Is it possible to generate schema from object literal?

[Gumichocopengin8]

I have an object literal like this.

const OBJ_LITERAL_ARRAY = [
  { name: 'name1', text: 'Text1' },
  { name: 'name'2, text: 'Text2' },
  // ... 1,000 more
  { name: 'nameN', text: 'TextN' },
] as const

It is painful to construct zod schema by hand.

Are there any ways to generate zod schema from the object literal like the following??

const OBJ_LITERAL_ARRAY_SCHEMA = z.generateFromObjectLiteral(OBJ_LITERAL_ARRAY);

// OBJ_LITERAL_ARRAY_SCHEMA should be like this
const OBJ_LITERAL_ARRAY_SCHEMA = z.array(
  z.union([
    z.object({
      code: z.literal('text1'),
      name: z.literal('Text1'),
    }),
    z.object({
      code: z.literal('regex2'),
      name: z.literal('Regex2'),
    }),
   // 1000 more...
  ])
);

[BrendanC23]

Could you dynamically build your schema based on the object? Something like this (untested):

const OBJ_LITERAL_ARRAY = [
  { name: 'name1', text: 'Text1' },
  { name: 'name'2, text: 'Text2' },
  // ... 1,000 more
  { name: 'nameN', text: 'TextN' },
] as const;

const zodObjects = OBJ_LITERAL_ARRAY.map((obj) => {
    return z.object({
        text: z.literal(obj.text),
        name: z.literal(obj.name),
    });
});

const OBJ_LITERAL_ARRAY_SCHEMA = z.array(z.union(zodObjects));

---

Alternatively, you could do something like (untested):

const OBJ_LITERAL_ARRAY_SCHEMA = z.array(
    z.object(
        text: z.string(),
        name: z.string(),
    ),
).superRefine((data, ctx) => {
    for (let i = 0; i < data.length; i++) {
        // If there's a pattern to name/text, you could calculate the expected value instead of comparing to OBJ_LITERAL_ARRAY
        if (data[0].text !== OBJ_LITERAL_ARRAY[i].text) {
            ctx.addIssue({ code: z.ZodIssueCode.custom, message: `Index ${i} text is invalid.` });
        }

        if (data[0].name !== OBJ_LITERAL_ARRAY[i].name) {
            ctx.addIssue({ code: z.ZodIssueCode.custom, message: `Index ${i} name is invalid.` });
        }
    }
});

[Gumichocopengin8]

Thanks, but both don't work.