如何在TypeScript中用AJV的JSONSchemaType定义unknown类型的JSON Schema

使用AJV的JSONSchemaType创建与TypeScript接口对应的JSON Schema时，因接口包含unknown类型属性，赋值时TypeScript抛出类型不兼容错误。

TypeScript接口及两次Schema尝试代码如下：

import { JSONSchemaType } from "ajv";

interface OptionArray {
  options?: {
    title?: string;
    value: unknown;
  }[];
}

const schemaAttempt1: JSONSchemaType<OptionArray> = {
  type: "object",
  properties: {
    options: {
      type: "array",
      items: {
        type: "object",
        properties: {
          title: { type: "string", nullable: true },
          value: {}
        },
        required: ["value"],
        additionalProperties: false 
      },
      nullable: true
    }
  },
  required: [],
  additionalProperties: false
};

const schemaAttempt2: JSONSchemaType<OptionArray> = {
  type: "object",
  properties: {
    options: {
      type: "array",
      items: {
        type: "object",
        properties: {
          title: { type: "string", nullable: true },
          value: true
        },
        required: ["value"],
        additionalProperties: false 
      },
      nullable: true
    }
  },
  required: [],
  additionalProperties: false
};

第一次尝试触发的TypeScript错误：

The types of 'properties.value' are incompatible between these types.
Type '{}' is not assignable to type '{ $ref: string; } | (UncheckedJSONSchemaType<unknown, false> & { nullable: true; const?: null | undefined; enum?: readonly unknown[] | undefined; default?: unknown; })'.

第二次尝试的错误：

The types of 'properties.value' are incompatible between these types.
Type 'boolean' is not assignable to type '{ $ref: string; } | (UncheckedJSONSchemaType<unknown, false> & { nullable: true; const?: null | undefined; enum?: readonly unknown[] | undefined; default?: unknown; })'.

需解决：如何在TypeScript中使用AJV的JSONSchemaType正确定义包含unknown类型属性的JSON Schema？

TypeScript的unknown类型对应JSON中任意合法值，JSON Schema没有直接的单一关键字匹配，需显式声明属性支持所有JSON基础类型，具体有两种可行方式：

方式1：枚举所有JSON基础类型 #

JSON的合法类型包括string、number、integer、boolean、object、array和null，将这些类型放入type数组中，即可匹配任意JSON值，同时通过TypeScript类型检查：

const schema: JSONSchemaType<OptionArray> = {
  type: "object",
  properties: {
    options: {
      type: "array",
      items: {
        type: "object",
        properties: {
          title: { type: "string", nullable: true },
          value: {
            type: ["string", "number", "integer", "boolean", "object", "array", "null"]
          }
        },
        required: ["value"],
        additionalProperties: false 
      },
      nullable: true
    }
  },
  required: [],
  additionalProperties: false
};

方式2：使用类型断言简化（适合快速场景） #

若不想枚举所有类型，可通过类型断言跳过TypeScript的严格检查，同时保证AJV的运行时逻辑正确：

const schema: JSONSchemaType<OptionArray> = {
  type: "object",
  properties: {
    options: {
      type: "array",
      items: {
        type: "object",
        properties: {
          title: { type: "string", nullable: true },
          value: {} as const
        },
        required: ["value"],
        additionalProperties: false 
      },
      nullable: true
    }
  },
  required: [],
  additionalProperties: false
};

注意：类型断言会绕过TypeScript的类型检查，仅在确认{}符合AJV的"任意值"匹配逻辑时使用。

内容的提问来源于stack exchange，提问作者mi105