Sequelize.js 继续子查询产品和评级

我有密码。我想得到：产品清单，订单数量，平均评分，评分数量。因此，我只得到一个产品-

Product.findAll({
      offset,
      limit: 15,
      order: [["amount", "desc"]],
      include: [
        {
          model: models.Rating,
          attributes: [
            [
              sequelize.fn("avg", sequelize.col("ratingValue")),
              "averageRating"
            ],
            [
              sequelize.fn("count", sequelize.col("ratingValue")),
              "amountOfRatings"
            ]
          ]
        }
      ]
    }).then(products => products));

如果我删除：

include: [
            {
              model: models.Rating,
              attributes: [
                [
                  sequelize.fn("avg", sequelize.col("ratingValue")),
                  "averageRating"
                ],
                [
                  sequelize.fn("count", sequelize.col("ratingValue")),
                  "amountOfRatings"
                ]
              ]
            }
          ]

我得到了我所需要的一切，没有评级。 完整项目的url:；

---

谢谢你的回答

问题似乎是sequelize默认情况下会进行内部连接。您需要一个等效的左外部联接，这样即使没有关联的评级，也会返回产品行

通过在include语句中指定required:false，可以更改默认行为：

      include: [
        {
          model: models.Rating,
          required: false,
          attributes: [
            [
              sequelize.fn("avg", sequelize.col("ratingValue")),
              "averageRating"
            ],
            [
              sequelize.fn("count", sequelize.col("ratingValue")),
              "amountOfRatings"
            ]
          ]
        }
      ]

---

sequelize中的sql查询：选择product.*，ratings.id作为ratings.id，avgratingValue作为ratings.Averaging，countratingValue作为ratings.AmountFratings，从选择product.id，product.price，product.title，product.amount，product.description，product.picture，product.createdAt，product.updatedAt FROM products AS product ORDER BY product.amount DESC LIMIT 0，15 AS product LEFT OUTER JOIN ratings AS ratings ON product.id=ratings.productId ORDER BY product.amount DESC；我是用sql编写查询的，它是有效的：使用omvapeshop；选择products.*，从productId=products.id的评级中选择avgratingValue作为平均值，从productId=products.id的评级中选择countratingValue作为产品左侧外部连接评级的数量。id=ratings.productId按产品排序。数量描述限制0,15；我讨厌sequelizealways需要一点调整，但一开始的主要问题是，即使没有评级，也会有左侧外部连接返回产品。问题不在连接中，我是舒尔