2016-08-21 123 views
1

我有一个情况,我从两个表中查询信息,这些表与“through”表有多对多关系。当我进行查询时,似乎通过不使用“through”表作为表连接引用并使用两个表属性接收输出记录,但我无法访问连接表的字段属性,因此我正确查询。这是输出的值。Sequelize查询多对多关系和访问对象

{"fullNameSlug":"Tester Test","email":"[email protected]","firstName":"Tester","lastName":"Test","teams":[{"teamName":"Sales","member":{"memberId":1,"memberEmail":"[email protected]","organizationId":1,"teamId":1,"userId":1,"created_at":"2016-08-21T21:15:19.000Z","updated_at":"2016-08-21T22:00:32.000Z","organization_id":1,"team_id":1,"user_id":1}}]} 

这里是我的查询,我是怎么设置的数据:

.get(function(req, res){ 
models.User.find({ 
       where: { 
        organizationId: organization.organizationId 
       }, attributes: ['email', 'firstName', 'lastName'], 
       include: [{ 
        model: models.Team, 
        attributes: ['teamName'] 
       }] 
      }); 
     }).then(function(currentUsers){ 
      res.jsonp(currentUsers); 
console.log(currentUsers); 
     }); 

这里是我试图访问teamName在我看来:{{currentUsers.teams.teamName}},它没有返回值,但{{currentUsers.email}}返回正确的用户电子邮件。

用户表:

module.exports = function(sequelize, DataTypes) { 

var User = sequelize.define('user', { 
    userId: { 
     type: DataTypes.INTEGER, 
     field:'user_id', 
     autoIncrement: true, 
     primaryKey: true 
    }, 
    firstName: { 
     type: DataTypes.STRING, 
     field: 'first_name' 
    }, 
    lastName: { 
     type: DataTypes.STRING, 
     field: 'last_name' 
    }, 
    email: { 
     type: DataTypes.STRING, 
     isEmail: true, 
     unique: true, 
     set: function(val) { 
      this.setDataValue('email', val.toLowerCase()); 
     } 
    }, 
    password: DataTypes.STRING, 
    organizationId: { 
     type: DataTypes.INTEGER, 
     field: 'organization_id', 
     allowNull: true 
    } 
}, { 
    underscored: true, 
    freezeTableName: true, 
    }, 
    classMethods: { 
     associate: function(db) { 
      User.belongsToMany(db.Organization, { through: 'member', foreignKey: 'user_id'}), 
      User.belongsToMany(db.Team, { through: 'member', foreignKey: 'user_id'}) 
     } 
}); 
    return User; 
} 

团队表:

module.exports = function(sequelize, DataTypes) { 

var Team = sequelize.define('team', { 
    teamId: { 
     type: DataTypes.INTEGER, 
     field: 'team_id', 
     autoIncrement: true, 
     primaryKey: true, 
     notNull: true 
    }, 
    teamName: { 
     type: DataTypes.STRING, 
     field: 'team_name' 
    }, 
    organizationId: { 
     type: DataTypes.INTEGER, 
     field: 'organization_id' 
    }, 
},{ 
    underscored: true, 
    freezeTableName: true, 
    classMethods: { 
     associate: function(db) { 
      Team.belongsToMany(db.User, { through: 'member', foreignKey: 'team_id' }); 
     }, 
    } 
}); 
    return Team; 
} 

会员表:

module.exports = function(sequelize, DataTypes) { 

var Member = sequelize.define('member', { 
    memberId: { 
     type: DataTypes.INTEGER, 
     field: 'member_id', 
     autoIncrement: true, 
     primaryKey: true 
    }, 
    memberEmail: { 
     type: DataTypes.STRING, 
     field: 'member_email', 
     isEmail: true, 
     unique: true 
    }, 
    organizationId: { 
     type: DataTypes.INTEGER, 
     field: 'organization_id', 
     allowNull: true 
    }, 
    teamId: { 
     type: DataTypes.INTEGER, 
     field: 'team_id', 
     allowNull: true 
    }, 
    userId: { 
     type: DataTypes.INTEGER, 
     field: 'user_id', 
     allowNull: true 
    } 
},{ 
    underscored: true, 
    freezeTableName: true, 
}); 
    return Member; 
} 

输出SQL:

SELECT `user`.*, `teams`.`team_id` AS `teams.teamId`, `teams`.`team_name` AS `teams.teamName`, `teams.member`.`member_id` AS `teams.member.memberId`, `teams.member`.`member_email` AS `teams.member.memberEmail`, `teams.member`.`organization_id` AS `teams.member.organizationId`, `teams.member`.`team_id` AS `teams.member.teamId`, `teams.member`.`user_id` AS `teams.member.userId`, `teams.member`.`created_at` AS `teams.member.created_at`, `teams.member`.`updated_at` AS `teams.member.updated_at`, `teams.member`.`organization_id` AS `teams.member.organization_id`, `teams.member`.`team_id` AS `teams.member.team_id`, `teams.member`.`user_id` AS `teams.member.user_id` FROM (SELECT `user`.`user_id` AS `userId`, `user`.`email`, `user`.`first_name` AS `firstName`, `user`.`last_name` AS `lastName` FROM `user` AS `user` WHERE `user`.`organization_id` = 1 LIMIT 1) AS `user` LEFT OUTER JOIN (`member` AS `teams.member` INNER JOIN `team` AS `teams` ON `teams`.`team_id` = `teams.member`.`team_id`) ON `user`.`userId` = `teams.member`.`user_id`; 

回答

1

考虑你的关系,用户有很多球队槽工作台部件,和您的查询返回预期很多团队(团队对象的数组)用户。您应该使用user.teams[0].teamName通过键获取特定团队,或者使用此阵列中的循环对象