Sequelize findAll() 在包含关联时抛出错误

Question

我对 Sequelize 还很陌生，并试图学习所有可以用来格式化查询的很酷的技巧，但遇到了一些困难。

基本上我有一张“发票”表。我有另一个表“invoice_items”，其中许多发票_项目与一张发票有关系。

目前我正在尝试在 findAll 查询中包含此一对多关联，以便发票_项目嵌套在每个发票对象的数组中。

大致如下：

[
  {
    name: invoice1
    invoiceItems: [
      itemOne,
      itemTwo,
    ]
  },
    {
    name: invoice2
    invoiceItems: [
      itemOne,
      itemTwo,
    ]
  }
]

当我尝试这个查询时（基本上是从文档中获取的）

db.invoice.findAll({
    where: {
        userId: 'ed8fdd40-b807-4e51-b1f5-90fb5b7f6e73'
    },
    include: { all: true, nested: true }
}).then(result => {
    console.log(result)
})

我遇到错误

c:\Users\brend\Desktop\projects\invoice-server\node_modules\sequelize\lib\model.js:1189
          let associations = result.get(include.association.as);
                                    ^

TypeError: result.get is not a function
    at c:\Users\brend\Desktop\projects\invoice-server\node_modules\sequelize\lib\model.js:1189:37
    at Array.reduce (<anonymous>)
    at c:\Users\brend\Desktop\projects\invoice-server\node_modules\sequelize\lib\model.js:1188:50
    at Array.map (<anonymous>)
    at Model._findSeparate (c:\Users\brend\Desktop\projects\invoice-server\node_modules\sequelize\lib\model.js:1186:39)
    at invoice.findAll (c:\Users\brend\Desktop\projects\invoice-server\node_modules\sequelize\lib\model.js:1153:24)
    at process.processTicksAndRejections (node:internal/process/task_queues:95:5)

Node.js v20.10.0

奇怪的是，当我尝试这样的操作时，查询成功，但未按预期格式化（每个关联的发票项目都有一个单独的发票对象）

db.invoice.findAll({
    where: {
        userId: 'ed8fdd40-b807-4e51-b1f5-90fb5b7f6e73'
    },
    include: "invoiceItems",
    raw : true ,
    nest : true
}).then(result => {
    console.log(result)
})

为了帮助识别问题，我将在下面添加更完整的代码（想知道我是否将关联弄乱了）：

发票型号

module.exports = (sequelize, Sequelize, DataType) => {
    const Invoice = sequelize.define('invoice', {
        id: {
         type: DataType.UUID,
         primaryKey: true
        },
        userId: {
         type: DataType.UUID,
         validate: {
            allowNull: false
         }
        },
        name: {
         type: DataType.STRING(150)
        }
//Additional fields left out here for simplicity
     },{
        underscored: true
     });
     Invoice.associate = models => {
      Invoice.hasMany(models.invoiceItem), {
         foreignKey: "invoiceId"
      }
      Invoice.belongsTo(models.user)
     };

     return Invoice
};

发票项目模型

module.exports = (sequelize, Sequelize, DataType) => {
    const InvoiceItem = sequelize.define('invoiceItem', {
        id: {
         type: DataType.UUID,
         primaryKey: true
        },
        invoiceId: {
         type: DataType.UUID,
         validate: {
            allowNull: false
         }
        },
        userId: {
         type: DataType.UUID,
         validate: {
            allowNull: false
         }
        },
        name: {
         type: DataType.STRING(150)
        }
//Additional fields left out here for simplicity
     },{
        underscored: true
     });
     InvoiceItem.associate = models => {
      InvoiceItem.belongsTo(models.invoice, {
         foreignKey: "invoiceId"
      })
      InvoiceItem.belongsTo(models.user), {
         foreignKey: "userId"
      }};

     return InvoiceItem
};

最后是我的 index.js 文件，它将所有内容组合在一起

'use strict';

const fs = require('fs');
const path = require('path');
const Sequelize = require('sequelize');
const basename = path.basename(__filename);
const sequelize = require('../config/database.js')

const db = {}
db.Sequelize = Sequelize;
db.sequelize = sequelize;

fs.readdirSync(__dirname)
  .filter(file => {
    return (
      file.indexOf('.') !== 0 &&
      file !== basename &&
      file.slice(-3) === '.js' &&
      file.indexOf('.test.js') === -1
    );
  })
  .forEach(file => {
    const model = require(path.join(__dirname, file))(db.sequelize, db.Sequelize, db.Sequelize.DataTypes)
    db[model.name] = model;
  });
Object.keys(db).forEach(modelName => {
  if (db[modelName].associate) {
    db[modelName].associate(db);
  }
});

module.exports = db;

非常感谢任何想法或建议，提前致谢！

Answer 1

到目前为止，您的设置似乎是正确的，问题可能来自您指定的选项（嵌套：true）：

db.invoice.findAll({
    where: {
        userId: 'ed8fdd40-b807-4e51-b1f5-90fb5b7f6e73'
    },
    include: { all: true, nested: true } <------ begin by removing nested: true
}).then(result => {
    console.log(result)
})

在上面的代码中，首先删除：nested: true，您可以查看的其他选项是 plain: true、raw: true 等

Answer 2

最后，当我第一次创建新的续集实例时，我通过从查询默认值中删除 raw: true 来消除了错误。部分问题似乎是原始的：真正的急切加载工作。然后我可以使用 get() 方法将结果解包为我想要的格式。有关格式的更多详细信息可以在我的相关问题中找到，以防有人在某些时候发现它有帮助。