从对象创建zod枚举

Question

我有一个对象

const properties = [
  { value: "entire_place", label: "The entire place" },
  { value: "private_room", label: "A private room" },
  { value: "shared_room", label: "A shared room" },
] as const;

我需要和佐德一起用它

1. Validate并解析我在后端
2. 上的数据，使用这些可能的值"entire_place" | "shared_room" | "private_room"

创建一个类型记录联合类型

根据zod文档，我可以这样做：

const properties = [
  { value: "entire_place", label: "The entire place" },
  { value: "private_room", label: "A private room" },
  { value: "shared_room", label: "A shared room" },
] as const;

const VALUES = ["entire_place", "private_room", "shared_room"] as const;
const Property = z.enum(VALUES);
type Property = z.infer<typeof Property>;

但是，我不想定义我的数据两次，一次用标签(标签用于ui目的)，另一次没有标签。

我只想使用properties对象定义一次，而不使用VALUES数组，并使用它创建zod对象并从zod对象推断类型。

有什么解决办法吗？

Answer 1

在这种情况下，我想我可以从Property直接推断出properties的类型。您可以避免重复使用如下代码：

import { z } from "zod";

const properties = [
  { value: "entire_place", label: "The entire place" },
  { value: "private_room", label: "A private room" },
  { value: "shared_room", label: "A shared room" }
] as const;

type Property = typeof properties[number]["value"];
// z.enum expects a non-empty array so to work around that
// we pull the first value out explicitly
const VALUES: [Property, ...Property[]] = [
  properties[0].value,
  // And then merge in the remaining values from `properties`
  ...properties.slice(1).map((p) => p.value)
];
const Property = z.enum(VALUES);