Node.js 加载sequelize中关系为空的项目

我是sequelize新手，我正在尝试加载任务关系为空的用户表中的所有条目。但它不起作用。以下是我尝试过的：

const express = require('express');
const app = express();

const Sequelize = require('sequelize');
const sequelize = new Sequelize('sequelize', 'mazinoukah', 'solomon1', {
  host: 'localhost',
  dialect: 'postgres',

  pool: {
    max: 5,
    min: 0,
    acquire: 30000,
    idle: 10000,
  },
});

const Task = sequelize.define('Task', {
  name: Sequelize.STRING,
  completed: Sequelize.BOOLEAN,
  UserId: {
    type: Sequelize.INTEGER,
    references: {
      model: 'Users', // Can be both a string representing the table name, or a reference to the model
      key: 'id',
    },
  },
});

const User = sequelize.define('User', {
  firstName: Sequelize.STRING,
  lastName: Sequelize.STRING,
  email: Sequelize.STRING,
  TaskId: {
    type: Sequelize.INTEGER,
    references: {
      model: 'Tasks', // Can be both a string representing the table name, or a reference to the model
      key: 'id',
    },
  },
});

User.hasOne(Task);
Task.belongsTo(User);

app.get('/users', (req, res) => {
  User.findAll({
    where: {
      Task: {
        [Sequelize.Op.eq]: null,
      },
    },
    include: [
      {
        model: Task,
      },
    ],
  }).then(function(todo) {
    res.json(todo);
  });
});

   app.listen(2000, () => {
      console.log('server started');
   });

---

如果我有三个用户，其中两个用户各有一个任务，我只想加载最后一个没有任务的用户。这在sequelize中可能吗？

经过多次调试，我找到了解决方案

app.get('/users', (req, res) => {
User.findAll({
    where: {
      '$Task$': null,
    },
    include: [
      {
        model: Task,
        required: false,
      },
    ],
  }).then(function(todo) {
    res.json(todo);
  });
});

加入本条

where: {
  '$Task$': null,
},

我可以只加载用户而不加载任务