用连接续集销毁记录

Question

在我的模型中有一个Users表和一个UserPhones表.User.id是UserPhones中的外键.

module.exports = (sequelize, DataTypes) => {
  const User = sequelize.define('User', {
      userid       : {
        type        : DataTypes.UUID,
        primaryKey  : true,
      },
      username : DataTypes.STRING,
    },
    {
      classMethods: {
        associate: (models) => {       
          models.User.hasOne(models.UserPhone, {foreignKey: 'userId'});
        }
      }
    });

  return User;
};


module.exports = (sequelize, DataTypes) => {
  const UserPhone = sequelize.define('UserPhone', {
      id              : {
        type        : DataTypes.UUID,
        primaryKey  : true,
      },
      userId   : {
        type      : DataTypes.UUID,
        references: {
          model     : require('.').User,
          key       : 'userid',
          deferrable: sequelize.Deferrable.INITIALLY_IMMEDIATE
        }
      },
      phoneNumber     : {
        type: DataTypes.STRING
      }
    },
    {
      classMethods: {
        associate: (models) => {
          models.UserPhone.belongsTo(models.User, {foreignKey: 'userId'});
        }
      }
    }
  );
  return UserPhone;
};

销毁电话号码很简单:

UserPhone.destroy({where: {phoneNumber: '123456789'}};

我想删除所有具有sequelize特定电话号码的用户.更好的是,删除所有拥有一组电话号码的用户.

Answer 1

我认为不可能同时执行DELETE和JOIN查询.

因此,

我想删除所有具有sequelize特定电话号码的用户.

UserPhone.findAll({attributes: ['userId'], where: {phoneNumber: '1234'}}
.then(function (userIds) {
    if (userIds.length === 0) 
       return Promise.resolve(true) //nothing to delete
    return User.destroy({where: {id: {$in: userIds}}});
})

删除具有一组电话号码的所有用户.

var array = ['123', '456'];
UserPhone.findAll({attributes: ['userId'], where: {phone: { $in: array }}}
.then(function (userIds) {
    if (userIds.length === 0) 
       return Promise.resolve(true) //nothing to delete
    return User.destroy({where: {id: {$in: userIds}}});
})