将 fastify 实例传递给顶层控制器

Question

与其将fastify instance 传递给每个函数，我怎样才能只在顶层传递一次，以便控制器中的所有函数都可以使用它？

// route file
import { FastifyPluginAsync } from 'fastify'

import RootController from '../root.controller'

const root: FastifyPluginAsync = async (fastify, opts): Promise<void> => {
  fastify.get('/', RootController.index(fastify))
  fastify.get('/show', RootController.show(fastify))
}

export default root

// controller file
import { FastifyInstance, FastifyRequest, FastifyReply } from 'fastify'

export default {
  index: (fastify: FastifyInstance) => async (request: FastifyRequest, reply: FastifyReply) => {
    const hugs = fastify.someSupport()

    reply.send({ status: 'ok', love: hugs })
  },

  show: (fastify: FastifyInstance) => async (request: FastifyRequest, reply: FastifyReply) => {
    const hugs = 'extra ' + fastify.someSupport()

    reply.send({ status: 'ok', love: hugs })
  }
}

Answer 1

答案是使用闭包进行简单的机械重构：

// controller file
import { FastifyInstance, FastifyRequest, FastifyReply } from 'fastify'

export default async function installRootController(fastify: FastifyInstance, opts): Promise<void> {
  fastify.get('/', async (request: FastifyRequest, reply: FastifyReply) => {
    const hugs = fastify.someSupport()

    reply.send({ status: 'ok', love: hugs })
  });

  fastify.get('/show', async (request: FastifyRequest, reply: FastifyReply) => {
    const hugs = 'extra ' + fastify.someSupport()

    reply.send({ status: 'ok', love: hugs })
  });
};

内部函数（/ 和 /show 的异步处理程序）可以访问 fastify，因为它可以从周围函数的词法范围访问。这使您可以将整个“控制器”封装为 Fastify 插件，并可以像这样安装：

fastify.register(installRootController);

请注意，“路由”文件现在没有用了。