens.'
  )
);

const title = v.getTitle(UsernameSchema); // 'Username'

Overriding inherited titles

Get the title of a Gmail schema with an overridden title.

const EmailSchema = v.pipe(v.string(), v.email(), v.title('Email'));

const GmailSchema = v.pipe(
  EmailSchema,
  v.endsWith('@gmail.com'),
  v.title('Gmail')
);

const title = v.getTitle(GmailSchema); // 'Gmail'

Related

The following APIs can be combined with getTitle.

Actions

• title