[question] How do we get string or string array from GET queries?

[tgfjt]

I expect the to be valid as an Array even if it has only one element like keys=foo.
Actually, I got an error: path":["keys"],"message":"Expected array, received string"}].

1. valid: /test?keys=foo&keys=bar
2. invalid: /test?keys=foo

How do I pass both of 1 & 2?
Thank you.

import { createRoute, z, OpenAPIHono } from '@hono/zod-openapi';

const app = new OpenAPIHono();

const ParamsSchema = z.object({
  keys: z
    .string()
    .array()
    .openapi({
      example: ['foo', 'bar'],
    }),
});

const route = createRoute({
  method: 'get',
  path: '/test',
  request: {
    query: ParamsSchema,
  },
  responses: {
    200: {
      content: {
        'application/json': {
          schema: ParamsSchema,
        },
      },
    },
  },
});

[yusukebe]

Hi @tgfjt

Try this:

const ParamsSchema = z.object({
  keys: z.union([
    z
      .string()
      .array()
      .openapi({
        example: ['foo', 'bar']
      }),
    z.string().openapi({
      example: 'foo'
    })
  ])
})

const testRoute = createRoute({
  method: 'get',
  path: '/test',
  request: {
    query: ParamsSchema
  },
  responses: {
    200: {
      content: {
        'application/json': {
          schema: ParamsSchema
        }
      },
      description: 'foo'
    }
  }
})

app.openapi(testRoute, (c) => {
  const { keys } = c.req.valid('query') ?? []
  return c.jsonT({
    keys
  })
})

[tgfjt]

@yusukebe
Understood, thank you so much!

Actually, I've tried not to use z.union as I heard that anyOf is not supported by OpenAPITools/openapi-generator.
That's can't be helped. I will try z.union. 👍
OpenAPITools/openapi-generator#10514