为什么sequelize查询返回一项，而原始SQL返回两项？

Question

以下是sequelize (6.3.5)中使用模型Artwork连接Artimage的查询

    const { Artwork, validateArtwork} = require('../models/artwork');
    const { Artimage, validateArtimage} = require('../models/artimage');
    artworks = await Artwork.findAll({
            where : { 
                id: {[Op.gt]:last_artwork_id},
            },
            include: [{
                model:Artimage,
                attributes:['width', 'height', ['name', 'img_name'], 'path', 'size_kb', ['label', 'img_label'], ['fileName', 'img_fileName']],
                
            }],
            attributes:{excludes:['fort_token']},
            order:[['id', 'DESC']]
        });

这里，当last_artwork_id为7时，将在artimage中选取artwork#8及其2个图像。但是，artworks.length是1而不是2。但是，当执行原始SQL代码时，它返回2个条目，因为artwork#8有2个图像。下面是上面代码中的原始SQL：

 SELECT "artwork"."id", "artwork"."name", "artwork"."auther", "artwork"."category", "artwork"."wt_g", "artwork"."production_year", "artwork"."dimension", 
 "artwork"."uploader_id", "artwork"."description", "artwork"."note", "artwork"."tag", "artwork"."deleted", "artwork"."status", "artwork"."artwork_data", 
 "artwork"."last_updated_by_id", "artwork"."fort_token", "artwork"."createdAt", "artwork"."updatedAt", "artimages"."id" AS "artimages.id", 
 "artimages"."width" AS "artimages.width", "artimages"."height" AS "artimages.height", "artimages"."name" AS "artimages.img_name", "artimages"."path" AS "artimages.path", 
 "artimages"."size_kb" AS "artimages.size_kb", "artimages"."label" AS "artimages.img_label", "artimages"."fileName" AS "artimages.img_fileName" FROM 
 "artworks" AS "artwork" LEFT OUTER JOIN "artimages" AS "artimages" ON "artwork"."id" = "artimages"."artwork_id" WHERE "artwork"."id" > 7 ORDER BY "artwork"."id" DESC

为什么带有sequelizejs的artworks.length是1个项目，而不是2个项目(正确)？

Answer 1

sequelizejs在具有表名为artimages：artworks.artimages[0] & artworks.artimages[1]的键的数组对象中返回联接的artimage