Typescript version of Quick Start example

[demian85]

I've been trying to compile any example provided in the Readme with no success.
Typescript is making my life miserable. I cannot decipher the types for this part of the code:

const inputSchema = { sql: z.string() }
const outputSchema = {
  rows: z.array(z.record(z.any(), z.any())),
  rowCount: z.number(),
}

type InputSchema = z.infer<typeof inputSchema>
type OutputSchema = z.infer<typeof outputSchema>

server.registerTool<InputSchema, OutputSchema>(
  'query',
  {
    title: 'SQL Query',
    description: 'Execute SQL queries on the database',
    inputSchema,
    outputSchema,
  },
  async ({ sql }) => {
      // ...
  }
)

I found out about z.infer by searching for a fix, but still it doesn't work.

Can someone please explain how to fix those types?

Thanks.

[demian85]

It even fails if I change it to:

const inputSchema = z.object({ sql: z.string() })
const outputSchema = z.object({
  rows: z.array(z.record(z.string(), z.any())),
  rowCount: z.number(),
})

[chipgpt]

Are you using zod v3? They are working on zod v4 support and release is imminent but I don't think it works now.

Additionally, I don't think you need to set the generics, they will be implied for you. This should work if on zod 3:

const inputSchema = { sql: z.string() }
const outputSchema = {
  rows: z.array(z.record(z.any(), z.any())),
  rowCount: z.number(),
}

server.registerTool(
  'query',
  {
    title: 'SQL Query',
    description: 'Execute SQL queries on the database',
    inputSchema,
    outputSchema,
  },
  async ({ sql }) => {
      // ...
      return {}
  }
)

[demian85]

Oh OK. I installed zod@3 and seems to work now, thanks.