Objection.js如何基于关系过滤模型

Question

因此，我一直试图找到一种在objection.js中执行以下sql操作的方法

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

这就是我到目前为止.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)

这就是承认的关系：

            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

必须使用joinRelation代替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()