Sequelize.js/Node.js/Express.js：Tasks.findAll() 返回一个 TypeError:Cannot read property 'findAll' of undefined答案

【问题标题】：Sequelize.js/Node.js/Express.js: Tasks.findAll()returns a TypeError:Cannot read property 'findAll' of undefinedSequelize.js/Node.js/Express.js：Tasks.findAll() 返回一个 TypeError:Cannot read property 'findAll' of undefined
【发布时间】：2020-05-26 03:52:34
【问题描述】：

代码应该在请求 /tasks 时返回一个带有空任务的 JSON 对象，而不是返回一条消息错误 - TypeError: cannot read property 'findAll' of undefined。根据消息的错误来源来自路由>tasks.js，请参见下面的屏幕截图或/和沙箱上的实时代码。

项目文件夹：

sandbox

一些代码： src>模型>tasks.js

module.exports = (sequelize, DataType) => {
  const Tasks = sequelize.define(
    "Tasks",
    {
      id: {
        type: DataType.INTEGER,
        primaryKey: true,
        autoIncrement: true
      },
      title: {
        type: DataType.STRING,
        allowNull: false,
        validate: {
          notEmpty: true
        }
      },
      done: {
        type: DataType.BOOLEAN,
        allowNull: false,
        defaultValue: false
      }
    },
    {
      classMethods: {
        associate: models => {
          Tasks.belongsTo(models.Users);
        }
      }
    }
  );
  return Tasks;
};

src>routes>tasks.js

module.exports = app => {
  const Tasks = app.db.models.tasks;
  app.get("/tasks", (req, res) => {
    Tasks.findAll({}).then(tasks => {//source of error as per error message
      res.json({ tasks: tasks });
    });
  });
};

src >db.js

var path = require("path");

var fs = require("fs");

var Sequelize = require("sequelize");
//const config = require("./libs/config.js");

var sequelize = null;
let db = null;

module.exports = app => {
  if (!db) {
    const config = app.libs.config;
    sequelize = new Sequelize(
      config.database,
      config.username,
      config.password,
      config.params
    );
    db = {
      sequelize,
      Sequelize,
      models: {}
    };
    const dir = path.join(__dirname, "models");
    fs.readdirSync(dir).forEach(file => {
      const modelDir = path.join(dir, file);
      const model = sequelize.import(modelDir);
      db.models[model.name] = model;
    });
    Object.keys(db.models).forEach(key => {
      db.models[key].options.classMethods.associate(db.models);
    });
  }
  return db;
};

src>index.js

var express = require("express");
var consign = require("consign");
var app = express();

consign({ cwd: "src" })
  .include("./libs/config.js")
  .then("db.js")
  .then("./libs")
  .then("./routes")
  .into(app);

【问题讨论】：

标签： javascript node.js express sequelize.js babeljs

【解决方案1】：

在文件 routes/tasks.js 第 2 行，在 task 上添加大写；

  const Tasks = app.db.models.Tasks;

那么它应该可以工作了。

【讨论】：