如何在ActiveRecord中合并两个字典条目？

Question

我在我的项目中使用cedict开源汉英词典。我把它放在一个ActiveRecord建模的postgresql数据库中，与其他模型没有关系。

当我运行DictionaryEntry.count(:id, :group => :simplified, :having => 'count(id) > 1')时，它显示2141个重复条目。这些条目中的简化字相同，但定义不同。

例如，注意放之四海而皆准有两个条目，但它有两个不同的定义：

irb(main):009:0> DictionaryEntry.find_all_by_simplified("放之四海而皆准")
  DictionaryEntry Load (193.6ms)  SELECT "dictionary_entries".* FROM "dictionary_entries" WHERE "dictionary_entries"."simplified" = '放之四海而皆准'
[
    [0] #<DictionaryEntry:0x007feb8750f9e0> {
                 :id => 42164,
        :traditional => "放之四海而皆准",
         :simplified => "放之四海而皆准",
             :pinyin => "fang4 zhi1 si4 hai3 er2 jie1 zhun3",
         :definition => "appropriate to any place and any time -idiom; universally applicable/a panacea",
         :created_at => Sat, 22 Dec 2012 03:07:44 UTC +00:00,
         :updated_at => Sat, 22 Dec 2012 03:07:44 UTC +00:00
    },
    [1] #<DictionaryEntry:0x007feb8750f378> {
                 :id => 42165,
        :traditional => "放之四海而皆準",
         :simplified => "放之四海而皆准",
             :pinyin => "fang4 zhi1 si4 hai3 er2 jie1 zhun3",
         :definition => "applicable anywhere -idiom",
         :created_at => Sat, 22 Dec 2012 03:07:44 UTC +00:00,
         :updated_at => Sat, 22 Dec 2012 03:07:44 UTC +00:00
    }
]

我想合并这两个条目，这样当我运行DictionaryEntry.find_all_by_simplified("放之四海而皆准")时，它将返回在/后面添加了已删除对象定义的对象，如下所示：

[
    [0] #<DictionaryEntry:0x007feb8750f9e0> {
                 :id => 42164,
        :traditional => "放之四海而皆准",
         :simplified => "放之四海而皆准",
             :pinyin => "fang4 zhi1 si4 hai3 er2 jie1 zhun3",
         :definition => "appropriate to any place and any time -idiom; universally applicable/a panacea/applicable anywhere -idiom",
         :created_at => Sat, 22 Dec 2012 03:07:44 UTC +00:00,
         :updated_at => Sat, 22 Dec 2012 03:07:44 UTC +00:00
    }
]

(我可能也想合并拼音，如果它碰巧不同，还不确定...嗯..。是的，我可能需要这样做。)

Answer 1

在rails控制台中输入以下命令似乎对我有效：

dups = DictionaryEntry.count(:id, :group => :simplified, :having => 'count(id) > 1')
dups.each do |d|
twins = DictionaryEntry.find_all_by_simplified(d[0])
if twins[0].pinyin == twins[1].pinyin
twins[1].definition = twins[1].definition + "/" + twins[0].definition
twins[1].save
twins[0].destroy
end
end

我不得不运行了几次。