未处理的拒绝类型错误:res.sendStatus 不是一个函数

Question

我在应用程序中遇到了这个奇怪的错误。正如您在package.json中看到的，express版本是> 4.x .

{
  "name": "MyAPI",
  "version": "1.0.0",
  "private": true,
  "scripts": {
    "start": "node ./bin/www"
  },
  "dependencies": {
    "bcrypt": "^0.8.6",
    "body-parser": "~1.13.2",
    "cookie-parser": "~1.3.5",
    "debug": "~2.2.0",
    "express": "~4.13.1",
    "jade": "~1.11.0",
    "morgan": "~1.6.1",
    "pg": "^4.5.5",
    "pg-hstore": "^2.3.2",
    "sequelize": "^3.23.2",
    "sequelize-cli": "^2.4.0",
    "serve-favicon": "~2.3.0",
    "validator": "^5.2.0"
  }
}

这是源代码

'use strict';
var express = require('express');
var router = express.Router();
var version = require('../package.json').version;
var sequelize = require('sequelize');
var userAccounts = require('../models').user_account;
router.post('/v' + version + '/register', function (res, req, next) {
    userAccounts.create(req.body).then(function () {
        next();
        return res.sendStatus(200).send({ message: ":D" });
    });
});
module.exports = router;

我还尝试将sendStatus更改为status但错误也说明了status同样的事情。 为什么我会收到这些错误？

Answer 1

快速处理程序函数的参数res和req顺序错误。它们应该是：

function (req, res, next) {}