Objection.js 如何根据关系过滤模型

Question

所以我一直在尝试找到一种方法来在 objection.js

中执行以下操作 sql


SELECT
    u.user_id, u.profile_photo_name, u.first_name,
    u.last_name, COUNT(*) AS count
FROM users AS u
    INNER JOIN post_users AS pu ON (u.user_id = pu.user_id)
WHERE
    u.organization_id = 686
GROUP BY user_id
ORDER BY count DESC
LIMIT 1

这是我目前所知道的...但不能以这种方式使用 $relatedQuery

return await User.$relatedQuery('recognitions')
.where('organization_id', organizationID)
.select(
    'user_id',
    'profile_photo_name',
    'first_name',
    'last_name',
    raw('COUNT(*) as count')
)
.groupBy('user_id')
.orderBy('count', 'desc')
.limit(1)

这是识别关系：

            recognitions: {
                relation: Model.ManyToManyRelation,
                modelClass: Post,
                join: {
                    from: 'users.user_id',
                    through: {
                        from: 'post_users.user_id',
                        to: 'post_users.post_id',
                    },
                    to: 'posts.post_id'
                }
            },

Answer 1

不得不使用 joinRelation 而不是像这样也使用 first() 而不是 limit(1) 来获取单个对象而不是长度为 1

的数组

return await User.query()
.joinRelation('recognitions')
.where('organization_id', organizationID)
.select(
    'user_id',
    'profile_photo_name',
    'first_name',
    'last_name',
    raw('COUNT(*) as count')
)
.groupBy('user_id')
.orderBy('count', 'desc')
.first()

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