Zod实例序列化反序列化报错：寻求更优解析/序列化方案

需求 #

• 创建Zod实例并序列化为JSON发送
• 接收JSON并解析回Zod实例

报错详情 #

ZodError: [
  {
    "code": "invalid_type",
    "expected": "string",
    "received": "object",
    "path": [
      0,
      "parents",
      0
    ],
    "message": "Expected string, received object"
  },

临时解决方案 #

发送端 #

手动将所有引用字段从对象映射为ID字符串数组：

export async function GET() {
    const dbPersons = await getCollection("person");
    const persons = dbPersons.map((dbPerson) => {
        // 直接返回dbPerson.data会触发上述错误
        // return dbPerson.data;
        // 可行的临时处理逻辑
        const parents = dbPerson.data.parents?.map(parent => parent.id);
        const children = dbPerson.data.children?.map(child => child.id);
        const spouses = dbPerson.data.spouses?.map(spouse => spouse.id);
        return Object.assign({}, {
            ...dbPerson.data,
            parents,
            children,
            spouses
        });
    });
    return new Response(JSON.stringify(persons));
}

接收端 #

const personAPIResp = await personsApi.json();
const persons = await ZPersonArray.parseAsync(personAPIResp);

关联的Schema定义 #

export const ZPerson = z.object({
    id: z.number({
        message: "Unique identifier of a person"
    }),
    // ...
    // 父母字段
    parents: z.array(reference("person")).optional(),
    // ...
});

疑问 #

是否存在更优解决方案？比如使用其他解析器或不同的序列化方式？

补充简化示例 #

config.ts

import { reference, z } from 'astro:content';
const ZPerson = z.object({
  id: z.number({
        message: "Unique identifier of a person"
    }),
    parents: z.array(reference("person")).optional(),
});
export const ZPersonArray = z.array(ZPerson);
export const collections = {
    person: ZPersonCollection
}

content/1.yaml

id: 1
parents: ["2"]

content/2.yaml

id: 2

测试代码

import { getCollection } from 'astro:content';
let dbPersons = await getCollection("person");
const jsonPerson = JSON.stringify(dbPersons);

const objectPerson = JSON.parse(jsonPerson);
const persons = await ZPersonArray.parseAsync(objectPerson);

内容的提问来源于stack exchange，提问作者MortalFool