sequelize.js - Sequelize 和创建新的持久化实例中的关联

Question

module.exports = (sequelize, DataTypes) => {
    let City = sequelize.define('City', {
        id: {type: DataTypes.INTEGER, primaryKey: true, autoIncrement: true},
        name: {type: DataTypes.STRING, allowNull: false, unique: true},
    });
    City.associate = (models) => {
        City.belongsTo(models.Country);
    };
    return City;
} 

module.exports = (sequelize, DataTypes) => {
    let Country = sequelize.define('Country', {
        id: {type: DataTypes.INTEGER, primaryKey: true, autoIncrement: true},
        code: {type: DataTypes.STRING, allowNull: false, unique: true },
        name: DataTypes.STRING
    });
    Country.associate = (models) => {
        Country.hasMany(models.City);
    };
    return Country;
}

我有以上型号。现在，当我创建 City 时，我正在这样做。

let country = await Country.findById(input.countryID);
if(country) {
  let city = await City.create({
    name: input.name
  });
  await city.setCountry(country);

  callback(null, city);  
} else {
  callback(new Error('Country does not exist'));
}

有没有一种方法可以在一个查询中直接创建与国家关联的城市，而不是为创建城市和在城市中设置国家进行多次查询？此外，我想在城市创建后填充响应以使其结果也包含国家。

Answer 1

你可以简单地做到这一点:

City.create({
    name: input.name ,
    country_id : input.countryID
}).then((data) => { // <----- Inserted if country_id exist in country table as per primary foreign key rule
    // Code for success
}).catch(err => { // Will throw error if condition doesn't match
    // Code for Error
})