2

我有一个查询*，结果如下：

#<ActiveRecord::Relation [
    #<BookRank id: 2, book_id: 2, list_edition_id: 1, rank_world: 5, rank_europe: 1>,
    #<BookRank id: 3, book_id: 1, list_edition_id: 1, rank_world: 6, rank_europe: 2>,
    #<BookRank id: 8, book_id: 2, list_edition_id: 3, rank_world: 1, rank_europe: 1>,
    #<BookRank id: 9, book_id: 1, list_edition_id: 3, rank_world: 2, rank_europe: 2
]>

我想要得到的是这样的哈希：

{
    book_id => {
        list_edition_id => {
            "rank_world" => value,
            "rank_europe" => value
        }
    }
}

（最重要的是按最低 list_edition_id 的 rank_world 值对哈希进行排序，但这可能太复杂了。）

ranks_relation.group_by(&:book_id)给了我一个哈希值，其中book_ids是键，但是排名数据仍然在数组中：

{
    2 => [
            #<BookRank id: 2, book_id: 2, list_edition_id: 1, rank_world: 5, rank_europe: 1>,
            #<BookRank id: 8, book_id: 2, list_edition_id: 3, rank_world: 1, rank_europe: 1>
    ],
    1 => [
            #<BookRank id: 3, book_id: 1, list_edition_id: 1, rank_world: 6, rank_europe: 2>
            #<BookRank id: 9, book_id: 1, list_edition_id: 3, rank_world: 2, rank_europe: 2>
    ]
}

我应该如何进行？

*编辑：这是模型结构和查询。另一个用户要求它：

class Book < ActiveRecord::Base
  has_many :book_ranks, dependent: :destroy
end

class List < ActiveRecord::Base
  has_many :list_editions, dependent: :destroy
end

class ListEdition < ActiveRecord::Base
  belongs_to :list
  has_many :book_ranks, dependent: :destroy
end

class BookRank < ActiveRecord::Base
  belongs_to :book
  belongs_to :list_edition

  has_one :list, through: :list_edition
end

对于查询，我已经使用了两个具有相关 ID 的数组Book和ListEdition：

BookRank.where(:book_id => book_ids, :list_edition_id => list_edition_ids)
4

3 回答 3

2

尝试这个

record = your_record

hash = {}

record.each do |record|
  hash[record.book_id] ||= {}
  hash[record.book_id][record.list_edition_id] = {
    'rank_world' => record.rank_world,
    'rank_europe' => record.rank_europe
  }
end
# hash will then be {2=>{1=>{"rank_world"=>5, "rank_europe"=>1}, 3=>{"rank_world"=>1, "rank_europe"=>1}}, 1=>{1=>{"rank_world"=>6, "rank_europe"=>2}, 3=>{"rank_world"=>2, "rank_europe"=>2}}}

这将仅遍历记录一次。

于 2018-07-04T11:38:52.573 回答
1

嘿@gibihmruby（顺便说一句好名字），

因此，既然您在评论中要求更具体地描述我使用 just and friends的丑陋方法，这是我的建议：group_by

rel.group_by(&:book_id).map do |k, v| 
  [k, v.group_by(&:list_edition_id)]
end.to_h

会产生类似的结构

{2=>
  {1=>
    [#<struct BookRank
      id=2,
      book_id=2,
      list_edition_id=1,
      rank_world=5,
      rank_europe=1>],
   3=>
    [#<struct BookRank
      id=8,
      book_id=2,
      list_edition_id=3,
      rank_world=1,
      rank_europe=1>]},
 1=>
  {1=>
    [#<struct BookRank
      id=3,
      book_id=1,
      list_edition_id=1,
      rank_world=6,
      rank_europe=2>],
   3=>
    [#<struct BookRank
      id=9,
      book_id=1,
      list_edition_id=3,
      rank_world=2,
      rank_europe=2>]}}

然后，您必须将最内部的对象映射到您想要的属性。如果您确定 和 的组合book_id是list_edition_id唯一的，则可以摆脱数组包装，然后映射到所需的属性。您可以将切片用于ActiveRecord对象。然后映射将是

rel.group_by(&:book_id).map do |book_id, grouped_by_book_id| 
  [
    book_id, 
    grouped_by_book_id.group_by(&:list_edition_id).map do |list_ed_id, grouped|
      [list_ed_id, grouped.first.slice(:rank_world, :rank_europe)]
    end.to_h
  ]
end.to_h

由于我没有创建模型，而只是使用了结构（如您在上面的示例中所见），因此我并没有真正自己测试最后一点。但它应该像这样工作，如果您发现错误或有更多问题，请发表评论。我仍然希望有人提出更好的解决方案，因为我现在经常自己寻找一个。

干杯:)

编辑：小修正

于 2018-07-04T11:33:11.517 回答
0 
ranks_relation.
  group_by(&:book_id).
  map do |id, books|
    [id, books.map do |book|
           [
             book.list_edition_id,
             {
               "rank_world" => book.rank_world,
               "rank_europe" => book.rank_europe
             }
           ]
         end.sort_by { |_, hash| hash["rank_world"] }.to_h
    ]
  end.to_h
于 2018-07-04T11:35:13.987 回答