使用 TypeScript 进行 Fastify，如何验证和键入路由请求？

Question

我的基本服务器，导入并注册路由join ，并添加 TypeProvider 到 fastify。

import Fastify from "fastify";
import { join } from "./routes/onboarding/join";
import { JsonSchemaToTsProvider } from "@fastify/type-provider-json-schema-to-ts";

const fastify = Fastify({
  logger: true,
}).withTypeProvider<JsonSchemaToTsProvider>();

fastify.register(join);

const start = async () => {
  try {
    await fastify.listen({ port: 3000 });
  } catch (err) {
    fastify.log.error(err);
    process.exit(1);
  }
};

start();

加入路线...

import {
  FastifyInstance,
  FastifyReply,
  FastifyRequest,
} from "fastify";

export async function join(fastify: FastifyInstance, _options: Object) {
  fastify.post(
    "/animals",
    {
      schema: {
        body: {
          type: "object",
          required: ["animal"],
          properties: {
            animal: { type: "string" },
          },
        },
      } as const,
    },
    async (request: FastifyRequest, _reply: FastifyReply) => {
      const { animal } = request.body;  <=== errors here on animal
      return animal;
    }
  );
}

我在const { animal }上用红色波浪线得到的错误

Property 'animal' does not exist on type 'unknown'.ts(2339)

文档在这里，但我想他们不是那么清楚

Answer 1

@fastify/type-provider-json-schema-to-ts类型提供程序导出一个插件类型FastifyPluginAsyncJsonSchemaToTs ，它帮助 TypeScript 从模式定义中确定类型。

在您的示例中， join声明为该类型并从插件 function 和处理程序中删除显式参数类型：

import { FastifyPluginAsyncJsonSchemaToTs } from "@fastify/type-provider-json-schema-to-ts";

export const join: FastifyPluginAsyncJsonSchemaToTs = async function (
  fastify,
  _options
) {
  fastify.post(
    "/animals",
    {
      schema: {
        body: {
          type: "object",
          required: ["animal"],
          properties: {
            animal: { type: "string" },
          },
        },
      } as const,
    },
    async (request, _reply) => {
      const { animal } = request.body; // animal is string
      return animal;
    }
  );
};

现在， animal的类型为string 。