在sequelize上，“findOne”的“包含”不起作用

我做了一个简单的测试，即搜索地址（id = 4）并检索链接到该地址的用户 .

这是我的模特：

user.js的

module.exports = function(sequelize, DataTypes) {
    return sequelize.define('User', {
        id: {
            type: DataTypes.INTEGER(10).UNSIGNED,
            allowNull: false,
            field: 'id',
            //primaryKey: true,
        },
        name: {
            type: DataTypes.STRING,
            allowNull: false,
            field: 'name',
        },
    }, {
        freezeTableName: true,
        tableName: 'user',
        createdAt: false,
        updatedAt: false,
        classMethods: {
            associate: function(models) {
                models.User.hasMany(models.UserAddress, { foreignKey: 'userId' });
            },
        },
    });
};

user_address.js

module.exports = function(sequelize, DataTypes) {
    return sequelize.define('UserAddress', {
        id: {
            type: DataTypes.INTEGER(10).UNSIGNED,
            allowNull: false,
            field: 'id',
        },
        userId: {
            type: DataTypes.INTEGER(10).UNSIGNED,
            allowNull: false,
            field: 'user_id',
        },
        title: {
            type: DataTypes.STRING,
            allowNull: true,
            field: 'title',
        },
        address: {
            type: DataTypes.STRING,
            allowNull: true,
            field: 'address',
        },
    }, {
        freezeTableName: true,
        tableName: 'user_address',
        createdAt: false,
        updatedAt: false,
        classMethods: {
            associate: function(models) {
                models.UserAddress.hasOne(models.User, { foreignKey: 'id' });
            },
        },
    });
};

这是我的测试文件：

db.UserAddress.findOne({
    where: { id: 4 },
    include: [db.User],
}).then(function(address) {
    console.log('------------------------------ Address by "include"');
    console.log('Address title: '+address.title);
    console.log('User id: '+address.userId);
    if(address.User !== null) {
        console.log('User name: '+address.User.name);
    } else {
        console.log('User name: NO USER');
    }

    console.log('');
    address.getUser().then(function(user) {
        console.log('------------------------------ Address by "getUser"');
        console.log('Address title: '+address.title);
        console.log('User id: '+address.userId);
        if(user !== null) {
            console.log('User name: '+address.user.name);
        } else {
            console.log('User name: NO USER');
        }
        console.log('');
    });
});

我用两个测试进行查询：

• 第一个目的是通过变量"user"直接恢复用户，所以感谢"include"的请求 .

• 另一个也检索用户，但这次是通过"getUser()" .

结果如下：

$ node test.js
Executing (default): SELECT `UserAddress`.`id`, `UserAddress`.`user_id` AS `userId`, `UserAddress`.`title`, `UserAddress`.`address`, `User`.`id` AS `User.id`, `User`.`name` AS `User.name` FROM `user_address` AS `UserAddress` LEFT OUTER JOIN `user` AS `User` ON `UserAddress`.`id` = `User`.`id` WHERE `UserAddress`.`id`=4;
------------------------------ Address by "include"
Address title: Test
User id: 3
User name: NO USER

Executing (default): SELECT `id`, `name` FROM `user` AS `User` WHERE (`User`.`id`=4);
------------------------------ Address by "getUser"
Address title: Test
User id: 3
User name: NO USER

可以观察到通过“include”和“getUser（）”检索结果是不可能的 . 该错误在SQL的日志中可见：

"include": LEFT OUTER JOIN `user` AS `User` ON `UserAddress`.`id` = `User`.`id`
and
"getUser()": SELECT `id`, `name` FROM `user` AS `User` WHERE (`User`.`id`=4);

虽然正确的答案应该是：

"include": LEFT OUTER JOIN `user` AS `User` ON `UserAddress`.`user_id` = `User`.`id`
and
"getUser()": SELECT `id`, `name` FROM `user` AS `User` WHERE (`User`.`id`=3);

所以我的问题是，在我的模型或我的请求中使用“include”和“getUser（）”来确定结果的正确性是什么？

谢谢 .

（另发布于：https://github.com/sequelize/sequelize/issues/1383）