javascript - 未处理的拒绝 SequelizeEagerLoadingError :Sequelize

Question

这是我的模型:sensor.js

   'use strict';
    module.exports = (sequelize, DataTypes) => {
      const Sensor = sequelize.define('Sensor', {
        sensorName: DataTypes.STRING,
        user: DataTypes.INTEGER
      }, {});
      Sensor.associate = function(models) {
        Sensor.hasMany(models.User, {
            foreignKey: 'user'
        })
      };
      return Sensor;
    };

用户.js

'use strict';
module.exports = (sequelize, DataTypes) => {
  const User = sequelize.define('User', {
    firstName: DataTypes.STRING,
    lastName: DataTypes.STRING,
    email: DataTypes.STRING
  }, {});
  User.associate = function(models) {
    User.belongsTo(models.Sensor, {
      foreignKey: 'user'
    })
  };
  return User;
};

在做的时候

Sensor.findAll({
  'raw':true,
  include: [{
    model: User
  }]
})
.then(function(result){
  console.log(result)
})

我收到以下错误:

Unhandled rejection SequelizeEagerLoadingError: User is not associated to Sensor!

Answer 1

您正在创建一个与对象关系同名的列 (user)，这可能会导致冲突。默认情况下，Sequelize 将自动创建 userId(或 user_id，具体取决于设置)列，您无需将其包含在 define() 或关联中，而您使用 user 作为 ID 关系。

使用 as 属性定义复数的 1:n 关系。

传感器.js

'use strict';
module.exports = (sequelize, DataTypes) => {
  const Sensor = sequelize.define('Sensor', {
    sensorName: DataTypes.STRING,
  }, {});
  Sensor.associate = function(models) {
    Sensor.hasMany(models.User, { as: 'users' })
  };
  return Sensor;
};

用户.js

'use strict';
module.exports = (sequelize, DataTypes) => {
  const User = sequelize.define('User', {
    firstName: DataTypes.STRING,
    lastName: DataTypes.STRING,
    email: DataTypes.STRING
  }, {});
  User.associate = function(models) {
    User.belongsTo(models.Sensor)
  };
  return User;
};

查询

Sensor.findAll({
  include: [{
    model: User,
    as: 'users,
  }],
})