當聚合 function 為 null 時，Sequelize 得到結果

Question

我有以下代碼返回客戶。 問題是我只為已下訂單並創建地址的客戶獲得結果。

CustomerModel.findAll({
    attributes: {
      include: [
        [db.sequelize.fn("COUNT", db.sequelize.col("orders.id")), "orderCount"],
        [db.sequelize.literal(`(
          SELECT o.createdDate
          FROM orders AS o
          WHERE
              o.customerId = customer.id
          ORDER BY o.createdDate DESC
          LIMIT 1
      )`),
      'latestOrderDate'],
      [db.sequelize.fn('sum', db.sequelize.col('orders.amount')), 'totalAmount'],
      ]
    },
    include: [
      { model: UserModel, as: 'user' },
      { model: AddressModel, required:false },
      { model: OrderModel, attributes: [], required:false }
    ]
  })

無論有訂單或地址，是否有可能獲得所有客戶？

謝謝！

Answer 1

我最終將兩個 sequelize.fn 更改為文字，它就像一個魅力！