javascript 在续集上，“findOne”的“包含”不起作用
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On sequelize, "include" of "findOne" not working
提问by SiteXw
I made a simple test that is to search for an address (id = 4) and retrieve the user who is linked to that address.
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。
Here are my Models:
这是我的模型：
user.js
用户.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' });
},
},
});
};
user_address.js
用户地址.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' });
},
},
});
};
And here is my test file:
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('');
});
});
I do a query with two tests:
我用两个测试做一个查询：
- The first aims to recover the user directly via the variable "user", so thanks to "include" of the request.
- And the other also retrieve the user but this time via "getUser()".
- 第一个旨在通过变量“user”直接恢复用户，因此感谢请求的“include”。
- 另一个也检索用户，但这次是通过“getUser()”。
Here is the result:
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
One can observe that it is impossible to retrieve the result via "include" and "getUser()". The error is visible in the log of SQL:
可以观察到，不可能通过“include”和“getUser()”检索结果。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);
While the correct answer should have been:
虽然正确答案应该是：
"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);
So my question is, what is the configuration to put in my Model or my request for the result to be correct with "include" and "getUser()" ?
所以我的问题是，要放入我的模型的配置是什么，或者我对“include”和“getUser()”结果正确的请求是什么？
Thank you.
谢谢你。
(Also posted on: https://github.com/sequelize/sequelize/issues/3182)
回答by jjbskir
Answer from the github page- need to use belongsToinstead of hasOne.
来自github 页面的回答- 需要使用belongsTo而不是hasOne.
user.js
用户.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' });
},
},
});
};
user_address.js
用户地址.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' });
},
},
});
};
