[How To?] Create a record in database on form submission...

export const carsRouter = router({
  all: publicProcedure.query(async ({ ctx }) => {
    const { db } = ctx
    const allCars = await db.select().from(CarTable).all()
    return allCars
  }),
  create: publicProcedure
    .input((raw) => parse(CarSchema, raw))
    .mutation(async ({ ctx, input }) => {
      const { db } = ctx
      await db.insert(CarTable).values(input).run()
    }),
})

export const CarTable = sqliteTable('Car', {
  id: integer('id').primaryKey({ autoIncrement: true }),
  uuid: binary('uuid', { length: 16 }).default('hex(randomblob(16))').unique(), // Generate a random 16-byte binary string
  make: text('make'),
  model: text('model'),
  year: integer('year'),
  color: text('color'),
  price: real('price'),
  mileage: integer('mileage'),
  fuelType: text('fuelType'),
  transmission: text('transmission'),
})

export const VirtualizedListScreen = (): React.ReactNode => {
  const [make, setMake] = useState("");

  const createCar = trpc.car.create.useMutation({
    onSuccess: () => {
      console.log("create post success")
      /* router.refresh();
      setName(""); */
    },
    onError: (e) => {
      console.log("Error", e)
    }
  });

  return (
    <YStack fullscreen f={1}>

      <Form
        onSubmit={() => {
          createCar.mutate({ make, uuid: 123456 });
        }}
      >
        <Input
          placeholder="Title"
          value={make}
          onChangeText={setMake}
        />
        <Form.Trigger asChild>

          <Button />
        </Form.Trigger>
      </Form>
      {carsListLayout}
    </YStack>
  )
}