Question

我刚开始使用sequelize并尝试重现以下查询。

我有以下模型结构：Review，Entity，ReviewThank

每个Entity可以包含多个Reviews，每个Review可以包含多个ReviewThanks。

每个评论的属性都是＆{39; thumbUp＆＃39; （boolean）评级。

我尝试生成以下查询以获得＆＃39; thankCount＆＃39;对于每个审核，以及每个thumbUpCount的实体评分 - totalCount和Review：

SELECT * FROM (
    SELECT COUNT("Review"."id") AS "totalCount", "Review"."EntityId", COUNT(CASE WHEN "Review"."thumbUp" THEN 1 END) AS "thumbUpCount"
    FROM "Reviews" AS "Review" GROUP BY "Review"."EntityId"
) AS "EntityRatingTable" LEFT JOIN (
    SELECT "Review"."id", "Review"."EntityId", "Review"."uid", "Review"."thumbUp", "Review"."caption", COUNT("ReviewThanks"."id") AS "thankCount"
    FROM "Reviews" AS "Review" LEFT OUTER JOIN "ReviewThanks" AS "ReviewThanks" ON "Review"."id" = "ReviewThanks"."ReviewId" 
    WHERE "Review"."UserId" IN (1) GROUP BY "Review"."id"
) AS "ReviewsTable" ON "ReviewsTable"."EntityId" = "EntityRatingTable"."EntityId";

是否有可能在续集中产生这个？我已经得到了＆＃34; ReviewsTable＆＃34;查询工作正常，但不确定如何（或如果可能）我可以加入＆＃34; EntityRatingTable＆＃34;？

这是我到目前为止所得到的：

models.Review.findAll({

    attributes: {
        include: [
            [models.sequelize.fn('COUNT', models.sequelize.col('ReviewThanks.id')), 'thankCount'],
            [models.sequelize.fn('COUNT', models.sequelize.col('Review.id')), 'reviewCount'],
        ],
        exclude: ["EntityId", "UserId"],
    },

    include: [
        {
            model: models.ReviewThank,
            attributes: [],
        }, {
            model: models.Entity,
        }
    ],

    group: ['"Review"."id"', '"Entity.id"'],

})

这个SQL在Sequelize中是否可行？

0 个答案: