即使在发送响应时也能在不发送响应错误的情况下解析 API答案

【问题标题】：Getting API resolved without sending a response error even when sending response即使在发送响应时也能在不发送响应错误的情况下解析 API
【发布时间】：2021-12-04 22:13:07
【问题描述】：

我在成功将数据保存在数据库中并保存上传图像后尝试发送res.json，但我不断收到API resolved without sending a response for /api/auth/registeration, this may result in stalled requests.。我还在 Next.js 中使用了强大的图像上传。

代码：

import connection from "../../../utils/connection/getConnection";
import formidable from "formidable";
const signupSchema = require("../../../models/signup");
import mkdirp from "mkdirp";
import bcrpt, { genSaltSync } from "bcrypt";
import fs from "fs";
export const config = {
  api: {
    bodyParser: false,
  },
};

const handlePost = async (req, res) => {
  const form = formidable.IncomingForm();

  form.parse(req, async function (err, field, files) {
    await fileSavour(field, files);
    return res.json({
      message: "success",
    });
  });
};

const fileSavour = async (fields, files) => {
  let { email, password } = fields;
  let imageName = files.image.name;

  let newPassword = await bcrpt.hash(password, genSaltSync(10));

  const newUser = new signupSchema({
    email,
    password: newPassword,
    image: imageName,
  });

  const Nuser = await newUser.save();

  if (Nuser) {
    await mkdirp("public/profileImages/" + Nuser._id);

    if (imageName) {
      const data = fs.readFileSync(files.image.path);

      const pathToSave = "public/profileImages/" + Nuser._id + "/" + imageName;

      fs.writeFileSync(pathToSave, data);
      await fs.unlinkSync(files.image.path);
      return;
    }
  }
};

const Register = async (req, res) => {
  req.method === "POST"
    ? handlePost(req, res)
    : req.method === "PUT"
    ? console.log("PUT")
    : req.method === "DELETE"
    ? console.log("DELETE")
    : req.method === "GET"
    ? console.log("GET")
    : res.status(404).send("");
};

export default Register;

【问题讨论】：

标签： javascript node.js next.js formidable

【解决方案1】：

您需要承诺form.parse 功能。目前，您的请求在表单被解析之前被返回。

让我们定义一个使用formidable 承诺表单解析的函数，并用它来解析所有请求。函数应该是这样的 -

const promisifiedParseForm = (req) => {
   const form = formidable.IncomingForm();
   return new Promise((resolve, reject) => {
      form.parse(req, function (err, field, files) {
         if (err) {
            reject(err)
         } else {
            resolve({ field, files })
         }
      });
   });
}

在handlePost函数中使用promisifiedParseForm函数，像这样-

const handlePost = async (req, res) => {
   const { field, files } = await promisifiedParseForm(req);
   
   await fileSavour(field, files);

   return res.json({
      message: "success",
   });
};

【讨论】：

【解决方案2】：

handlePost 处理函数不会等待form.parse 回调执行（以及随后调用res.json），而是立即返回。为了防止这种情况，您可以用 Promise 包装 form.parse 以确保处理程序等待回调函数执行。

const handlePost = async (req, res) => {
    const form = formidable.IncomingForm();
    
    await new Promise(function(resolve, reject) {
        form.parse(req, async function(err, fields, files) {
            await fileSavour(field, files);
            resolve();
        });
    });

    res.json({ message: "success" });
};

【讨论】：