在續集上，“findOne”的“include”不起作用

Question

我做了一個簡單的測試，即搜索地址（id = 4）並檢索鏈接到該地址的用戶。

這是我的模型：

用戶.js

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' });
            },
        },
    });
};

用戶地址.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' });
            },
        },
    });
};

這是我的測試文件：

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('');
    });
});

我用兩個測試做一個查詢：

• 第一個旨在通過變量“user”直接恢復用戶，因此感謝請求的“include”。
• 另一個也檢索用戶，但這次是通過“getUser()”。

結果如下：

$ 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

可以觀察到，不可能通過“include”和“getUser()”檢索結果。 錯誤在SQL的日志中可見：

"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);

雖然正確答案應該是：

"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);

所以我的問題是，放入我的模型的配置是什么，或者我對“include”和“getUser()”的結果正確的要求是什么？

謝謝你。

（也發布在： https : //github.com/sequelize/sequelize/issues/3182 ）

Answer 1

來自github 頁面的答案 - 需要使用belongsTo而不是hasOne 。

用戶.js

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' });
            },
        },
    });
};

用戶地址.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.belongsTo(models.User, { foreignKey: 'userId' });
            },
        },
    });
};