我有一个案例,我正在从两个与“直通”表具有多对多关系的表中查询信息。当我进行查询时,我似乎通过不使用“通过”表作为表联接引用并接收具有两个表属性的输出记录来正确查询,但我无法访问联接表的字段属性。这是输出的值。
{"fullNameSlug":"Tester Test","email":"test@test.com","firstName":"Tester","lastName":"Test","teams":[{"teamName":"Sales","member":{"memberId":1,"memberEmail":"test@test.com","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}}]}
这是我的查询以及我如何设置数据:
.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);
});
以下是我在 View 中尝试访问 teamName 的方法:{{currentUsers.teams.teamName}}
,它不返回值,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;
}
成员表:
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;
}
输出的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`;
最佳答案
考虑您的关系,用户在表成员中拥有许多团队,并且您的查询按预期返回具有许多团队(团队对象数组)的用户。您应该使用 user.teams[0].teamName
按键获取特定团队,或循环此数组中的对象
关于node.js - Sequelize 查询多对多关系并访问对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39069707/