How can I add z.record(z.any()) to z.object()?

[dolfandringa]

Hi,

I want to create a schema equivalent to this type definition:

type T = {
    a: number;
    [key: string]: any;
}

Which could be successfully used as type for

{a: 1} and {a: 1, b: 'test', c:{ d: 'e'}} but not {b: 'test'}. I have seen z.record(z.any()), but I can't find a way to use it in an object to get the type definition above. I have tried

const TS = z.object({a: z.number()}, z.record(z.any()));
type T = z.infer<typeof TS>;
const t1: T = { a:1};
const t2: T = { a:1, b:2};

which results in

**<repl>.ts:30:25 - error TS2322: Type '{ a: number; b: number; }' is not assignable to type '{ a: number; }'.
  Object literal may only specify known properties, and 'b' does not exist in type '{ a: number; }'.**

and

const TS = z.object({a: z.number()}).merge(z.record(z.any()));

resulting in

<repl>.ts:5:44 - error TS2345: Argument of type 'ZodRecord<ZodString, ZodAny>' is not assignable to parameter of type 'AnyZodObject'.

and

const TS = z.union([z.object({a: z.number()}), z.record(z.any())]);
type T = z.infer<typeof TS>;
const t1: T = { a:1};
const t2: T = { a:1, b:2};
const t3: T = { b:2};

which doesn't result in an error but should.

How can I achieve this?
Cheers

[dolfandringa]

Lol, as usual, after trying everything and forcing myself to write the question down, I realize there is one thing I didn't try yet: catchall

const TS = z.object({a: z.string(), b: z.number()}).catchall(z.any());
type T = z.infer<typeof TS>;
const t: T = {a: 'test', b:2};
const t2: T = {a: 'test', b:2, c: {a:4, d:'e'}, f:'test'};
const t3: T = {b:2, c: {a:4, d:'e'}, f:'test'};

last line resulting in

<repl>.ts:10:7 - error TS2741: Property 'a' is missing in type '{ b: number; c: { a: number; d: string; }; f: string; }' but required in type '{ [x: string]: any; a: string; b: number; }'.

The z.record() sent me on a wild goose change because it almost did what I want, but it turned out to be much easier. Just documenting the answer for others.