我有 3 个表 User、Cars 和 UserCars
User{id, name, phone, email}
Cars{id, name, manufacturer}
UserCars{id, car_id, user_id, role}
User have many cars(through UserCars)
Cars have many users(through UserCars)
我正在使用 express js
router.get('/', async (req, res) => {
try {
let car = await Car.findOne({
where: {
id: req.car_id
}});
let users = await car.getUsers({joinTableAttributes: ['role']})
res.send(users)
} catch (e) {
console.log(e)
res.status(400).send(e)
}
})
这是我的回应
[
{
"id": 1,
"name": "test",
"email": null,
"phone": null,
"createdAt": "2019-07-09T09:38:11.859Z",
"updatedAt": "2019-07-12T04:34:20.922Z",
"User_car": {
"role": "driver"
}
}
]
但是知道如何在用户对象中包含角色,而不是在 User_car 表中单独指定它, 有没有办法让我得到下面的输出
[
{
"id": 1,
"name": "test",
"email": null,
"phone": null,
"role": 'driver'
"createdAt": "2019-07-09T09:38:11.859Z",
"updatedAt": "2019-07-12T04:34:20.922Z"
}
]
最佳答案
您可以在获取属性时使用 sequelize.literal
来获取该字段。
attributtes: [
// define your other fields
[sequelize.literal('`users->User_car`.`role`'), 'role'],
]
现在,我不确定这是否适用于 car.getUsers
。我通常使用 include
进行单个查询并定义“join”表,这样我就可以知道在 sequelize 上它是如何命名的。让我举个例子。
module.exports = (sequelize, DataTypes) => {
const UserCar = sequelize.define('UserCar', {
// id you don't need and id field because this is a N:M relation
role: {
type: DataTypes.STRING
},
carId: {
field: 'car_id',
type: DataTypes.INTEGER
},
userId: {
field: 'user_id',
type: DataTypes.INTEGER
},
}, {
tableName: 'User_car',
underscored: true,
createdAt: 'created_at',
updatedAt: 'updated_at',
});
UserCar.associate = (models) => {
models.user.belongsToMany(models.car, { as: 'cars', through: User_car, foreignKey: 'user_id' });
models.car.belongsToMany(models.user, { as: 'users', through: User_car, foreignKey: 'car_id' });
};
return UserCar;
};
router.get('/', async (req, res) => {
try {
const users = await User.findAll({
include: [{
model: Car,
as: 'cars',
where: { id: req.car_id }
}],
attributtes: [
'id',
'name',
'email',
'phone',
[sequelize.literal('`cars->User_car`.`role`'), 'role'],
]
})
res.send(users)
} catch (e) {
console.log(e)
res.status(400).send(e)
}
});
关于node.js - 如何合并 Node js中其他表的响应属性,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57001995/