错误：在急切加载过程中，尝试使用Sequelize.cast或Sequelize.fn选择属性而未指定结果别名

Question

数据：一个Office有许多OfficeLocations ，每个Office都有许多Ratings 。 我需要编写一个查询时，只获取Offices ，有在免得一个Rating 。 我的查询：

let condition = {
    include: [{
        model: OfficeLocation.unscoped(),
        attributes: [
            '"Office"."id" as "Office.id"',
            '"OfficeLocations"."id" AS "OfficeLocation.id"'
        ],
        include: [
            {
                model: Rating.unscoped(),
                attributes: [
                    '*',
                    sequelize.fn('COUNT', sequelize.col('"OfficeLocations->Ratings"."id"'))
                ]
            }
        ],
        group: '"Office.id", "OfficeLocation.id"',
        having: sequelize.where(
            sequelize.fn('COUNT', sequelize.col('"OfficeLocations->Ratings"."id"')),
            '>',
            0
        )
    }]
}

Office.findAll(condition).then(data => {
    res.send(data);
}).catch(e => {
    console.log(e);
});

但是我在控制台中出错：

Error: Tried to select attributes using Sequelize.cast or Sequelize.fn without specifying an alias for the result, during eager loading. This means the attribute will not be added to the returned instance
at include.attributes.map.attr (/Users/.../node_modules/sequelize/lib/dialects/abstract/query-generator.js:1307:17)
at Array.map (<anonymous>)
at Object.generateInclude (/Users/.../node_modules/sequelize/lib/dialects/abstract/query-generator.js:1287:52)
at Object.generateInclude (/Users/.../node_modules/sequelize/lib/dialects/abstract/query-generator.js:1355:39)

我也尝试过

where: sequelize.literal('COUNT(DISTINCT(`OfficeLocations`.`Ratings`.`id`)) > 0'),

而不是属性/组/具有属性，但是它也不起作用。

谢谢。

更新

此SQL查询可根据需要运行：

sequelize.query(`
    SELECT
    "Office"."id" as "Office.id",
    "Office"."name",
    "Office"."website",
    "OfficeLocations"."id" AS "OfficeLocations.id",
    COUNT("OfficeLocations->Ratings"."id") as "RatingsCount"

    FROM "Companies" AS "Office"
    LEFT OUTER JOIN ( "OfficeLocations" AS "OfficeLocations"
        INNER JOIN "Ratings" AS "OfficeLocations->Ratings"
        ON "OfficeLocations"."id" = "OfficeLocations->Ratings"."OfficeLocationId"
    )
    ON "Office"."id" = "OfficeLocations"."OfficeId"

    GROUP BY "Office.id", "OfficeLocations.id"
    HAVING COUNT("OfficeLocations->Ratings"."id") > 0
`)

除了我要获取所有数据。

Answer 1

请参阅Model.findAll语法了解options参数。

1. group和having options对象的属性。
2. 对于选定列/ expr（ attributes选项）的别名，请使用数组： [expr, alias] 。
3. 您可以传递attributes.include和/或attributes.exclude数组。
4. 您可以传递include[].attributes选项以供参考，以包含的Model的属性。
5. 您也可以使用include[].required选项在INNER和OUTER JOIN之间选择。

您的情况：

let options = {
        include: [
            {
                model: OfficeLocation,
                required: false, //false for OUTER JOIN, but I think that you can use INNER JOIN
                attributes: [
                    "id", //this is OfficeLocation.id, see 4th item above. 
                    [Sequelize.fn("COUNT", Sequelize.col('`OfficeLocations->Ratings`.`id`')), "RatingsCount"]
                ],
                include: [
                    {
                        model: UserRating,
                        attributes: [],
                        required: true
                    }
                ]
            }
        ],
        group: [
            `Office.id`,
            `OfficeLocations.id`
        ],
        having: Sequelize.where(Sequelize.fn("COUNT", Sequelize.col('`OfficeLocations->Ratings`.`id`')), ">", 0)
    };

请注意，由Sequelize生成的别名可能会更改，因此您应该为Sequelize.col更新它。