無法與Sequelize建立一對一關系

Question

我宣布：

Item.hasOne(Item, { foreignKey: 'parentAid', as: 'Parent' })

我詢問：

Item.find({where : {aid: aid}, include: [Item]}).complete(function(err, article) { .. };

我得到：

Error: Item is not associated to Item!

我究竟做錯了什么？

.................................................. ............................................

更新1

感謝Jan Aargaard Meier的有用回答，我得以將其更改為：

ItemModel.belongsTo(ItemModel, { foreignKey: 'parentAid', as: 'Parent', foreignKeyConstraint: true });
ItemModel.hasMany(ItemModel, { as: 'Children', constraints: false });

this.articleRelations.push({
    model: ItemModel,
    as: 'Parent'
});

this.articleRelations.push({
    model: ItemModel,
    as: 'Children'
});

// ...

我的查詢現在是：

{where : {aid: aid}, include: this.articleRelations}

但是我收到以下錯誤：

{
code : "ER_BAD_FIELD_ERROR",
errno : 1054,
sqlState : "42S22",
index : 0,
sql : "SELECT `item`.*, `Parent`.`aid` AS `Parent.aid`, `Parent`.`gid` AS `Parent.gid`, `Parent`.`title` AS `Parent.title`, `Parent`.`type` AS `Parent.type`, `Parent`.`parentAid` AS `Parent.parentAid`, `Parent`.`createdAt` AS `Parent.createdAt`, `Parent`.`updatedAt` AS `Parent.updatedAt`, `Parent`.`itemId` AS `Parent.itemId`, `Parent`.`aid` AS `Parent.aid`, `Parent`.`aid` AS `Parent.aid`, `Parent`.`aid` AS `Parent.aid`, `Children`.`aid` AS `Children.aid`, `Children`.`gid` AS `Children.gid`, `Children`.`title` AS `Children.title`, `Children`.`type` AS `Children.type`, `Children`.`parentAid` AS `Children.parentAid`, `Children`.`createdAt` AS `Children.createdAt`, `Children`.`updatedAt` AS `Children.updatedAt`, `Children`.`itemId` AS `Children.itemId`, `Children`.`aid` AS `Children.aid`, `Children`.`aid` AS `Children.aid`, `Children`.`aid` AS `Children.aid` FROM (SELECT `item`.* FROM `item` WHERE `item`.`aid`=2 LIMIT 1) AS `item` LEFT OUTER JOIN `item` AS `Parent` ON `Parent`.`aid` = `item`.`parentAid` LEFT OUTER JOIN `item` AS `Children` ON `item`.`aid` = `Children`.`itemId`;"

}

注意：*表名是item *查詢包含itemId ，我在任何地方都沒有定義。 那似乎是個錯誤？

供參考，這是我的模型：

ItemModel = sequelize.define('ExerciseItem', {
        aid: {type: Sequelize.INTEGER.UNSIGNED, primaryKey: true, autoIncrement: true},
        gid: {type: Sequelize.INTEGER.UNSIGNED},
        title: Sequelize.STRING(100),
        type: Sequelize.INTEGER.UNSIGNED,
        parentAid: Sequelize.INTEGER.UNSIGNED
    },{
        freezeTableName: true,
        tableName: 'item'
});

Answer 1

Item.find({where : {aid: aid}, include: [{ model: Item, as: 'Parent' }])

如果為關系提供別名，則在進行急切加載時必須提供該別名（就像在項目實例上調用getParent而不是getItem ）

這是因為別名（使用as ）使您可以創建與同一模型的多個關聯，因此，當您僅提供模型時，就沒有辦法讓續集知道您實際要加載的模型。

我們一直在談論使用關系調用的返回值的能力，例如：

var itemParentRelation = Item.hasOne(Item, { foreignKey: 'parentAid', as: 'Parent' })
Item.find({where : {aid: aid}, include: [itemParentRelation])
// or
Item.find({where : {aid: aid}, include: [item.relations.parent])

但是現在您必須使用帖子開頭提供的代碼：)