0

读完这篇文章(@thomasfedb 答案)后,我认为这可行:

@categories = Category.joins(:category_users).where("category_users.user_id = ? AND category_users.interested_learning IS TRUE", current_user.id)
@search = Resource.joins(:categories).where(category_id: @categories.subtree_ids).search_filter(params[:search_filter]).paginate(:per_page => 20, :page => params[:page]).search(params[:q])

但相反,我收到此错误

undefined method `subtree_ids' for #<ActiveRecord::Relation:0x007fce832b1070>

我也从这里尝试了@Rahul 的回答

没有祖先的后代,它是这样工作的

@search = Resource.joins(:category_users).where("category_users.user_id = ? AND category_users.interested_learning IS TRUE", current_user.id).search_filter(params[:search_filter]).paginate(:per_page => 20, :page => params[:page]).search(params[:q])

虽然我想在找到后代后进行 Ransack 搜索,但没有 Ransack 也会出现错误。出于这个原因,我没有将它包含在标题中。如果没有 Ransack,它将是这样的:

@categories = Category.joins(:category_users).where("category_users.user_id = ? AND category_users.interested_learning IS TRUE", current_user.id)
@resources = Resource.joins(:categories).where(category_id: @categories.subtree_ids)

我会很感激任何关于这方面的建议可以工作

4

1 回答 1

1

@categories.subtree_ids将不起作用,因为subtree_ids它是一个实例方法,而您正在调用ActiveRecord::Relation. 试试这个:

@categories.map(&:subtree_ids).flatten.uniq

这可能不是特别高效,但祖先存储和解析类似于“10/1/3”的列来管理层次结构。以上可能会触发 N+1 查询。另一种选择是自己解析列。我不确定祖先是否提供了更好的方法来做到这一点,但这很简单:

arr = @categories.pluck(:ancestry)
#> ["10/1", "5/6/1", "2", nil, nil]

arr.compact
#> ["10/1", "5/6/1", "2"]

arr.map { |ids| ids.split('/') }
#> [["10","1"],["5","6","1"],["2"]]

arr.flatten
#> ["10","1","5","6","1","2"]

arr.uniq
#> ["10","1","5","6","2"]

arr.map(&:to_i)
#> [10,1,5,6,2]

把它们放在一起(我建议在一个方法中使用多行):

@categories.pluck(:ancestry).compact.map { |ids| ids.split('/') }.flatten.uniq.map(&:to_i)
#> [10,1,5,6,2]
于 2013-12-12T21:35:14.620 回答